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Tamil Nadu Board of Secondary EducationHSC Science Class 11th
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# Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide chapter 2 - Kinematics [Latest edition]

#### Chapters

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## Chapter 2: Kinematics

EvaluationExercises
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Evaluation [Pages 98 - 99]

### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide Chapter 2 KinematicsEvaluation [Pages 98 - 99]

#### Multiple Choice Questions

Evaluation | Q I. 01. | Page 98

Which one of the following Cartesian coordinate systems are not followed in physics?

•

•

Evaluation | Q I. 02. | Page 98

Identify the unit vector in the following.

• hati + hatj

• hati/sqrt2

• hatk - hatj/sqrt2

• (hati + hatj)/sqrt2

Evaluation | Q I. 03. | Page 98

Which one of the following physical quantities cannot be represented by a scalar?

• Mass

• length

• momentum

• magnitude of acceleration

Evaluation | Q I. 04. | Page 98

Two objects of masses m1 and m2 fall from the heights h1 and h2 respectively. The ratio of the magnitude of their momenta when they hit the ground is ______

• sqrt(h_1/h_2)

• sqrt((m_1h_1)/(m_2h_2))

• m_1/m_2sqrt(h_1/h_2)

• m_1/m_2

Evaluation | Q I. 05. | Page 98

If a particle has negative velocity and negative acceleration, its speed ______

• increases

• decreases

• remains same

• zero

Evaluation | Q I. 06. | Page 98

If the velocity is vecv = 2hati + t^2hatj - 9veck, then the magnitude of acceleration at t = 0.5 s is ______

• 1m s-2

• 2 m s-2

• zero

• -1 m s-2

Evaluation | Q I. 07. | Page 98

If an object is dropped from the top of a building and it reaches the ground at t = 4 s, then the height of the building is ______

(ignoring air resistance) (g = 9.8 ms–2)

• 77.3 m

• 78.4 m

• 80.5 m

• 79.2 m

Evaluation | Q I. 08. | Page 98

A ball is projected vertically upwards with a velocity v. It comes back to the ground in time t. Which v-t graph shows the motion correctly?

•

•

•

•

Evaluation | Q I. 09. | Page 98

If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant t is ______

• 1

• 2

• 4

• 0.5

Evaluation | Q I. 10. | Page 99

A ball is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?

•

•

•

•

Evaluation | Q I. 11. | Page 99

If a particle executes uniform circular motion in the xy plane in a clockwise direction, then the angular velocity is in ______

• +y direction

• +z direction

• -z direction

• -x direction

Evaluation | Q I. 12. | Page 99

If a particle executes uniform circular motion, choose the correct statement.

• The velocity and speed are constant.

• The acceleration and speed are constant.

• The velocity and acceleration are constant.

• The speed and magnitude of acceleration are constant.

Evaluation | Q I. 13. | Page 99

If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to the ground is ______

• u^2/(2g)

• u^2/(g)

• u/(2g)

• (2u)/g

Evaluation | Q I. 14. | Page 99

Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects is denoted as R30° and R60°. Choose the correct relation from the following.

• R30° = R60°

• R30° = 4R60°

• R30° = R_{60°}/2

• R30° = 2R60°

Evaluation | Q I. 15. | Page 99

An object is dropped in an unknown planet from a height of 50 m, it reaches the ground in 2 s. The acceleration due to gravity in this unknown planet is ______

• g = 20 m s-2

• g = 25 m s-2

• g = 15 m s-2

• g = 30 m s-2

Evaluation [Pages 99 - 100]

### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide Chapter 2 KinematicsEvaluation [Pages 99 - 100]

#### Short Answer Questions

Evaluation | Q II. 1. | Page 99

Explain what is meant by Cartesian coordinate system?

Evaluation | Q II. 2. | Page 99

Define a vector. Give examples?

Evaluation | Q II. 3. | Page 99

Define a scalar. Give examples?

Evaluation | Q II. 4. | Page 99

Write a short note on the scalar product between two vectors.

Evaluation | Q II. 5. | Page 100

Write a short note on the vector product between two vectors.

