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# Gradient schemes for the Stefan problem

3 COFFEE - COmplex Flows For Energy and Environment
CRISAM - Inria Sophia Antipolis - Méditerranée , JAD - Laboratoire Jean Alexandre Dieudonné : UMR7351
JAD - Laboratoire Jean Alexandre Dieudonné
Abstract : We show in this paper that the gradient schemes (which encompass a large family of discrete schemes) may be used for the approximation of the Stefan problem $\partial_t \bar u - \Delta \zeta (\bar u) = f$. The convergence of the gradient schemes to the continuous solution of the problem is proved thanks to the following steps. First, estimates show (up to a subsequence) the weak convergence to some function $u$ of the discrete function approximating $\bar u$. Then Alt-Luckhaus' method, relying on the study of the translations with respect to time of the discrete solutions, is used to prove that the discrete function approximating $\zeta(\bar u)$ is strongly convergent (up to a subsequence) to some continuous function $\chi$. Thanks to Minty's trick, we show that $\chi = \zeta(u)$. A convergence study then shows that $u$ is then a weak solution of the problem, and a uniqueness result, given here for fitting with the precise hypothesis on the geometric domain, enables to conclude that $u = \bar u$. This convergence result is illustrated by some numerical examples using the Vertex Approximate Gradient scheme.
Document type :
Journal articles

Cited literature [16 references]

https://hal.archives-ouvertes.fr/hal-00751555
Contributor : Robert Eymard <>
Submitted on : Wednesday, November 14, 2012 - 10:44:30 AM
Last modification on : Monday, March 4, 2019 - 2:04:24 PM
Document(s) archivé(s) le : Friday, February 15, 2013 - 2:55:09 AM

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### Identifiers

• HAL Id : hal-00751555, version 1

### Citation

Robert Eymard, Pierre Féron, Thierry Gallouët, Raphaèle Herbin, Cindy Guichard. Gradient schemes for the Stefan problem. International Journal on Finite Volumes, Institut de Mathématiques de Marseille, AMU, 2013, Volume 10 special. ⟨hal-00751555⟩

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Elementary Algebra

In order to add rational numbers with a common denominator, we simply add the numerators and place the result over the common denominator. This works because fractions with the same denominator express different amounts of parts out of the same total amount of parts. For example, one slice, or $\frac{1}{8}$ of a pizza plus two slices, or $\frac{2}{8}$ of a pizza is equal to three slices, or $\frac{3}{8}$ of a pizza.


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# How do you multiply (2c - 3) ( 4c - 3)?

Apr 9, 2018

$8 {c}^{2} - 18 c + 9$

#### Explanation:

To multiply these two parentheses you have to FOIL

This means to multiply the FIRST two values in each

Then multiply the OUTSIDE two values in each

Then multiply the INSIDE two values in each

Then multiply the LAST two values in each

First:
$2 c \cdot 4 c = 8 {c}^{2}$

Outside:
$2 c \cdot \left(- 3\right) = - 6 c$

Inside:
$- 3 \cdot 4 c = - 12 c$

Last:
$- 3 \cdot - 3 = 9$

Now combine all the terms:

$8 {c}^{2} - 6 c - 12 c + 9$

Now combine like terms:

$8 {c}^{2} - 18 c + 9$

Apr 9, 2018

$8 {c}^{2} - 6 c - 12 c + 9$

#### Explanation:

You multiply the first two terms $\left(2 c \times 4 c\right)$, then the outer terms $\left(2 c \times \left(- 3\right)\right)$, then the inner terms $\left(- 3 \times 4 c\right)$, then the last 2 terms $\left(2 c \times \left(- 3\right)\right)$.


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# zbMATH — the first resource for mathematics

Pontryagin duality for metrizable groups. (English) Zbl 0899.22001
An abelian topological group $$G$$ is called reflexive if the canonical mapping from $$G$$ to the second dual group $$\nu\colon G^{\wedge\wedge}$$ is an isomorphism of topological groups, where $$G^\wedge$$ is the group of continuous characters of $$G$$ equipped with the compact-open topology. The classical result by Pontryagin-van Kampen says that every locally compact abelian (LCA) group is reflexive. A lot of effort has been made over the past decades to try and find natural classes of reflexive non-LCA topological abelian groups. Perhaps the best known achievements are those by Smith who proved the reflexivity of additive groups of Banach spaces [M. Smith, Ann. Math., II. Ser. 56, 248-253 (1952; Zbl 0047.10701)], by Kaplan who proved that the class of reflexive groups is closed under some natural operations not preserving local compactness [S. Kaplan, Duke Math. J. 15, 649-658 (1948; Zbl 0034.30601)], and by Banaszczyk, who had introduced the class of so-called nuclear topological groups and shown that every complete metrizable nuclear group is reflexive [W. Banaszczyk, Additive subgroups of topological vector spaces, Lect. Notes Math. 1466 (Berlin etc. 1991; Zbl 0743.46002)]. The present reviewer had proved in [V. Pestov, Bull. Aust. Math. Soc. 52, 297-311 (1995; Zbl 0841.22001)] that free topological abelian groups over some topological spaces are reflexive, and asked whether every complete and Čech-complete topological group having sufficiently many characters is reflexive. The present article provides, among many other things, a negative answer to this question: it is shown that every complete locally bounded non locally convex topological vector space admitting a weaker locally convex topology (such as $$l_p$$ with $$0<p<1$$) supplies a counter-example. The central result of the note under review is however this: a metrizable separable (that is, Polish) abelian group is reflexive if and only if $$\nu\colon G^{\wedge\wedge}$$ is an algebraic isomorphism. It is also shown that a metrizable topological abelian group is reflexive if and only if it satisfies the duality for convergence groups introduced and studied by Binz and Butzmann [E. Binz, Continuous convergence on $$C(X)$$, Lect. Notes Math. 469 (Berlin etc. 1975; Zbl 0306.54003)], even if in general the two dualities are known to be different [M. J. Chasco and E. Martin-Peinador, Arch. Math. 63, 264-270 (1994; Zbl 0806.43003)].

##### MSC:
 22A05 Structure of general topological groups 46A16 Not locally convex spaces (metrizable topological linear spaces, locally bounded spaces, quasi-Banach spaces, etc.)
Full Text:


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# All Questions

23 views

### Super classical r-matrices and Poisson Lie supergroups

In classical case, given an r-matrix $r$ for $sl_n$, we can compute the corresponding Poisson bracket on $SL_n$ by using the formula $\{L \otimes L\} = [r, L \otimes L]$. For example, let $g=sl_2$ and ...
158 views

### Artin conjecture on L-functions

Artin conjecture on Artin $L$-functions asserts that the Artin $L$-function $L(\rho,s)$ of a non-trivial irreducible representation $\rho$ of the Galois group $\Gamma$ of a number field admits ...
33 views

60 views

### On some examples of critical families

I'm reading the book on Injective choice functions by Holz, Podewski and Steffens, and I find it to be at the same time well written and quite difficult. It has almost no examples - and in quite a few ...
147 views

### Does complex multiplication for higher dimensional abelian varieties give some generalization of class field theory?

I am currently learning some aspects of the theory of complex multiplication for elliptic curves, and the relationship with class field theory. As I understand it, there is a very special class of ...
74 views

### Minimize matrix distance to tensor product

Minimize the following function: $f(V) = || V \otimes V - U_1 \otimes U_2 ||$ where $U_1, U_2 \in SU(n)$ are fixed and we minimize over all $V \in SU(n)$. The norm is from the trace inner product. ...
47 views

### A generalization of the Tucker circle theorem and the Thomsen theorem associated with a conic

I gave a generalization of the Tucker circle theorem and the Thomsen theorem at here. Now, I give a more generalization of these theorems as following: Problem: Let $A_1A_2A_3A_4A_5A_6$ be a hexagon, ...
34 views

### Software for matching theorems to inputted conditions/hypotheses

Many times I find myself going through analysis books, wikipedia and papers, looking for what is known for my functions/objects at hand. So is there any software that at least tries to move in that ...
113 views

### Does this PDE have a name?

I'm looking for any and all information that might be known about the following second-order PDE for one function $u(x,y)$: $u_{xy} = u_x e^u + u_y e^{-u}$ e.g., Does it have a name? Is it known ...
Let $\delta_1,...,\delta_n$ be $n$ independent identically distributed Bernoulli random variables with $\mathbb{P}(\delta_1=1)=p$. We consider a set $\Omega = \{\mathbf{a}:=(a_1,...,a_n)~|~a_i\in [0,c/... 3answers 195 views ### When can the Cayley graph of the symmetries of an object have those symmetries? Let$P$be an object in$\mathbb{R}^n$with symmetry group$G$. Let$C$be the a Cayley graph of$G$. When can$C$be embedded in$\mathbb{R}^m$so that the embedded graph has the same symmetry ... 0answers 85 views ### example of fuchsian groups acting on 2-sphere by G. Martin Currently I am reading a paper "Infinite group actions on spheres" by Gaven Martin. I am a first year graduate students and I got lots of questions, so one of them is about the following example: (... 3answers 279 views ### How to sample a uniform random polyomino? A polyomino is formed by joining finitely many unit squares edge to edge. It may be regarded as a finite subset of the regular square tiling with a connected interior. In particular, for us, ... 2answers 521 views ###$x_1 = 2$,$x_{n + 1} = {{x_n(x_n + 1)}\over2}$, what can we say about$x_n \text{ mod }2$? This question was asked on MathStackexchange here, but there was no answer, so I am asking it here. Let$$x_1 = 2, \quad x_{n + 1} = {{x_n(x_n + 1)}\over2}.$$What can we say about the behavior of$x_n ...
For two finite groups $G_1, G_2$ if for every integer $n\geq 0$, $|G_1^n| = |G_2^n|$, then is it true that $G_1\cong G_2$? By $G^k$ we mean set $\{g^k|g\in G\}$.


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# Proving Two Hermitian Matrices have the same eigenvectors

I am currently stuck on the following proof.

Suppose that a (n by n) unitary matrix U can be written as U=M+iN where M and N are Hermitian matrices.

Now assuming that M and N have n distinct eigenvalues it can be shown that they have the same eigenvectors.