Evaluation | Q II. 6. | Page 100

How do you deduce that two vectors are perpendicular?

Evaluation | Q II. 7. | Page 100

Define displacement.

Evaluation | Q II. 7. | Page 100

Define distance.

Evaluation | Q II. 8. | Page 100

Define velocity.

Evaluation | Q II. 8. | Page 100

Define speed.

Evaluation | Q II. 9. | Page 100

Define acceleration.

Evaluation | Q II. 10. | Page 100

What is the difference between velocity and average velocity?

Evaluation | Q II. 11. | Page 100

Define a radian?

Evaluation | Q II. 12. | Page 100

Define angular displacement.

Evaluation | Q II. 12. | Page 100

Define angular velocity.

Evaluation | Q II. 13. | Page 100

What is non-uniform circular motion?

Evaluation | Q II. 14. | Page 100

Write down the kinematic equations for angular motion.

Evaluation | Q II. 15. | Page 100

Write down the expression for angle made by resultant acceleration and radius vector in the non-uniform circular motion.

Evaluation [Page 100]

### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide Chapter 2 KinematicsEvaluation [Page 100]

#### Long Answer Questions

Evaluation | Q III. 1. | Page 100

Explain in detail the triangle law of addition.

Evaluation | Q III. 2. | Page 100

Discuss the properties of scalar and vector products.

Evaluation | Q III. 3. | Page 100

Derive the kinematic equations of motion for constant acceleration.

Evaluation | Q III. 4. | Page 100

Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.

Evaluation | Q III. 5. | Page 100

Derive the equation of motion, range, and maximum height reached by the particle thrown at an oblique angle θ with respect to the horizontal direction.

Evaluation | Q III. 6. | Page 100

Derive the expression for centripetal acceleration.

Evaluation | Q III. 7. | Page 100

Derive the expression for total acceleration in the non-uniform circular motion.

Exercises [Pages 100 - 102]

### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide Chapter 2 KinematicsExercises [Pages 100 - 102]

Exercises | Q IV. 1. | Page 100

The position vector particle has a length of 1m and makes 30° with the x-axis. What are the lengths of the x and y components of the position vector?

Exercises | Q IV. 2. | Page 100

A particle has its position moved from vecr_1 =  3hati + 4hatj to vecr_2 =  hati + 2hatj. Calculate the displacement vector (∆vecr) and draw the vecr_1,  vecr_2 and Deltavecr vector in a two dimensional Cartesian coordinate system.

Exercises | Q IV. 3. | Page 100

Calculate the average velocity of the particle whose position vector changes from vecr_1 = 5hati + 6hatj  to vecr_2 = 2hati + 3hatj in a time 5 second.

Exercises | Q IV. 4. | Page 100

Convert the vector vecr = 3hati + 2hatj into a unit vector.

Exercises | Q IV. 5. | Page 100

What are the resultants of the vector product of two given vectors given by vecA = 4hati - 2hatj + hatk and vecB = 5hati + 3hatj - 4hatk?

Exercises | Q IV. 6. | Page 101

An object at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?

Exercises | Q IV. 7. | Page 101

The following graphs represent the velocity-time graph. Identify what kind of motion a particle undergoes in each graph.

 (a) (b) (c) (d)
Exercises | Q IV. 8. | Page 101

The following velocity-time graph represents a particle moving in the positive x-direction. Analyze its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.

Exercises | Q IV. 9. | Page 101

A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion.

(a) vx – decreases and increases
(b) vy – remains constant
(c) Acceleration – varies
(d) Position vector – remains downward

Exercises | Q IV. 10. | Page 101

A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountain is v. Calculate the total area around the fountain that gets wet.

Exercises | Q IV. 11. | Page 101

The following table gives the range of a particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity, (g value).