My attempts to show it have been unsuccessful. I was wondering whether any one could give me advice on how to do this because I am getting nowhere at the moment.

Many Thanks

## 1 Answer

Since $I = U^*U = UU^*$, the following two operators are both equal to the identity:

1. $M^2 - N^2 + iMN - iNM$
2. $M^2 - N^2 -iMN + iNM$

This leads to $iMN - iNM = -iMN +iNM$ which gives us $$MN - NM = -MN + NM\\ 2MN = 2NM\\ MN = NM$$ so the two matrices $M$ and $N$ commute and must therefore have the same eigenvectors.

Commuting $n\times n$ matrices $M$ and $N$, each with with $n$ distinct eigenvalues will share eigenvectors because if $v$ is an eigenvector of $N$ with eigenvalue $l$, then so is $Mv$: $$N(Mv) = (NM)v = (MN)v = M(Nv) = M(lv) = l(Mv)$$ and since only scalar multiples of $v$ has that property, $Mv$ must be a scalar multiple of $v$, and therefore $v$ is an eigenvector of $M$ as well.


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# Faces, Edges and Vertices of a Dodecahedron

Dodecahedrons are three-dimensional figures formed by 12 pentagonal faces. Dodecahedrons are one of the five platonic solids. In total, dodecahedrons have 12 faces, 30 edges, and 20 vertices. Three pentagonal faces meet at each vertex.

In this article, we will learn about the faces, vertices, and edges of dodecahedrons in more detail.

##### GEOMETRY

Relevant for

Learning about the faces, vertices, and edges of dodecahedrons.

See faces

##### GEOMETRY

Relevant for

Learning about the faces, vertices, and edges of dodecahedrons.

See faces

## Faces of a dodecahedron

The faces of dodecahedrons are the two-dimensional flat surfaces that form the three-dimensional dodecahedron. Dodecahedrons have a total of twelve pentagonal faces.

Another way to define the faces of the dodecahedrons is as the two-dimensional pentagons that are formed by the vertices and the edges.

Each face of the dodecahedron connects with five other pentagonal faces.

Since dodecahedrons are regular figures, their twelve faces have the same shape and the same dimensions. Therefore, all twelve faces of the dodecahedron are congruent.

This means that to calculate the surface area of a dodecahedron, we have to find the area of one of the faces and multiply by 12. Thus, we have:

$latex A_{s}=12A_{p}$

where, $latex A_{p}$ is the area of one of the pentagonal faces of the dodecahedron.

Alternatively, we can use the standard formula for the Surface Area of a Dodecahedron:

$$A_{s}=3\sqrt{25+10\sqrt{5}}~{{a}^2}$$

## Vertices of a dodecahedron

The vertices of a dodecahedron are the points where the edges meet. Specifically, three edges meet at each vertex of a dodecahedron.

Alternatively, we can also define the vertices of a dodecahedron as the points where three pentagonal faces of the dodecahedron meet.

In total, we have 20 vertices in dodecahedrons. Since the faces are pentagonal, each face is made up of 5 vertices and each vertex is shared by three faces.

## Edges of a dodecahedron

The edges of a dodecahedron can be defined as the line segments joining two vertices. The edges form the outline of each pentagonal face of an octahedron.

Another way to define edges is as the line segments where two pentagonal faces of the dodecahedron meet. Therefore, we know that the edges lie on the limits of the dodecahedron.

In total, dodecahedrons have 30 edges. Since the faces are pentagonal, each face of the dodecahedron has five edges, as we see in the diagram below.

In the diagram, we see that each face of the dodecahedron has five edges, and each edge is shared by another face.


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Prev

# Log Law of the Wall

Next

Use the slider to choose varying values of $$R_m$$; the values of $$R_\tau$$, $$n$$, and $$s$$ will simultaneously update.

The panel below will plot the same flow expressed in inner wall variables $$y^+$$ and $$u^+$$. The inner wall variable serves to magnify the three distinct regimes in a wall bounded turbulent flow.

Superimposed on the panel are two functions $$u^+ = y^+$$ (in red) and $$u^+ = \frac{1}{\kappa}\ln y^+ + c$$ (in green) where $$\kappa = 0.41$$ and $$c = 5.5$$ are empirical constants (e.g. [10]).

The first relationship which holds in the range $$0 < y^+ < 5$$ is called the "viscous sublayer region" and the second relationship which holds in the range $$30 < y^+ < 100$$ is called the "logarithmic law of the wall". Beyond $$y^+ > 100$$ is the "velocity defect region" where $$\frac{u_m - \bar{u}}{u_\tau} = F\left(\eta\right)$$ where $$F$$ is an empirical function (e.g. [11]).

$$R_\tau =$$ ,
$$n =$$ ,
$$s =$$


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# Linearization and Tangent Line Approximation

Here's some vocabulary:
The line tangent to a function (that's differentiable) at is also called the linearization of at .
You can use the linearization of a function at to approximate values of near . This technique is also called tangent line approximation. Here's an example. Suppose someone asks you to estimate (without a calculator).

Without using any calculus, you know is a number between which two consecutive integers? Why?

But with calculus, we can do much better than that, even in our heads (or at least with paper and pencil only). We're going to use the linearization of at (because is the nearest perfect square to ). Determine this linearization and input your answer in the applet below as the function .

How does what you observed in the applet related to Good Definition #2 of a tangent?

In other words, plugging into (which you can do without a calculator) should yield a result that is extremely close to because . Use the scientific calculator below to compute these two quantities.

Use the linearization of at to approximate .

Use tangent line approximation to estimate the cube root of . (This means you'll have to determine what function and linearization to use.)


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Q

# A block of mass m is moving initially with velocity u on a horizontal rough surface.

And due to friction, it comes to rest after covering a distance S. Then find S.
Views

For the below figure

S is given by

$S = \frac{u^{2}}{2\mu g}$

where

$\mu=$ coefficient of friction

S = distance traveled

g = gravity

u=initial velocity

Exams
Articles
Questions


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# What is e^(ln(x)?

Oct 23, 2015

#### Answer:

It's $x$.

#### Explanation:

The logarithm and the exponential are inverse function, which means that if you combine them, you obtain the identity function, i.e. the function $I$ such that $I \left(x\right) = x$.

In terms of definitions, it becomes obvious. The logarithm $\ln \left(x\right)$ is a function which tells you what exponent you must give to $e$ to obtain $x$. So, ${e}^{\log \left(x\right)}$, literally means:

"$e$ to a power such that $e$ to that power gives $x$".


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What is the solution to the exercise 3.11 in the RL book by Sutton and Barto

I'm trying to solve the exercise 3.11 from the second-edition from the Sutton and Barto's book

Exercise 3.11 If the current state is $$S_t$$ , and actions are selected according to a stochastic policy $$\pi$$, then what is the expectation of $$R_{t+1}$$ in terms of $$\pi$$ and the four-argument function $$p$$ (3.2)?

Here's my attempt.

For each state $$s$$, the expected immediate reward when taking action $$a$$ is given in terms of $$p$$ by eq 3.5 in the book:

$$r(s,a) = \sum_{r \in R} r \, \sum_{s'\in S} p(s',r| s,a) = E[R_t | S_{t-1} = s, A_{t-1} = a]$$ (1)

The policy $$\pi(a | s)$$ on the other hand gives the probability of taking action $$a$$ given state $$s$$.

Is it possible to express the expectation of the immediate reward over all actions $$A$$ from the state $$s$$ using (1) as

$$E[R_t | S_{t-1} = s, A] = \sum_{a \in A} \pi(a|s) r(a,s) \ \ \ \ \ \ \ \ (2) ?$$

If this is valid, is this also valid in the next time step

$$E[R_{t+1} | S_{t} = s, A] = \sum_{a \in A} \pi(a|s) r(a,s) \ \ \ \ \ \ \ \ (3) ?$$

If (2) and (3) are OK, then

$$E[R_{t+1} | S_{t} = s, A] = \sum_{a \in A} \pi(a|s) \sum_{r \in R} r \, \sum_{s'\in S} p(s',r| s,a)$$

First note that $$\mathbb{E}[R_{t+1} |S_t=s] = \sum_{s',r}rm(s',r|s)$$ where $$m(\cdot)$$ is the mass function for the joint distribution of $$S_{t+1},R_{t+1}$$.

If you are currently in state $$S_t$$ and we condition on taking action $$a$$ then the expected reward at time $$t+1$$ is given as follows:

\begin{align} \mathbb{E}[R_{t+1} | S_t = s, A_t=a] & = \sum_{s',r}rp(s'r|s,a)\;. \end{align}

However, action $$A_t$$ is taken according to some stochastic policy $$\pi$$ so we need to marginalise this out of our expectation by using the tower law - i.e. we take

$$\mathbb{E}_{A_t \sim \pi}[\mathbb{E}[R_{t+1} | S_t = s, A_t=a]|S_t = s] = \sum_a \pi(a|s)\sum_{s',r}rp(s'r|s,a) = \mathbb{E}[R_{t+1} | S_t = s]\;.$$

To see why this holds we can re-write using some arbitrary mass functions $$f(\cdot),h(\cdot),g(\cdot),m(\cdot)$$ as

$$\pi(a|s)p(s'r|s,a) = \frac{f(a,s)}{g(s)} \times \frac{h(s',r,a,s)}{f(a,s)} = m(s',r,a|s)\;,$$ so when we would end up with (after re-arranging the summations)

$$\sum_{s',r}r \sum_{a}m(s',r,a|s) = \sum_{s',r}r m(s',r|s) = \mathbb{E}[R_{t+1}|S_t = s]\;;$$ as required.

NB: what you have done is mostly correct except be careful when going from (2) to (3). They are exactly the same equation except for the time stamps, which means you would have to change the time stamps in your $$r(s,a)$$. Note that when you are at time step $$t$$ you take your action $$A_t$$ from your current $$S_t$$ to transition into state $$S_{t+1}$$ and then receive reward $$R_{t+1}$$ (and the next state).


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### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

# Square Overlap

##### Age 14 to 16 ShortChallenge Level

Two squares of side length 2 cm are placed on top of each other so that the angle between the squares is 60$^\text{o}$.

What is the area of the overlap?

This problem is adapted from the World Mathematics Championships
You can find more short problems, arranged by curriculum topic, in our short problems collection.


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# Orthogonalization method

The basic authors of the description: Daria Inzhelevskaya(text), Alexey Frolov(editing)

The Gram--Schmidt orthogonalization is a method that constructs a set of orthogonal vectors ${\displaystyle \mathbf {b}_{1},\;\ldots ,\;\mathbf {b} _{N}}$ or a set of orthonormal vectors ${\displaystyle \mathbf {e} _{1},\;\ldots ,\;\mathbf {e}_{N}}$ from a given set of linearly independent vectors ${\displaystyle \mathbf {a} _{1},\;\ldots ,\;\mathbf {a} _{N}}$. This is done in such a way that each vector ${\displaystyle \mathbf {b} _{j}}$ or ${\displaystyle \mathbf {e} _{j}}$ is a linear combination of the vectors ${\displaystyle \mathbf {a} _{1},\;\ldots ,\; \mathbf {a} _{j}}$. The process may be used for obtaining the QR decomposition, where the system of original vectors is the columns of a given matrix, while the columns of Q are the result of orthogonalization. Thus, unlike the Givens (rotation) and Householder (reflection) methods, which are based on the left unitary/orthogonal reduction to triangular form, the orthogonalization method reduces the original matrix by right non-orthogonal (triangular) transformations to a unitary/orthogonal matrix.

## Mathematical foundations of the method

The classical orthogonalization method for the QR decomposition of a square matrix (real version) is fairly simple. However, due to its instability, which manifests itself in the non-orthogonality of resulting systems, the method is very rarely used in practice.

Let $\mathbf{a}_1,\;\ldots,\;\mathbf{a}_N$ be linearly independent vectors. Define the projection of a vector $\mathbf{a}$ on (the direction of) a vector $\mathbf{b}$ by the formula $\mathbf{proj}_{\mathbf{b}}\,\mathbf{a} = {\langle \mathbf{a}, \mathbf{b} \rangle \over \langle \mathbf{b}, \mathbf{b}\rangle} \mathbf{b} ,$

where $\langle \mathbf{a}, \mathbf{b} \rangle$ is the scalar product of the vectors $\mathbf{a}$ and $\mathbf{b}$.

In the k-dimensional real space, the scalar product of the vectors $\mathbf{ a= [a_1, a_2, ...,a_k]}$ and $\mathbf{ b= [b_1, b_2, ..., b_k]}$ is defined as

$\langle \mathbf{a}, \mathbf{b} \rangle=\sum_{i=1}^k a_ib_i=a_1b_1+a_2b_2+\cdots+ a_kb_k$.

The above operator projects the vector $\mathbf{a}$ on the direction of the vector $\mathbf{b}$.

The orthogonality of the first two vectors in the orthogonalization process below is attained at Step (2).

The classical Gram--Schmidt process is performed as follows:

${\begin{array}{lclr} {\mathbf {b}}_{1}&=&{\mathbf {a}}_{1}&(1)\\ {\mathbf {b}}_{2}&=&{\mathbf {a}}_{2}-{\mathbf {proj}}_{{{\mathbf {b}}_{1}}}\,{\mathbf {a}}_{2}&(2)\\ {\mathbf {b}}_{3}&=&{\mathbf {a}}_{3}-{\mathbf {proj}}_{{{\mathbf {b}}_{1}}}\,{\mathbf {a}}_{3}-{\mathbf {proj}}_{{{\mathbf {b}}_{2}}}\,{\mathbf {a}}_{3}&(3)\\ {\mathbf {b}}_{4}&=&{\mathbf {a}}_{4}-{\mathbf {proj}}_{{{\mathbf {b}}_{1}}}\,{\mathbf {a}}_{4}-{\mathbf {proj}}_{{{\mathbf {b}}_{2}}}\,{\mathbf {a}}_{4}-{\mathbf {proj}}_{{{\mathbf {b}}_{3}}}\,{\mathbf {a}}_{4}&(4)\\ &\vdots &&\\{\mathbf {b}}_{N}&=&{\mathbf {a}}_{N}-\displaystyle \sum _{{j=1}}^{{N-1}}{\mathbf {proj}}_{{{\mathbf {b}}_{j}}}\,{\mathbf {a}}_{N}&(N) \end{array}}$

From each vector $\mathbf{b}_j \;(j = 1 \ldots N)$, one can obtain a normalized vector: $\mathbf{e}_j = {\mathbf{b}_j\over \| \mathbf{b}_j \|}$ (the normalized vector has the same direction as the original one, while its norm is 1 ). Here, the norm is consistent with the scalar product: $\| x \| = \sqrt{\langle x, x \rangle}$

The output of the Gram--Schmidt process:

The system $\mathbf{b}_1,\;\ldots,\;\mathbf{b}_N$ of orthogonal vectors or the system $\mathbf{e}_1,\;\ldots,\;\mathbf{e}_N$ of orthonormal vectors.

In practice, the most often used form of the method is the orthogonalization method with re-orthogonalization.


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In DepthAstronomy

# Cheap balloon-borne telescopes aim to rival space observatories

See allHide authors and affiliations

Science  28 Feb 2020:
Vol. 367, Issue 6481, pp. 967
DOI: 10.1126/science.367.6481.967

## Summary

No instrument has revealed more about the heavens than NASA's Hubble Space Telescope, which cost nearly $5 billion to build and launch. Yet a team of researchers thinks it can match some of Hubble's capabilities with a telescope costing millions of dollars, hoisted to the edge of space on a balloon. During an 18-hour flight over Canada in September 2019, the$2 million Superpressure Balloon-borne Imaging Telescope (SuperBIT) held steady enough to image distant the stars with a resolution of 260 milliarcseconds, only a factor of five less sharp than Hubble and at the theoretical limit for its 0.5-meter mirror. To compete with space telescopes, however, a balloon-borne telescope must also stay aloft for many nights. To achieve that, the team hopes to fly SuperBIT on one of NASA's new superpressure balloons, which can stay aloft for many weeks. SuperBIT researchers hope to do an observing run in 2021 and are working on a larger, 1.5-meter telescope that they hope will reproduce some of Hubble's capabilities for a fraction of the cost.

View Full Text


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Legacy Documentation

Sloth and Darknet

Since we use Darknet, there are special configurations needed. After having cloned the vision_dev repository, add the following line to your .bashrc, replacing the dummy path with the path to the vision_dev/sloth directory so that sloth can create darknet files.

export PYTHONPATH=/path/to/vision_dev/sloth:\$PYTHONPATH

Using sloth output with Darknet

When attempting to use tagged images with darknet, the annotation file will need to be in the correct format. To convert to this format, use the following command

sloth convert <original file> <new_filename>.darknet

This will ask sloth to convert an annotation file in one format to the correct *.darknet format.

Once the file is in this format, you will need to run the sloth_to_darknet.py script which will generate the multiple files darknet requires based upon the compact information in the *.darknet file.

The following is an example command to generate said files.

./sloth_to_darknet.py -f /path/to/label/file -o /path/to/annotation/dir/ (optional)[-t training_list_filename.txt]

Getting Data to Label

We use the download_data.sh and download_data_current.sh scripts from the vision_dev repository to get data from the server. Make sure that you always do a git pull before getting new data because the scripts change frequently.

First claim one or more batches to label from the current spreadsheet of data. The current spreadsheet is located here. Note which numbers you want.

Once you have claimed some batches run the following command in vision_dev to download and extract the data:

./download_data.sh bag_prefix_0 ## ## ##

For example if you want batch numbers 00 04 and 12 of the path_marker batch you would run:

./download_data.sh path_marker_0 00 04 12


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# Natural Number Ordering is Transitive

## Theorem

Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.

Let $<$ be the relation defined on $\N$ such that:

$m < n \iff m \in n$

where $\N$ is defined as the minimal infinite successor set $\omega$.

Then:

$k < m, m < n \implies k < n$

That is: $<$ is a transitive relation.

## Proof

Let $k < m, m < n$.

By definition it follows that $k \in m, m \in n$.

We have from Element of Finite Ordinal iff Subset that:

$k \in m \iff k \subseteq m$
$m \in n \iff m \subseteq n$

It follows from Subset Relation is Transitive that $k \subseteq n$.

Hence the result.

$\blacksquare$


text:
Statistical inference on restricted partially linear additive errors-in-variables models

Author Info

• Chuanhua Wei

()

• Qihua Wang
Registered author(s):

Abstract

As a useful extension of partially linear models and additive models, partially linear additive model has been paid considerable attention in recent years. This paper considers statistical inference for the semiparametric model when the covariates in the linear part are measured with additive error. To test hypothesis on the parametric component, we propose a novel test statistic based on the difference between the corrected residual sums of squares under the null and alternative hypotheses, and show that its limiting distribution is that of a weighted sum of independent standard $\chi_{1}^{2}$ . We also develop an adjusted test statistic, which has an asymptotically standard chi-squared distribution. Some simulation studies are conducted to illustrate our approaches. Copyright Sociedad de Estadística e Investigación Operativa 2012

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Article provided by Springer in its journal TEST.

Volume (Year): 21 (2012)
Issue (Month): 4 (December)
Pages: 757-774

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Handle: RePEc:spr:testjl:v:21:y:2012:i:4:p:757-774

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Keywords: Errors-in-variables; Partially linear additive model; Corrected-profile least-squares approach; Restricted estimator; 62G05; 62G10;

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## anonymous 3 years ago please help? Evaluate the definite integral (if it exists) intergral (e^1/3)/-6x^2 dx from 1 to 2 ?

• This Question is Closed
1. amistre64

e^1/3 looks like a variable, but its not; its just a constant

2. anonymous

$\int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx$

3. amistre64

$\huge \int\limits\limits_{1}^{2} \frac{ e^\frac{ 1 }{ x } }{ -6x^2 } dx$ is that 1/3 or 1/x? need to verify

4. anonymous

thats what it looks like properly :)

5. anonymous

$\frac{ 1 }{ x }$

6. amistre64

well, if this is going to be simple than with any luck it will come from something like:$\Large \frac d{dx}e^{1/x}$

7. amistre64

so lets start by taking the derivative of that

8. hartnn

yeah, just put u= 1/x du=... ?