 Planet Range Jupiter 50 m Earth 75 m Mars 90 m Mercury 95m

Exercises | Q IV. 12. | Page 101

The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is

• 30°

• 45°

• 150°

• 120°

Exercises | Q IV. 13. | Page 101

Compare the components for the following vector equations

(a) "T"hatj - "mg"hatj = "ma"hatj
(b) vecT + vecF = vecA + vecB
(c) vecT - vecF = vecA - vecB
(d) "T"hatj + "mg"hatj = "ma"hatj

Exercises | Q IV. 14. | Page 102

Calculate the area of the triangle for which two of its sides are given by the vectors vecA = 5hati - 3hatj, vecB = 4hati + 6hatj

Exercises | Q IV. 15. | Page 102

If Earth completes one revolution in 24 hours, what is the angular displacement made by Earth in one hour? Express your answer in both radian and degree.

Exercises | Q IV. 16. | Page 102

An object is thrown with an initial speed of 5 m s-1 with an angle of projection of 30°. What is the height and range reached by the particle?

Exercises | Q IV. 17. | Page 102

A football player hits the ball with a speed of 20 m s-1 with an angle of 30° with respect to horizontal direction as shown in the figure. The goal post is at a distance of 40 m from him. Find out whether the ball reaches the goal post?

Exercises | Q IV. 18. | Page 102

If an object is thrown horizontally with an initial speed of 10 ms -1 from the top of a building of height 100 m. What is the horizontal distance covered by the particle?

Exercises | Q IV. 19. | Page 102

An object is executing uniform circular motion with an angular speed of pi/12 radians per second. At t = 0 the object starts at an angle θ = 0 What is the angular displacement of the particle after 4 s?

Exercises | Q IV. 20. | Page 102

Consider the x-axis as representing east, the y–axis as north, and the z-axis as vertically upwards. Give the vector representing each of the following points.

a) 5 m northeast and 2 m up

b) 4 m southeast and 3 m up

c) 2 m northwest and 4 m up

Exercises | Q IV. 21. | Page 102

The Moon is orbiting the Earth approximately once in 27 days, what is the angle transversed by the Moon per day?

Exercises | Q IV. 22. | Page 102

An object of mass m has angular acceleration α = 0.2 rad s-2. What is the angular displacement covered by the object after 3 seconds? (Assume that the object started with angle zero with zero angular velocity).

## Chapter 2: Kinematics

EvaluationExercises

## Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide chapter 2 - Kinematics

Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Physics Volume 1 and 2 Answers Guide chapter 2 (Kinematics) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Tamil Nadu Board of Secondary Education Class 11th Physics Volume 1 and 2 Answers Guide solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Tamil Nadu Board Samacheer Kalvi textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 11th Physics Volume 1 and 2 Answers Guide chapter 2 Kinematics are Introduction to Kinematics, Concept of Rest and Motion, Components of Vectors, Elementary Concept of Vector Algebra, Multiplication of Vector by a Scalar, Distance and Displacement, Position Vectors, Differential Calculus, Integral Calculus, Motion Along One Dimension, Projectile Motion.

Using Tamil Nadu Board Samacheer Kalvi Class 11th solutions Kinematics exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Tamil Nadu Board Samacheer Kalvi Solutions are important questions that can be asked in the final exam. Maximum students of Tamil Nadu Board of Secondary Education Class 11th prefer Tamil Nadu Board Samacheer Kalvi Textbook Solutions to score more in exam.

Get the free view of chapter 2 Kinematics Class 11th extra questions for Class 11th Physics Volume 1 and 2 Answers Guide and can use Shaalaa.com to keep it handy for your exam preparation

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lilypond-user
[Top][All Lists]

## Re: Chord from a variable

 From: David Kastrup Subject: Re: Chord from a variable Date: Sun, 07 Aug 2011 21:15:01 +0200 User-agent: Gnus/5.13 (Gnus v5.13) Emacs/24.0.50 (gnu/linux)

> Then I must ask the other way:
> Why is the c in the chord one octave higher?
> I.e., why is the c in \chordmode { c } interpreted as c'?

Because that's a more natural pitch to start a chord.  I think that
chordmode, for whatever reason, ignores octave specifications, either
just on the bass note for inversions or altogether.