9. anonymous

exactly, @hartnn, this substution does the trick.

10. anonymous

do we not have to do something with the derivavtive of -6x^2?

11. amistre64

we should verify that if its going to be simple first and see if we can modifiy it with a useful form of "1"

12. anonymous

the integral exists anyway! i know that much

13. amistre64

$\frac d{dx}e^{1/x}=\frac{e^{1/x}}{-2x^2}$agree?

14. anonymous

where did the -2x^2 come from? is the derivative of -6x^2 not 12x?

15. amistre64

your confusing your rules .... its best to see these things on a more holistic level :)

16. anonymous

oh ok :)

17. amistre64

$\frac d{dx}e^{u(x)}=\frac{d}{dx}u(x)~e^{u(x)}=u' ~e^u$

18. amistre64

let u= 1/x du/dx = -1/2x^2 du = -1/2x^2 dx

19. anonymous

how do i get the derivative of 1/x?

20. amistre64

1/x = x^-1 which is then just the power rule .... which means i got a bad 2 in there :)

21. amistre64

im thinking of a sqrt function ..... silly me

22. anonymous

what do you mean a bad 2? :)

23. amistre64

x^-1 goes to -x^(-2) 1/x goes to -1/x^2 NOT -1/2x^2

24. anonymous

oh ok i get ya now

25. amistre64

so just pull out the 1/6 and integrate the easy way

26. anonymous

wheres the 1/6?

27. amistre64

.....

28. amistre64

$\int\frac{e^{1/x}}{-6x^2}dx$ $\int\frac16 \frac{e^{1/x}}{-x^2}dx$ $\frac16\int \frac{e^{1/x}}{-x^2}dx$

29. anonymous

oh yes i see :) got confused for a minute!

30. amistre64

just becasue you were doing vector calculus the other day doesnt mean you have to forget the fundamentals ;)

31. amistre64

you got it from here?

32. anonymous

$\frac{ e^\frac{ 1 }{ x } }{ 6 }$

33. anonymous

something like that right?

34. amistre64

yes, but apply your limits

35. anonymous

what do you mean?

36. hartnn

e^u/6

37. hartnn

not, e^(1/x) /6

38. amistre64

i mean ....$\int_{a}^{b}f(x)~dx=F(b)-F(a)$

39. hartnn

plus you need to change limits beforehand, u=1/x when x=1,u=1 when x=2, u=1/2

40. anonymous

but u=1/x?

41. amistre64

if you do a usub, then you need to do hartnns route otherwise its just e^(1/x)/6 applied at 1 and 2

42. anonymous

so i put 1 and 2 in for x?

43. amistre64

3 methods that im aware of; complete usub changes limits to us partial usub where you undo the u back to x and just working thru a normal integration were nothing is changed about.... which is what i did

44. anonymous

$\frac{ e^u }{ 6 }$

45. amistre64

$\int_{1}^2f(x)dx=F(2)-F(1)$ and we determined that $F(x)=\frac16e^{1/x}$

46. anonymous

and then i got that^

47. anonymous

so i sub in 2 and 1 for x then subtract them? is that it?

48. amistre64

yes$\frac16(e^{1/2}-e^{1/1})$ yes$\frac16(e^{1/2}-e)$

49. anonymous

if i work that out it gives me something like -0.782?

50. anonymous

if i just leave it as $\frac{ 1 }{ 6 } (e^\frac{ 1 }{ 2 } - e$

51. hartnn

did u mean -0.1782 ? and you can keep your answer as 1/6(e^(1/2)-e)

52. anonymous

when i worked it out it was -0.782 but if i leave it as the 1/6... it works out just the same :)

53. anonymous

thanks for the help :)

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## Introduction

It is commonly known that stable self-confined modes, such as solitons, may be produced by the balance between nonlinear and dispersive effects in the medium1. Solitons have been observed in diverse contexts, including water waves2, nonlinear fiber optics3,4 (as temporal solitons), Bose-Einstein condensates (BECs)5,6,7,8,9,10,11, plasmas12,13 and plasmonics14,15, proteins16 and DNA17, etc. Optical spatial solitons were created too in a great variety of settings, such as cells filled by vapors of alkali metals18, photorefractive crystals19,20, waveguides made of liquid dielectrics21,22, silica23 and second-harmonic-generating materials24, nematic liquid-crystal planar cells25, semiconductor waveguides26, arrayed waveguides27, and others.

While solitons have been originally introduced as exact solutions of integrable models1,28,29, nonintegrable systems provide for more generic and more realistic description of various physical settings. In particular, numerous dissipative systems, although lacking integrability, readily give rise to robust localized dissipative structures (LDSs), alias dissipative solitons30,31. In optics, an important example of a nonlinear dissipative medium which supports LDSs is provided by an optical resonator filled with a dispersive loss material featuring the Kerr nonlinearity, which is pumped by a coherent light beam (injected signal). This system is well modeled by the Lugiato-Lefever (LL) equation, originally introduced in32. As a mean-field equation, it applies to other settings too, such as Fabry-Perot resonators and ring cavities, fully or partially filled with nonlinear materials32, crystalline whispering-gallery-mode disk resonators33, and photonic-crystal-fiber resonators pumped by a coherent continuous-wave input beam34,35. In these contexts, the LL equation has been widely used to model Kerr frequency combs36,37,38,39,40, with applications to optical metrology41, high-precision spectroscopy42,43, optical atomic clocks44,45, phase evolution in pulse trains46,47, optical communications48, synthesis of arbitrary optical waveforms49,50, and radio-frequency photonics51. A review of the development of various applications of the LL equations has been published recently52. Soliton-like LDSs in the 1D LL equation are important modes too, in a broad range of values of the respective physical parameters53.

In many physically relevant contexts, especially as concerns realizations in optics, one- and two-dimensional (1D and 2D) LDSs supported by localized gain were studied in detail in the context of complex Ginzburg-Landau (CGL) equations54,55,56,57,58,59,60,61,62, see also reviews in63 and64. Specifically, by considering a 1D model with the tightly localized gain in the form of a delta-function, placed on top of the spatially uniform linear loss, analytical solutions for pinned LDSs pinned to the delta-function were found in54,60 (see also a review in64). Stable LDSs pinned to one or two gain-carrying “ hot spots”, shaped as narrow Gaussians, were reported too58,62. Further, stable 2D LDSs, including ones with an intrinsic vortex structure, supported by hot spots in the 2D geometry, were predicted in works55,56,57,59,61.

The objective of the present work is to introduce localized pump in the framework of the 1D and 2D LL equations, and find stable confined modes, which may be supported by the spatially focused pump. The difference from the previous works, which were dealing with the CGL equations54,55,56,57,58,59,60,61,62,64, is that the pump is represented by free terms in the LL equations, which do not multiply the field variable, while in the models of the CGL type the gain terms provide the parametric pump, i.e., they multiply the field variable.

We report analytical solutions of the 1D and 2D LL equation with the localized external pump, using a possibility to find exact analytical solutions for 1D modes pinned to the pump represented by the delta-function, and a variational approach, respectively. In the case of weak pump and loss, a simple perturbation theory for 1D modes is developed too. In a systematic form, the results for confined modes, including the analysis of their stability, are produced by means of numerical methods. The results demonstrate good agreement between the analytical predictions and numerical findings. In particular, while the exact analytical solutions for pinned modes in 1D are available under a special condition, we demonstrate that very similar stable numerical solutions exist when this condition does not hold. The predicted confined stable modes, pinned to the “ hot spots”, may be used, in particular, for the design of pixels placed at required positions, cf. the formation of pixels predicted by the LL equation in other contexts65.

## Results

### The one-dimensional Lugiato-Lefever equation. The 1D model equations

In the 1D setting, the scaled LL equation for amplitude ϕ(x, t) of the electromagnetic field in a nonlinear lossy cavity driven by a real localized pump E(x) is

$$i(\gamma +\frac{\partial }{\partial t})\varphi =[-\frac{1}{2}\frac{{\partial }^{2}}{\partial {x}^{2}}+{\rm{\Delta }}+\sigma |\varphi {|}^{2}]\varphi +E(x),$$
(1)

where γ > 0 is the dissipation rate, Δ is detuning of the pump with respect to the cavity, while σ = −1 and +1 corresponds to the self-focusing and defocusing nonlinearity, respectively. Note that dissipative solitons in the model of a fiber cavity, based on the 1D LL equation written in the temporal domain, with a pump in the form of a period train of Gaussians pulses, placed on top a nonzero background, were recently considered in work66. Accordingly, the LL equation (1) may also be considered in the temporal domain, with t and x replaced, respectively, by the propagation distance (z) and the temporal coordinate (usually denoted τ).

Stability of various patterns produced by Eq. (1) and its 2D counterpart considered below may be enhanced if an extra cubic lossy term, which represents the two-photon absorption, is added to the model. Then, Eq. (1) is replaced by

$$i(\gamma +{\rm{\Gamma }}|\varphi {|}^{2}+\frac{\partial }{\partial t})\varphi =[-\frac{1}{2}\frac{{\partial }^{2}}{\partial {x}^{2}}+{\rm{\Delta }}+\sigma |\varphi {|}^{2}]\varphi +E(x),$$
(2)

with γ, Γ > 0.

Before proceeding to analysis of confined modes supported by the tightly localized pump, it is relevant to mention that spatial localization may also be provided, in the presence of the usual uniform pump, by a confining (typically, harmonic-oscillator) potential67. On the other hand, results reported below demonstrate that the use of the narrow pump region does not imply that modes supported by it must necessarily be narrow too. Note that effects of local defects on LDSs in similar settings were previously studied in works68,69.

### The perturbative treatment

In the case of the self-focusing nonlinearity (σ = −1) and positive detuning, Δ > 0, one can develop a perturbation theory for the case of small γ and small E(x) in Eqs (1) and (2). In the zero-order approximation, a localized solution is given by the usual nonlinear-Schrödinger soliton28,

$$\varphi (x)={e}^{-i\zeta }\,\sqrt{2{\rm{\Delta }}}\,\text{sec}h(\sqrt{2{\rm{\Delta }}}x)$$
(3)

(as the zero-order approximation for localized patterns in the LL equation with E = const, the soliton waveform was used before70). The constant phase shift in ansatz (3), ζ, for stationary modes is then determined by the balance condition for the integral power,

$$P={\int }_{-\infty }^{+\infty }{|\varphi (x)|}^{2}dx\mathrm{.}$$
(4)

Indeed, it follows from condition dP/dt = 0 that

$$\gamma P+\Gamma {\int }_{-\infty }^{+\infty }{|\varphi (x)|}^{4}dx=-{\int }_{-\infty }^{+\infty }E(x)\text{Im}\{\varphi (x)\}dx.$$
(5)

Substituting the sech approximation (3) into Eq. (5) predicts the value of the phase shift:

$$\sin \,\zeta =\frac{2[\gamma +(4/3){\rm{\Gamma }}{\rm{\Delta }}]}{{\int }_{-\infty }^{+\infty }\,E(x)\,\sec \,h(\sqrt{2{\rm{\Delta }}}x)dx},$$
(6)

which is written for the generalized LL equation (2), that includes the cubic loss Γ. This result makes sense if it yields |sinζ| ≤ 1, which implies that the LDS of the prsent type exists if the pump’s strength exceeds a threshold value, which is a combination of dissipation coefficients γ and Γ. In fact, a mode pinned to the localized pump exists at all values of its strength, as demonstrated by the exact solution displayed below, the threshold being an artifact following from the assumption of the rigid form of the perturbative ansatz (3).

Note that, even for $$E(x)={\rm{const}}\equiv { {\mathcal E} }_{0}$$, integral $${\int }_{-\infty }^{+\infty }E(x)\rm{sech} (\sqrt{2{\rm{\Delta }}}x)dx$$ converges, hence the approximation based on Eqs (3)–(6) may correctly predict a state sitting on top of a small-amplitude CW background, with amplitude $${\varphi }_{0}\approx { {\mathcal E} }_{0}/(\Delta +i\gamma )$$, under the condition that the LDS’s amplitude, (2Δ)1/2, is much larger than ϕ 0, i.e., $${ {\mathcal E} }_{0}^{2}\ll {{\rm{\Delta }}}^{3}$$. Detailed comparison of predictions of the perturbation theory with numerical results will be presented elsewhere.

### A particular exact solution and states close to it

In the case when the gain is localized in a very narrow region, it may be approximated by the Dirac’s delta-function, cf. a similar approximation adopted for a strongly localized gain in the CGL model54:

$$E(x)={E}_{0}\,\delta (x)$$
(7)

(a similar model including a localized gain, with an LDS pinned to it, was also formulated in terms of the Swift-Hohenberg equation71). This means that the homogeneous version of Eq. (2),

$$i(\gamma +{\rm{\Gamma }}|\varphi {|}^{2}+\frac{\partial }{\partial t})\varphi =[-\frac{1}{2}\frac{{\partial }^{2}}{\partial {x}^{2}}+{\rm{\Delta }}+\sigma |\varphi {|}^{2}]\varphi ,$$
(8)

must be solved with the boundary condition at x = 0 which determines the jump of the first derivative induced by δ(x) in Eq. (7):

$$\frac{d\varphi }{dx}{|}_{x=+0}-\frac{d\varphi }{dx}{|}_{x=-0}=2{E}_{0}.$$
(9)

In this case, one can find a particular exact solution to the generalized LL equation (8) in the form of

$$\varphi (x)=\frac{A{e}^{i\zeta }}{{[\sinh (\lambda (|x|+\xi ))]}^{1+i\mu }},$$
(10)

with parameters (μ is called the chirp)

$$\mu =-\gamma /{\lambda }^{2},$$
(11)
$${A}^{2}=3\gamma /(2{\rm{\Gamma }}),$$
(12)
$${\lambda }^{2}=\frac{\gamma }{4\Gamma }(\sqrt{9{\sigma }^{2}+8{{\rm{\Gamma }}}^{2}}+3\sigma ).$$
(13)

This particular solution is a non-generic one, as it exists at the single value of the mismatch parameter,

$${\rm{\Delta }}=\frac{{\lambda }^{2}}{2}(1-{\mu }^{2})\equiv \frac{1}{2}(\frac{3\sigma \gamma }{{\rm{\Gamma }}}-{\lambda }^{2})$$
(14)

(in other words, it is a codimension-one type of the exact solution, with “ one” referring to constraint (14), which must be adopted to produce the analytical expression). Note that the solution given by Eq. (13) exists (i.e., it gives λ 2 > 0) for both σ = −1 and +1. The presence of the cubic-loss coefficient, Γ > 0, is necessary for the existence of the solution. Indeed, in the limit of Γ → 0 Eq. (13) leads to divergence:

$${\lambda }^{2}\approx \{\begin{array}{c}3\gamma /(2{\rm{\Gamma }})\,{\rm{at}}\,\sigma =+1,\\ (1/3){\rm{\Gamma }}\gamma \,{\rm{at}}\,\sigma =-1.\end{array}$$
(15)

Finally, parameters ξ and ζ in expression (10) are obtained by its substitution in jump condition (9):

$$A\lambda (1+i\mu ){e}^{i\zeta \lambda }\frac{\cosh (\lambda \xi )}{{[\sinh (\lambda \xi )]}^{2+i\mu }}=-{E}_{0}.$$
(16)

An explicit result, following from Eq. (16), is

$$\xi =\frac{1}{2\lambda }{\rm{arcosh}}(1+\frac{\chi }{{E}_{0}^{2}}+\sqrt{4+\frac{\chi }{{\lambda }^{2}{E}_{0}^{4}}}),$$
(17)
$$\chi \equiv {A}^{2}{\lambda }^{2}(1+{\mu }^{2}),$$
$$\zeta =\pi -\arctan \,\mu +\mu \,\mathrm{ln}(\sinh (\lambda \xi )),$$
(18)

where arcosh(Z) ≡ ln(Z + (Z 2 −1)1/2).

In the CGL model with localized gain (rather than pump), exact pinned states are also codimension-one solutions, the difference being that, in the latter case they coexist with the zero state, which may or may not be stable solutions54.

In Fig. 1(a,b) we display typical examples of the analytically found modes pinned to the delta-function for focusing and defocusing nonlinearities, by choosing σ = −1 and σ = 1, respectively, along with their numerically found counterparts. In this case, we set parameters as γ = Γ = E 0 = 1, and took Δ as per Eq. (14). The numerical counterparts were produced by using the naturally regularized delta function in the form given by Eq. (31) (see Methods), for three different values of width w. It is relevant to mention that, while the regularized delta-function approaches the standard delta-function in the limit of w → 0, the use of a finite stepsize Δx in the numerical procedure gives rise to a critical value $${w}_{cr}\simeq {\rm{\Delta }}x/2$$ of w, the numerical solution getting drastically different from the analytical one at w < w cr. With the increase of the cubic-loss strength, Γ, w cr blows up (increases very fast) at $$\Gamma \gtrsim 3$$.

Further, in Fig. 2 we present systematic results for the 1D modes produced by analytical solution (10) and its numerical counterparts. These are the peak local power, max[|ϕ|2], the integral power, P (see Eq. (4)), and the mean squared width,

$$\langle {x}^{2}\rangle ={P}^{-1}{\int }_{-\infty }^{+\infty }{x}^{2}{|\varphi (x)|}^{2}dx,$$
(19)

shown in the left and right panels of the figure, as functions of the two nonlinearity coefficients, viz., the cubic-loss strength Γ and self-interaction strength σ (in the right panel, σ is considered as a continuously varying parameter, while in the left panel it is fixed to be σ = ±1 for the self-defocusing and focusing cases). Note that, as predicted by the analytical solutions (see Eqs (10) and (15)), the integral power P vanishes at Γ → 0. On the other hand, the results pertaining to σ = +1 and −1 tend to converge at large values of Γ, as the dissipative nonlinearity is dominant in this limit. The numerically generated findings are very close to the analytical predictions.

It is worthy to note conspicuous maxima of the peak local power and integral power, observed in Fig. 2(b) at σ = 0 and σ ≈ −1, respectively. Further, the bottom panel in Fig. 2(b) reveals a counter-intuitive feature of the pinned states: they shrink (〈x 2〉 → 0) in the limit of large σ > 0, i.e., strong self-defocusing (the same is also demonstrated by Eq. (13), which predicts $${\lambda }^{2}\sim \mathrm{1/}\langle {x}^{2}\rangle \to \infty$$ at σ → +∞). Usually, self-confined modes shrink in the opposite limit, of strong self-focusing. This surprising finding may be explained by the effect introduced by the cubic loss term ~Γ. Indeed, as mentioned above, the exact solution for the pinned state does not exist at Γ = 0, and, in the presence of Γ > 0, the shape of the mode is essentially affected by its chirp, which is produced by Eq. (11).

Further, Fig. 3 displays the effect of the variation of the pump’s amplitude E 0 and dissipation coefficient γ on the peak local power (max[|ϕ|2]) and integral power P (see Eq. (4)) of numerical solutions obtained from Eq. (2), for both the self-defocusing and focusing signs of the nonlinearity, i.e., σ = +1 and σ = −1, respectively, along with the counterparts predicted by the above analytical solutions. Naturally, the peak local and integral powers increase with the growth of E 0, and decrease with the growth of γ. These properties can be used for an effective control of the localized modes by means of parameters E 0 and γ.

Comparing the results obtained for the self-defocusing (σ = +1) and self-focusing (σ = −1) signs of the nonlinearity, we again observe a “ counter-intuitive” phenomenon, similar to that mentioned above, i.e., the solution is more localized in the case of the self-defocusing case than in the self-focusing case. Note that numerical results closely follow their analytical counterparts in Fig. 3 too.

Because the above codimension-one analytical solution is valid only under condition (14) imposed on the parameters, it is necessary to investigate the structural stability of the pinned modes against departure from this condition. To this end, in Fig. 4 we compare the solutions (both analytical and numerically found ones) obtained with the value of Δ selected as per Eq. (14), and their numerical counterparts obtained with this Δ replaced by 0.75Δ and 1.25Δ. We conclude that these considerable variations of Δ produce a weak effect on the solutions, i.e., they are structurally stable, effectively representing generic pinned modes, rather than specially selected ones.