--
David Kastrup

format: plain


text:
Quadratic Equation – JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. If the roots of the equation $b x^{2}+c x+a=0$ be imaginary, then for all real values of x, the expression $3 b^{2} x^{2}+6 b c x+2 c^{2}$ is :-

(1) Greater than –4ab (2) Less than –4ab (3) Greater than 4ab (4) Less than 4ab

[AIEEE-2010]

Sol. (1)

since the roots of $\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{a}=0$ are imaginary

$\because c^{2}-4 a b<0 \Rightarrow c^{2}<4 a b$

for exp. $3 b^{2} x^{2}+6 b c x+2 c^{2}$

Min. value $=\frac{-\mathrm{D}}{4 \mathrm{a}}=\frac{\left(36 \mathrm{b}^{2} \mathrm{c}^{2}-24 \mathrm{b}^{2} \mathrm{c}^{2}\right)}{12 \mathrm{b}^{2}}=\frac{-12 \mathrm{b}^{2} \mathrm{c}^{2}}{12 \mathrm{b}^{2}}=-\mathrm{c}^{2}$

$\therefore-\mathrm{c}^{2}>-4 \mathrm{ab}$

So exp. is greater than (– 4ab)

Q. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+1=0,$ then $\alpha^{2009}+\beta^{2009}=$

(1) –2 (2) –1 (3) 1 (4) 2

[AIEEE-2010]

Sol. (3)

Roots of equation $\mathrm{x}^{2}-\mathrm{x}+1=0 \mathrm{ar}$

$\alpha=-\omega, \quad \beta=-\omega^{2}$

$\alpha^{2009}+\beta^{2009}=(-\omega)^{2009}+\left(-\omega^{2}\right)^{2009}$c

$=-\left(\omega^{2}+\omega\right)=1$

Q. Let for $a \neq a_{1} \neq 0, f(x)=a x^{2}+b x+c, g(x)=a_{1} x^{2}+b_{1} x+c_{1}$ and $p(x)=f(x)-g(x)$

If $\mathrm{p}(\mathrm{x})=0$ only for $\mathrm{x}=-1$ and $\mathrm{p}(-2)=2,$ then the value of $\mathrm{p}(2)$ is:

(1) 18 (2) 3 (3) 9 (4) 6

[AIEEE-2011]

Sol. (1)

$\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}+1)^{2}$

$\mathrm{P}(-2)=2=\mathrm{k}(-1)^{2}$

$\Rightarrow \mathrm{k}=2$

$\therefore \mathrm{P}(\mathrm{x})=2(\mathrm{x}+1)^{2}$

$\Rightarrow \mathrm{P}(2)=18$

Aliter :

$\mathrm{P}(\mathrm{x})=\left(\mathrm{a}-\mathrm{a}_{1}\right) \mathrm{x}^{2}+\left(\mathrm{b}-\mathrm{b}_{1}\right) \mathrm{x}+\left(\mathrm{c}-\mathrm{c}_{1}\right)=0$

only x = –1

P(x) = 0

So roots are equal.

it means D = 0

$\left(b-b_{1}\right)^{2}=4\left(a-a_{1}\right)\left(c-c_{1}\right) \Rightarrow q^{2}=4 p r$

Let

$\mathrm{a}-\mathrm{a}_{1}=\mathrm{p}$

$\mathrm{b}-\mathrm{b}_{1}=\mathrm{q}$

$\mathrm{c}-\mathrm{c}_{1}=\mathrm{r}$

P(–1) = 0

p – q + r = 0 …… (1)

4p – 2q + r = 2 ….. (2)

4p + 2q + r = ?

4p + 2q + r = ?

$(p+r)^{2}-4 p r=0$

$(p-r)^{2}=0$

from eq. (1) q = 2r

So from eq. (2) 4r – 4r + r = 2

r = 2

So 4p + 2q + r = 4r + 4r + r = 9r = 18

Q. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are:

(1) –4, –3 (2) 6, 1 (3) 4, 3 (4) –6, –1

[AIEEE-2011]

Sol. (2)

Q. The equation $\mathrm{e}^{\mathrm{sinx}}-\mathrm{e}^{-\mathrm{sinx}}-4=0$ has :

(1) exactly four real roots.