While changes in the profiles of the solutions produced by the variation of Δ are relatively small, it is relevant to mention that the solutions are more sensitive to the variation in the case of the self-focusing than in the defocusing case.

Finally, systematic simulations of the perturbed solutions corroborate the stability of all the numerical solutions emulating the analytically predicted modes pinned to the delta-function. In fact, all the solutions are strong attractors, as direct simulations demonstrate that Eq. (2) readily generates precisely these states, starting from the zero input, ϕ(x, 0) = 0. This numerical result is important, because the stability of the analytically found solutions cannot be explored in an analytical form.

### The two-dimensional Lugiato-Lefever equation

The 2D version of the 1D LL equation (1) is

$$i(\gamma +\frac{\partial }{\partial t})\varphi =[-\frac{1}{2}{\nabla }_{\perp }^{2}+{\rm{\Delta }}+\sigma |\varphi {|}^{2}]\varphi +E(x,y),$$
(20)

where $${\nabla }_{\perp }^{2}=\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\partial {y}^{2}}$$, and the cubic loss is not included here (Γ = 0), as, unlike the exact 1D solutions, this term is not necessary for finding 2D solutions reported here. Further, one may fix here γ = 1 by means of rescaling. Below, we consider the Gaussian 2D shape of the pump, given by

$$E(x,y)=\frac{{P}_{0}}{\sqrt{\pi }\eta }{\rm{e}}{\rm{x}}{\rm{p}}(-\frac{{x}^{2}+{y}^{2}}{2{\eta }^{2}}),$$
(21)

where P 0 is the pump’s integral intensity, and parameter η controls its width.

In Fig. 5 we display the existence diagram of stable solutions produced by direct simulations of Eq. (20) in the plane of the mismatch and nonlinearity coefficients, (Δ, σ), for fixed pump’s parameters, P 0 = 10 and η = 1. Light yellow boxes denote values of parameters at which stable 2D solutions are crater-shaped (see Fig. 9 below), while red boxes correspond to the single-peak (bell-shaped) solutions, as shown below in Fig. 7. As concerns variational equations (28)–(30) (see Methods), their physically relevant solutions, corresponding to B > 0, have been found, by means of the relaxation method, for all values of parameters covering the range of σ [−5, 5] and Δ  [−20, 20], the respective picture essentially coinciding with one displayed on the basis of the full numerical solution in Fig. 5(a).

In Fig. 6(a) and (b) we display the integral power of the confined 2D modes, P, defined as per Eq. (25), as a function of the pump’s amplitude P 0 and nonlinearity strength σ, respectively (see Eq. (21)). Note that in Fig. 6(a) the analytical results, produced by the variational ansatz (26), and their numerical counterpart, obtained from direct simulations of Eq. (20), are very close to each other. We observe that, in the self-focusing case (σ = −1, shown by the line with circles), the integral power is slightly larger than in the self-defocusing case (σ = 1, shown by the line with boxes). In Fig. 6(b), the abrupt growth of the power for Δ = −10 at σ > 4 make the numerical solutions unstable.

Generic examples of the local-power profiles, |ϕ|2, for the 2D modes, obtained from direct simulations (at t = 100) for two different values of Δ, and the comparisons with the corresponding approximate analytical solutions, based on ansatz (26), are displayed in Fig. 7. Actually, the numerical and analytical profiles are indistinguishable at these values of Δ, in accordance with the above results which also demonstrated very good agreement of the analytical predictions with the numerical counterparts at large values of Δ. However, at small values of Δ, the numerical solutions feature a strong increase in the norm and may become unstable. In this case, the analytical approximation is not relevant.

The situation in a parameter region where stable stationary modes are absent (see Fig. 5) is illustrated by numerically generated solutions (at t = 10) displayed in Fig. 8(a,b) for σ = −1 and Δ = −1, and in Fig. 8(c)–(d) for σ = −1 and Δ = +1. Due to the instability of the numerical solutions, the analytical predictions are not relevant in this case.

As mentioned above, in a small part of their existence region (covered by yellow squared in Fig. 5), numerically found stable 2D modes feature a crater-like shape, with the maximum of the local power attained at a finite difference from the center, see an example in Fig. 9. Obviously, the analytical approximation based on ansatz (26) cannot reproduce this shape.

## Discussion

The modifications of the well-known 1D and 2D LL (Lugiato-Lefever) equation introduced in this work, with tightly localized pump, make it possible to create new stable confined modes, which are of interest in terms of the use of the LL equations as models of the pattern formation in nonlinear dissipative media. They may also be used to design compact pixels that can be created in cavities modeled by the LL equations. The present work is based on the combination of analytical and numerical methods, in the 1D and 2D geometries alike, the analytical parts helping to achieve a deeper insight into the variety of steady-state confined modes produced by the LL equations.

In the 1D geometry, we have first developed a simple perturbation theory, based on the usual sech ansatz (3), in the case of weak pump and loss. Other results have produced a family of exact analytical solutions, assuming that the tightly focused gain is represented by the delta-function, while the self-interaction may have both focusing and defocusing signs (σ < 0 and σ > 0, respectively). The analytical form of the solution is given by Eqs (10)–(13), under the condition that the mismatch, Δ, takes the specially selected value (14), and the cubic nonlinear term, which represents two-photon losses in the optical medium (with rate Γ), is present. Furthermore, numerical results, displayed in Fig. 4, corroborate the structural stability of the codimension-one analytical solutions, because the deviation of Δ from the spacial value (14) leads to weak variation of the stable pinned solutions. Most essential parameters which control the shape of the 1D pinned modes are two nonlinearity coefficient, σ and Γ. Characteristic features of the solution is the cusp at the center, and the phase structure (chirp). A remarkable fact is that the exact solutions are very close to their numerical counterparts, produced by the localized pump shaped as a regularized delta-function, and the family of the so generated 1D modes is entirely stable. In fact, the proximity of the numerical and analytical solutions additionally confirms the structural stability of the latter. A noteworthy (and counter-intuitive) feature of the 1D modes is that they shrink with the increase of the strength of the self-defocusing nonlinearity. The 1D solution produced by the analysis may help to find similar states in more general pattern-formation models.

In the 2D geometry with the pump applied at a small Gaussian-shape “ hot spot”, systematic numerical results are reported in the combination with approximate analytical findings produced by the variational approximation. A vast stability area in the system’s parameter space has been found, the most essential parameters being the above-mentioned mismatch and nonlinearity coefficients, Δ and σ (the 2D system is considered without the two-photon loss, Γ = 0, as its presence is not a necessary condition for finding the relevant solutions). In most cases, the 2D modes pinned to the “ hot spot” feature a single-peak (bell-shaped) structure, which is stable, and is well approximated by the variational ansatz. In a small part of the parameter space, 2D stable modes feature a crater-like shape, with the maximum local power found at a finite distance from the center. In another small part of the parameter space, 2D modes are unstable.

As an extension of the analysis, it may be interesting to use numerical methods to construct modes pinned to a set of two mutually symmetric 1D or 2D hot spots, cf. a similar configuration elaborated for the 1D CGL equation in ref.72. In particular, in the case of the self-focusing nonlinearity, σ < 0, one may expect spontaneous symmetry breaking between peaks attached to the two pump maxima. On the other hand, in the 2D geometry it may also be interesting to introduce ring-shaped pump, which may give rise to confined modes with a vortex structure, cf. a similar consideration for the 2D CGL equation in ref.61. A possibility of spontaneous breaking of the axial symmetry in vortex modes may be addressed too, following the pattern of the analysis performed in the framework of the CGL equation74.

## Methods

### The variational approach

Firstly, we define ϕ(x, y, t) ≡ Φ(x, y, t) exp (−γt), casting Eq. (20) in the form of

$$i\frac{\partial }{\partial t}{\rm{\Phi }}=[-\frac{1}{2}{\nabla }_{\perp }^{2}+{\rm{\Delta }}+\sigma {e}^{-2\gamma t}|{\rm{\Phi }}{|}^{2}]{\rm{\Phi }}+E{e}^{\gamma t},$$
(22)

which can be derived from a real time-dependent Lagrangian,

$$\begin{array}{rcl}L & = & \int \int dxdy\{\frac{i}{2}({{\rm{\Phi }}}_{t}^{\ast }{\rm{\Phi }}-{{\rm{\Phi }}}^{\ast }{{\rm{\Phi }}}_{t})+\frac{1}{2}(|{{\rm{\Phi }}}_{x}{|}^{2}+|{{\rm{\Phi }}}_{y}{|}^{2})\\ & & +\Delta |{\rm{\Phi }}{|}^{2}+\frac{\sigma }{2}{e}^{-2\gamma t}|{\rm{\Phi }}{|}^{4}+E{e}^{\gamma t}({{\rm{\Phi }}}^{\ast }+{\rm{\Phi }})\}\,\mathrm{.}\end{array}$$
(23)

Note that the following exact power-balance equation is produced by Eq. (20):

$$\frac{dP}{dt}=-2\gamma P-2\iint \text{Im}\{\varphi (x,y,t)\}E(x,y)dxdy,$$
(24)

for the integral power defined as

$$P=\int \int |\varphi (x,y,t){|}^{2}dxdy,$$
(25)

cf. the 1D counterpart given by Eq. (5).

For the variational approximation, we use the 2D isotropic Gaussian ansatz 73,

$${\rm{\Phi }}={e}^{\gamma t}A(t)\exp [-(B(t)-iC(t))({x}^{2}+{y}^{2})],$$
(26)

where A, B and C are real variational parameters, subject to obvious constraint B > 0. Next, substituting the ansatz in Eq. (23) and performing the integration, we arrive at the following effective Lagrangian:

$$\begin{array}{rcl}{L}_{{\rm{eff}}} & = & \frac{\pi }{2}{e}^{2\gamma t}\{\frac{{A}^{2}}{2{B}^{2}}\frac{dC}{dt}+[1+\frac{{\rm{\Delta }}}{B}+\frac{{C}^{2}}{{B}^{2}}]{A}^{2}\\ & & +\frac{\sigma {A}^{4}}{4B}+\frac{8\eta {P}_{0}(1+2B{\eta }^{2})A}{1+(4{B}^{2}+4{C}^{2}){\eta }^{4}+4B{\eta }^{2}}\}.\end{array}$$
(27)

The variational (Euler-Lagrange) equations following from Lagrangian (27), ∂L eff/∂(A, B, C) = 0, are

$$\sigma {A}^{3}+\frac{[4B(B+{\rm{\Delta }})+4{C}^{2}]}{2B}A+\frac{8\eta {P}_{0}{B}^{2}(1+2B{\eta }^{2})}{\sqrt{\pi }[4{B}^{3}{\eta }^{4}+4{B}^{2}{\eta }^{2}+(4{\eta }^{4}{C}^{2}+1)B]}=0,$$
(28)
$$\sigma B{A}^{3}+4({\rm{\Delta }}B+2{C}^{2})A+\frac{8{\eta }^{3}{P}_{0}{B}^{3}[1+4({B}^{2}-{C}^{2}){\eta }^{4}+4B{\eta }^{2}]}{\sqrt{\pi }{(1+4({B}^{2}+{C}^{2}){\eta }^{4}+4B{\eta }^{2})}^{2}}=0,$$
(29)
$$\frac{\pi (2C-\gamma )A}{2{B}^{2}}-\frac{32{\eta }^{5}{P}_{0}(1+2B{\eta }^{2})C}{\sqrt{\pi }{[1+4({B}^{2}+{C}^{2}){\eta }^{4}+4B{\eta }^{2}]}^{2}}=0.$$
(30)

### Numerical simulations

To solve the one-dimensional equation (2) numerically, we made use of the regularized delta-function based on the usual Gaussian expression (see ref64. and references therein):

$$\tilde{\delta }(x)={(\sqrt{\pi }w)}^{-1}\exp (-{x}^{2}/{w}^{2}),$$
(31)

with finite width w.

We employed a fourth-order split-step method to solve Eqs (2) and (20), starting from the zero input, ϕ(x, 0) = 0. An output was categorized as a stable mode if it maintained a static profile for a long time ($$ts\sim 1000$$, which corresponds $$\gtrsim 100$$ characteristic diffraction times). In most simulations, the spatial and temporal steps were fixed as Δx = 0.04 and Δt = 0.001. To produce the numerical results for the 1D LL equation, shown in Fig. 2, we chose values of w such that the resultant integral power was different from the analytical counterpart, corresponding to exact solution (10), by no more than 3%.

To solve Eqs (28)–(30) for A, B, and C, produced by the variational approximation, we used a relaxation method with a fixed error constraint of 10−6. Then, the so found values were inserted in ansatz (26) to produce the full variational approximation for the 2D modes. Lastly, the above-mentioned scaled value of the dissipation parameter, γ = 1, was set in all the simulations.


text:
# 17-28-043

ignitor

device intended, either by itself or in combination with other components, to generate voltage pulses to start a discharge lamp without providing for the preheating of the electrodes

Note 1 to entry: The element that releases the starting voltage pulse can be either triggered or non-triggered.

Note 2 to entry: This entry was numbered 845-08-33 in IEC 60050-845:1987.

Note 3 to entry: This entry was numbered 17-549 in CIE S 017:2011.

Publication date: 2020-12


text:
Generalization of Borsuk-Ulam

Let $n$ be a positive intger. Is the following true? For continuous maps $f: \mathbb S^n \rightarrow \mathbb S^n$ and $g: \mathbb S^n \rightarrow \mathbb R^n$, there exists a point $x \in \mathbb S^n$ such that $g(x) = g(f(x))$.

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OK, so I've been thinking about your problem a little bit, and I'm still a little bit unclear. Why do you call this statement a generalization of Borsuk-Ulam? – Thierry Zell Apr 9 '11 at 17:05
@Thierry: If f is the antipodal map of the sphere, then this is exactly the Borsuk-Ulam Theorem. – Bill Kronholm Apr 9 '11 at 17:12
Thanks Bill, I guess I'm not very awake this morning! – Thierry Zell Apr 9 '11 at 17:16

This is false for all $n \geq 2$, but true for $n=1$.

A map $g \colon X \to Y$ of topological space is said to be free if there is a map $f \colon X \to X$ such that $g(x) \neq g(f(x))$ for all $x\in X$. This definition appears to be due to Hopf, and was given in

H. Hopf, Freie Überdeckungen und freie Abbildungen, Fund. Math. 28 (1937), 33--57.

In that same paper, Hopf apparently proved that:

1. There is a free map $g \colon \mathbb S^n \to \mathbb R^n$ for every $n \geq 3$.
2. There is no free map from $\mathbb S^n$ into $\mathbb R$ if $n \geq 1$.

A generalization was obtained by Pannwitz in

E. Pannwitz, Eine freie Abbildung der n-dimensionalen Sphäre in die Ebene. Math. Nachr. 7 (1952), 183–-185.

According to the MR, she proved that, for every $n \geq 0$, there exists a free map $g \colon \mathbb S^n \to \mathbb R ^2$.

Unfortunately, I'm having trouble accessing the papers, so I can't provide any more details.

Update. Still no luck getting Pannwitz's paper. However, I have managed to find the following related paper

C. Biasi, D. de Mattos, E. dos Santos, Applications of the non-standard version of the Borsuk--Ulam theorem, JP J. Geom. Topol. 9 (2009), no. 3, 273--284.

(available online here), which mentions the above papers of Hopf and Pannwitz in its Introduction. More interesting however is Theorem 2.1, which I believe can be applied to Sergei's construction to yield Pannwitz's result for $n\geq 2$. This would show that the question in the OP has a negative answer if $n\geq 2$.

It remains to show that the answer is "yes" if $n=1$. (As Harry notes, the answer is also "yes" if $n=0$.) The following is Hopf's argument. Let $g \colon \mathbb S^1 \to \mathbb R$ and $f \colon \mathbb S^1 \to \mathbb S^1$ be given. Since $\mathbb S^1$ is compact, $g$ attains its maximum, resp. minimum, at some $x$, resp. $y$, in $\mathbb S^1$. In particular, $g(x) \geq g(f(x))$ and $g(y) \leq g(f(y))$. The intermediate value theorem then asserts that there is a $z \in \mathbb S^1$ for which $g(z) = g(f(z))$.

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It's also trivially true if n=0. :) – Harry Altman Apr 10 '11 at 15:31

This is not true for $n=2$.

Represent $\mathbb S^2$ as the cylinder $\mathbb S^1\times[-1,1]$ with two discs $D_+$ and $D_-$ attached to the boundary components $\mathbb S^1\times\{1\}$ and $\mathbb S^1\times\{-1\}$, resp. Denote by $\mathbb S^2_+$ and $\mathbb S^2_-$ the "positive" and "negative" hemispheres: $\mathbb S^2_+=(\mathbb S^1\times[0,1])\cup D_+$ and $\mathbb S^2_-$ is the opposite.

First we define $g:\mathbb S^2\to\mathbb R^2$. Consider a smooth $\infty$-shaped loop $\gamma:\mathbb S^1\to\mathbb R^2$, namely $$\gamma(t) = (\sin t,\tfrac12\sin2t)$$ (here $\mathbb S^1=\mathbb R/2\pi\mathbb Z$). Note that the velocity of $\gamma$ is separated away from the vertical vector $e_2=(0,1)$, namely $\angle(\dot\gamma(t),e_2)\ge\pi/4$ for all $t\in\mathbb S^1$. Choose $\varepsilon>0$ so small that $\angle (\gamma(t+\varepsilon)-\gamma(t),e_2)> \pi/5$ for all $t$.

For $x=(t,s)$ from the cylinder, define $$\begin{cases} g(x) =\gamma(t)+1000s\cdot\overrightarrow{(1,10)} , &\qquad s\ge 0 \cr g(x) =\gamma(t)+1000|s|\cdot\overrightarrow{(-1,10)} , &\qquad s\le 0 . \end{cases}$$ (its image of the cylinder consists of two almost vertical strips above the $\infty$-figure). Observe that the image of each boundary component $\mathbb S^1\times\{\pm 1\}$ is separated away from the image of the other half of the cylinder (by distance at least 100). Extend $g$ to $D_+$ and $D_-$ so as to fill these boundary components within their neighborhoods of radius 2. Then $g(D_+)\cap g(\mathbb S^2_-)=\emptyset$ and $g(D_-)\cap g(\mathbb S^2_+)=\emptyset$.

Since $\gamma(t+\varepsilon)-\gamma(t)$ never forms a small angle with $e_2$, the construction guarantees that $g(t,s)\ne\gamma(t+\varepsilon)$ for all $t\in\mathbb S^1$, $s\in[-1,1]$.

Now we define $f:\mathbb S^n\to\mathbb S^n$. For $x=(t,s)$ from the cylinder, define $f(x)=(t+\varepsilon,0)$, so the cylinder is projected to its equator and slightly rotated. Extend $f$ to $D_+$ and $D_-$ so that $f(D_+)\subset \mathbb S^2_-$ and $f(D^-)\subset \mathbb S^2_+$.

For these $f$ and $g$, we have $g(x)\ne g(f(x))$ for all $x\in\mathbb S^2$. Indeed, if $x=(t,s)\in\mathbb S^1\times[-1,1]$, then $f(g(x))=\gamma(t+\varepsilon)\ne g(x)$ as noted above. For $x\in D^+$ this follows from the fact that $g(D^+)\cap g(f(D^+))=\emptyset$, and similarly for $D^-$.

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Minor note, you seem to have clockwise and counterclockwise backwards. – Harry Altman Apr 10 '11 at 15:10
Thanks, corrected this by making the example more explicit. – Sergei Ivanov Apr 10 '11 at 22:05

On the other hand, it is known to be true if f is an involution, i.e. f(f(x))=x. This was proved in a paper by C.T. Yang in Annals of Mathematics in 1954.

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You mean this paper? jstor.org/stable/1969632 – Erel Segal-Halevi May 18 '15 at 9:22


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# Tag Info

## Hot answers tagged hedge

4

To continue from uness' answer (edit: just seen the OP was very old, but will leave here anyway!) . The greeks will be every element of market risk to which the the CVA is sensitive. Writing in words for celerity: A CVA is a credit linked option on the underlying instrument. You are sensitive to the credit default- (specifically the swap obligation payment ...