(2) infinite number of real roots.

(3) no real roots.

(4) exactly one real root.

[AIEEE-2012]

Sol. (3)

Q. If the equations $\mathrm{x}^{2}+2 \mathrm{x}+3=0$ and $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$, have a common root, then a : b : c is :

(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2

[JEE-MAIN-2013]

Sol. (1)

$\mathrm{x}^{2}+2 \mathrm{x}+3=0$

$\mathrm{D}<0$

$\therefore \quad \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 \mathrm{has}$

both roots common

$\therefore \quad a: b: c=1: 2: 3$

Q. Let  and $\beta$ be the roots of equation $\mathrm{x}^{2}-6 \mathrm{x}-2=0 .$ If $\mathrm{a}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}},$ for $\mathrm{n} \geq$ 1, then the value of $\frac{\mathrm{a}_{10}-2 \mathrm{a}_{8}}{2 \mathrm{a}_{9}}$ is equal to :

(1) 3 (2) – 3 (3) 6 (4) – 6

[JEE-MAIN-2015]

Sol. (1)

Q. The sum of all real values of x satisfying the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is :

(1) 5 (2) 3 (3) –4 (4) 6

[JEE-MAIN-2016]

Sol. (2)

$\mathrm{x}^{2}-5 \mathrm{x}+5=1 \Rightarrow \mathrm{x}=1,4$

$x^{2}-5 x+5=-1 \Rightarrow x=2,3$

but 3 is rejected

$\mathrm{x}^{2}+4 \mathrm{x}-60=0 \Rightarrow \mathrm{x}=-10,6$

Sum = 3

Q. If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to-

If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to-

[JEE-MAIN-2018]

Sol. (2)


text:
AAS 199th meeting, Washington, DC, January 2002
Session 130. Surveys, Surveys
Display, Thursday, January 10, 2002, 9:20am-4:00pm, Exhibit Hall

## [130.06] Chandra Multiwavelength Plane Survey (ChaMPlane): Photometry Results from the First Two Years

P. Zhao, J. E. Grindlay, P. Edmonds, M. R. Garcia, J. E. McClintock (Harvard-Smithsonian CfA), H. Cohn, P. Lugger, S. Slavin (Indiana U.), D. W. Hoard, S. Wachter (CTIO)

We present the photometric results of the Chandra Multiwavelength Plane Survey (ChaMPlane), which is a project to identify a large sample of serendipitous X-ray sources in the galactic plane, in order to determine the populations of accretion-powered binaries in the Galaxy. Deep Chandra field observations have sufficient sensitivity to reach objects with Lx = 1031~erg/s out to about 7~kpc, i.e. a substantial fraction of the galactic disk. The ChaMPlane project has been chosen as one of the NOAO Long Term Survey Program. We have successfully conducted the ChaMPlane survey for the first two years and obtained deep images in the V, R, I and \bf H\alpha bands for more than 20 Chandra fields from the CTIO and KPNO 4-m telescopes, which covers more than eight square degrees in the galactic plane. Cataclysmic Variables (CVs) and quiescent Low Mass X-ray Binaries (qLMXBs) are identified by their ubiquitous \bf H\alpha excess as blue'' objects in the R vs. \bf (H\alpha - R) diagram. We describe our data analysis methods and present initial photometry results for this first significant sample of ChaMPlane optical data. Initial X-ray results for a subset of these fields are presented by Grindlay et al. This work is supported by NASA/SAO grant AR1-2001X and NSF grant AST-0098683.


text:
# Umm, cauchy-schwarz

1. Sep 6, 2005

### misogynisticfeminist

I'm self taught though, so please bear with any questions i have. One side of the cauchy schwarz innequality (CSI, nice acronym) is

l u.v l

Firstly what's the difference between ll u ll and l u l . I thought the norm was the length.

Also, what does it mean by the length of the dot product of u and v? I thought the dot product was a number itself, and not a tuple or something.

Last edited: Sep 6, 2005
2. Sep 6, 2005

### quasar987

llull = norm/lenght of the vector u.

|u| = ABSOLUTE VALUE of the scalar u.

3. Sep 7, 2005

### TD

$$\left| {\vec x \cdot \vec y} \right| \leqslant \left\| {\vec x} \right\|\left\| {\vec y} \right\|$$

As quasar987 said, " | | " is for the absolute value. So as you said correctly, the inner product is a scalar so the inequality states that the absolute value of the inner product of two vectors is always less than (or equal too) the product of the norms of both vectors.

4. Sep 7, 2005

### matt grime

However, many people use |x| and ||x|| interchangably since there is no real difference between them; a 1-d vector is a scalar so the scalar norm and the vector norm are the same here. And most (pure) mathematicians do not distinguish symbolically between vectors and scalars; they let the context make it clear which is meant.

5. Sep 7, 2005

### misogynisticfeminist

ahh, that cleared the doubts. Thanks alot.

6. Sep 12, 2005

### Dr Avalanchez

I do not agree on that. ||.|| is a norm and | . | is the absolute value. norm() applies to vectors, abs() applies to scalars. abs() is a well defined function for real numbers (and the complex analogon), whereas norm() is not. A function is a norm in some sort of vector space iff it satisfies the parallellogram identity. The well-known frobeniusnorm is just an example, just as the 1-norm, 2-norm, maxnorm, sylvesternorm,... In fact, just like inner products on some vector space, you can "invent" a norm (as long as it satisfies the parallellogram identity).

However, if explicitly stated that the vector space is euclidean/unitarian, with standard norm, then I agree with you. But not that mathematicians do not distinguish any difference between them.

7. Sep 12, 2005

### matt grime

But we are talking about vector spaces and the euclidean norm.

In other cases this won't be true: say in the theory of elliptic curves the synmbol | | will often be taken to be the p-adic valuation.

8. Sep 12, 2005

### Hurkyl

Staff Emeritus
norm() it is too defined for the real numbers... Or are you asserting that your textbooks define the Euclidean norm specifically to exclude vector spaces of dimension 1 (and of dimension 0)?

9. Sep 14, 2005

### Dr Avalanchez

No, I was just pointing out that you can have all sorts of norms, so it should be stated what vector space we're talking about, and which norm. It was a critique on Matt that mathematicians do make a distinction (unless it's clear what we're talking about, like here).


text:
# Finding the inverse of a quadratic function?

Hello, I am sorry for removing the template but it is not an actual problem that I need help with. Well it is, but it should be pretty quick and straight forward.

The problem is located here: http://www.cdli.ca/courses/math3103/unit05_org01_ilo04/images/3-less9.gif [Broken]

(It's a bit difficult to type out.)

I'm having difficulty with only 1 step that I need straightened out.

On the second last line, where did the 1/2 outside of the square root function come from? And Why does it appear there? I am thinking the 1/2 just got moved out of it, because it was under x+3.

Last edited by a moderator:

Dick
Homework Helper
They wrote (x+3)/-2=(x+3)*2/((-2)*2)=(2x+6)/(-4)=(1/4)*((2x+6)/(-1))=(1/4)*(-2x-6). Then they took sqrt(1/4)=1/2 and moved that out.

Multiply both numerator and denominator by -2 inside the square root

$$y + 1 = \pm \sqrt{\frac{-2(x + 3)}{- 2 (-2)}$$

Multiply both numerator and denominator by -2 inside the square root

$$y + 1 = \pm \sqrt{\frac{-2x-6}{4}$$


text:
Algebra 1 : How to use the quadratic function

Example Questions

Example Question #1 : Understand Functions: Ccss.Math.Content.8.F.A.1

Solve the equation:

Explanation:

To solve the quadratic equation, , we set the equation equal to zero and then factor the quadratic, . Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations  and      to obtain the answers  and

Example Question #1 : Functions

Solve for :

The solution is undefined.