4

CVA is a price. Just like any price, you compute its sensitivities (greeks) and then use financial products to bring them as close to zero as possible. It's not possible to derive a hedging strategy just by looking at the CVA figure, it's like asking what the hedging strategy of a product is if its price is USD 1M... You need the CVA greeks. The ...

4

An Investment Bank earns a profit by selling you an option at a slightly higher price than the theoretical price, or buying it back from you at a slightly lower price. They call this "earning a spread". Then they hedge the option, so as not to make any [further] gains or losses on it (other than the risk free rate). Another way they could earn a profit is ...

4

I'm no special expert on options and their Greeks. However, I have had a decade plus experience of almost-daily discussions with a bank derivatives desk, on pin risk and the behaviour of autocallable, cliquet etc. structures. You are correct that a gamma hedge would require an options as opposed to underlying hedge. However, the traders' obsession with gamma ...

4

Step 1: Know your distribution Since $\int_0^t W_s\mathrm{d}s\sim N\left(0,\frac{1}{3}t^3\right)$, we have \begin{align*} S_t &= S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \int_0^t W_s\mathrm{d}s \right) \\ &\overset{d}{=} S_0 \exp\left( rt-\frac{1}{6}\sigma^2 t^3 + \sigma \sqrt{\frac{1}{3}t^3} Z \right) \\ &\overset{d}{=} S_0 \exp\left( ...

3

Short answer: Do not use BS for AC Long answer: There are plenty of question here about this already. Typically it is priced via Monte Carlo but that is not a model, just an implementation of some model (LV, SV, SLV). Standard would be to use SLV but even there are issues with decorrelation and bilocality if you look at basket AC (which are very common). ...

3

I will work through a detailed example. I hope it helps. Suppose for simplicity that you are trying to hedge the interest rate risk of one simplistic debt instrument. Suppose that the obligor owes you USD 10 million and promises to pay fixed 10% annual coupon, and to repay the principal in equal installments in years 4 through 7. So the cash flows look like ...

3

Let's use the following returns matrix, X 2Y 5Y 10Y -------------------------- 0.0143 0.0910 0.1451 0.1791 0.3505 0.4588 0.0572 0.1358 0.0120 0.0357 0.1809 0.2884 -0.0571 -0.1096 -0.0719 0.0286 0.0710 0.1319 0.0429 0.1806 0.2754 -0.0357 -0.0579 -0.1075 0.0714 0.2513 0.4304 -0.0214 -0....

3

This is just like any other option. For example, if you are trading an IBM option, you hedge with IBM stock, which doesn't expire at all, (obviously). You then sell your hedge in the gamma-storm that enuses at expiry. For longer term options where you have an expiry that goes way past the futures you have two choices: Trade the rolls periodically and/or ...

2

This is a resource you may want to look at. https://personal.vanguard.com/pdf/ISGHC.pdf Additionally, this books seems good for this particular topic: Risk Without Reward: The Case for Strategic FX Hedging. Also, take a look at Advanced Bond Portfolio Management: Best Practices in Modeling and Strategies edited by Frank J. Fabozzi, Lionel Martellini, ...

2

I would put the answer a bit differently: In the end you care about the price, don't you? If you sell the bond then it is bad if you can sell it for less. No matter what the yield is. E.g. if you have assets in a mutual fund then investors enter and leave the fund and you probably have to sell and buy assets (and there are more clever ways of cash ...

2

If the aim of the hedge is to make the portfolio insensitive (as much as possible) to small movements in the yield, then the question that needs answering is the following: If the yield of the hedge moves by $x$, by how much did the yield of the bond move? The answer is given by the correlation between yield movements between bond and hedging instrument, ...

2

The concept of covered interest rate parity (CIP) dictates that the forward price should equal the spot price multiplied by the ratio between domestic and foreign interest rates: $\ F = S*(1+i_d)/(1+i_f) \\$ In practice CIP means that the outcome of buying an FX forward should be equal to borrowing money in domestic currency (at the domestic interest rate),...

2

I'll try to give some views on this, I hope it helps bringing some closure to your question. You seem to relate consensus to "theoretical prices". I think this is a bit misleading. I view consensus as nothing more than the average view across the street for market factors, e.g. the correlation between the Korean KOSPI and the Spanish IBEX, the ...

2

Achieving gamma neutrality refers to your whole portfolio situation. If you have a portfolio $P$ made up of $n_S$ shares of stock $S$, and of $n_1$, $n_2$ option calls $C_1$, $C_2$ on $S$ (options 1 and 2 differ in strike price or expiration), pursuing a gamma hedging strategy would imply to achieve neutrality on the Delta $\Delta_P$ and Gamma $\Gamma_P$ ...

2

re 1 and 2: You might want to Google something like 'volatility decay' or 'volatility drag' for leveraged ETFs. re 3: a quick look at http://pages.stern.nyu.edu/~adamodar/New_Home_Page/datafile/histretSP.html suggests the long-term average return for US stocks is ~11.6% pa. You're proposing giving up ~5% pa to hedge. Not clear that's a good idea. also re 3: ...

2

\begin{align} d \left(e^{-rt} \left(V^i - V^a \right)\right) &= \left(d e^{-rt} \right) \left(V^i-V^a \right) + e^{-rt} d(V^i - V^a)\\ &= (-e^{-rt} r dt) (V^i - V^a) + e^{-rt} (dV^i - dV^a) \\ &= e^{-rt} [ -r (V^i - V^a)dt + (dV^i - dV^a) ] \end{align} So multiplying everything by $e^{rt}$ gives the result.

1

St. Louis Fed Financial Stress Index - https://fred.stlouisfed.org/series/STLFSI2 There are others (e.g. from the sellside) but this is publicly available. EDIT: looks like some of the sellside indicators are published on Bloomberg (no permissioning required), e.g. GFSI Index from BAML.

1

If we work through this practically, a hypothetical derivative under IFRS9 cash flow hedge should be representative of the underlying, henceforth the MTM CCS must be mark to market on the currency side opposite to that of the hedged instrument. For example, a EUR corporate might issue a yankee bond in 100mm USD and cross currency swap it back to EUR under a ...

1

Without a risk free investment, the efficient frontier is described by a hyperbola, as you have already suggested. Efficient Frontier: Tour de force Given asset covariance matrix $\Sigma$ and the full-investment condition $w_1+\ldots w_N=1$, it can be traced out by optimising \min_w w^T\Sigma w \quad s.t. \quad w^T\mathbf{1}=1 \quad \mathrm{and} \quad ... 1 Note that \begin{align*} Call_{\rm BTC}=\frac{1}{S}Call_{\rm USD}. \end{align*} The premium adjusted delta Delta_{PA} is defined as the change of Call_{\rm BTC} with respect to the change of the spot in BTC, that is, \begin{align*} Delta_{PA} &= \lim_{\Delta S\rightarrow 0}\frac{\Delta Call_{\rm BTC}}{\frac{\Delta S}{S}}\\ &=\lim_{\Delta S\... 1 So, basically, the answer is no. For capital requirements Basel has three categories: a) Counterparty Credit Risk b) Market Risk c) Operation Risk All RWA calculations are additive. If your hedge is with the same counterparty then it likely offsets a) and b) and possibly c). If your hedge is only a market hedge then it will only offset b) and possibly ... 1 Most of the literature in Finance assumes continuous hedging which is just practically impossible. Minimum variance hedge ratio assumes the same. Unless you’re a bank or HF that can cost effectively re-hedge portfolios regularly, you’ll never get anything near a perfect hedge. For a retail investor I would say to avoid time evolving betas like with Kalman ... 1 If you purchase a Stock today in USD you will model that it has some value in USD in the future, S_{t, usd} = S_{0, usd} + W_{t, usd} $$where W_{t, usd} is some random motion, possibly with drift, such that E[W_{t, usd}] = \mu. Ideally you would exchange S_{t, usd} at maturity, so this is the amount of notional that should be translated to ... 1 The two formulations seem to be exactly the same. If I take the equations from the first method: w_{D1}*\Delta_{D1} + w_{D2}*\Delta_{D2} = -2 w_{D1}*\Gamma_{D1} + w_{D2}*\Gamma_{D2} = -3 And substitute for delta and gamma of the two options: -w_{D1}+ 5 w_{D2}= -2 2w_{D1} -2w_{D2} = -3 which after shifting the constants to the left becomes ... 1 Intuitively, this is the "coupon effect" at work – when the yield curve is upward sloping, lower coupon bonds have higher yield and their yields move up more when the overall curve shifts up (all else equal). The opposite is true when the yield curve is downward sloping. We'll focus on when the curve is upward sloping below. I think it's probably best to ... 1 1) You didn't specify whether you're interested in ex post or ex ante risk analysis. But in either case, beta / correlation to the relevant stock market index is really important for long/short equity funds. They could be just leveraged beta plays and providing no alpha, or they could be market neutral and providing only alpha. Looking at exposure or ... 1 So after much calculations, this is the approach: In 1 year you need €1,000,000, how much do you need currently (x)? If the euro interest rate is at 6%,$$ x\ \times\ (1+6\%) = x(1.06) = €\ 1,000,000 \begin{align}x\ = \frac{€\ 1,000,000}{1.06} \newline \therefore\ x\ = €\ 943,396.22\end{align}$$With the current spot rate, we can convert € 943,... 1 Ito's lemma gives$$dF = \left(\frac{\partial F}{\partial t}+\frac{1}{2}\frac{\partial^2F}{\partial S^2}\sigma^2 S^2\right)dt + \frac{\partial F}{\partial S}dS = adt + bdS $$Using the usual rules, e.g. dz^2 = dt, we get$$ dS^2 = \sigma^2S^2dt,dF^2 = b^2dS^2 = b^2\sigma^2S^2dt,$$and$$dSdF = bdS^2 = b \sigma^2S^2dt,$$so this gives$$dP^2 = dS^2 +...

1

What you call additional basis risk is unpredictable. It may win or lose in rolling strategy against buying 1 year futures once. But what is measurable is bid/offer spread. In 1 y contract it might be significantly wider that in quarter futures, even considering that you sell 4 times and buy 3 (lose 7 half spreads).

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