Explanation:

To factor this equation, first find two numbers that multiply to 35 and sum to 12.  These numbers are 5 and 7.  Split up 12x using these two coefficients:

Example Question #1 : How To Use The Quadratic Function

Given , find .

Explanation:

Plug in a for x:

Next plug in (a + h) for x:

Therefore f(a+h) - f(a) =  .

Example Question #1 : How To Use The Quadratic Function

Which of the following is the correct solution when    is solved using the quadratic equation?

Explanation:

Example Question #141 : Functions And Lines

Give the minimum value of the function .

This function does not have a minimum.

Explanation:

This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula , setting :

Now evaluate the function at :

Example Question #91 : Algebraic Functions

Quadratic equations may be written in the following format:

In the equation ,  what is the value of ?

Explanation:

For the given equation below:

The values of each coefficient are:

Example Question #1 : How To Use The Quadratic Function

Solve for x.

Explanation:

The quadratic formula is as follows:

We will start by finding the values of the coefficients of the given equation, but first we must simplify.

Move all the terms to one side and set the equation equal to .

Rearrange.

We can then find the values of the coefficients of the equation:

Quadratic equations may be written in the following format:

In our case, the values of the coefficients are:

Substitute the coefficient values into the quadratic equation:

After simplifying we are left with:

Example Question #1 : Functions

Solve for :

Explanation:

To find , we must factor the quadratic function:

Example Question #2 : Understand Functions: Ccss.Math.Content.8.F.A.1

Solve for :

Explanation:

To find , we want to factor the quadratic function:


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# High‐Energy Gamma Rays from the Massive Black Hole in the Galactic Center

,

The Astrophysical Journal

University of Chicago Press

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There is no author summary for this article yet. Authors can add summaries to their articles on ScienceOpen to make them more accessible to a non-specialist audience.

### Most cited references35

• Record: found

### High-Energy Electromagnetic Conversion Processes in Intense Magnetic Fields

(1966)
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### Opacity of the Universe to High-Energy Photons

(1967)
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Is Open Access

### Observability of Gamma Rays from Dark Matter Neutralino Annihilations in the Milky Way Halo

(1997)
Recent advances in N-body simulations of cold dark matter halos point to a substantial density enhancement near the center. This means that, e.g., the $$\gamma$$ ray signals from neutralino dark matter annihilations would be significantly enhanced compared to old estimates based on an isothermal sphere model with large core radius. Another important development concerns new detectors, both space- and ground-based, which will cover the window between 50 and 300 GeV where presently no cosmic $$\gamma$$-ray data are available. Thirdly, new calculations of the $$\gamma$$-ray line signal (a sharp spike of $$10^{-3}$$ relative width) from neutralino annihilations have revealed a hitherto neglected contribution which, for heavy higgsino-like neutralinos, gives an annihilation rate an order of magnitude larger than previously predicted. We make a detailed phenomenological study of the possible detection rates given these three pieces of new information. We show that the proposed upgrade of the Whipple telescope will make it sensitive to a region of parameter space, with substantial improvements possible with the planned new generation of Air Cherenkov Telescope Arrays. We also comment on the potential of the GLAST satellite detector. An evaluation of the continuum $$\gamma$$-rays produced in neutralino annihilations into the main modes is also done. It is shown that a combination of high-rate models and very peaked halo models are already severely constrained by existing data.
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### Author and article information

###### Journal
The Astrophysical Journal
ApJ
University of Chicago Press
0004-637X
1538-4357
January 20 2005
January 20 2005
: 619
: 1
: 306-313
###### Article
10.1086/426426


text:
# Competitive Lotka–Volterra equations

The competitive Lotka–Volterra equations are a simple model of the population dynamics of species competing for some common resource. They can be further generalised to include trophic interactions.

## Overview

The form is similar to the Lotka–Volterra equations for predation in that the equation for each species has one term for self-interaction and one term for the interaction with other species. In the equations for predation, the base population model is exponential. For the competition equations, the logistic equation is the basis.

The logistic population model, when used by ecologists often takes the following form:

${dx \over dt} = rx\left(1-{x \over K}\right).$

Here x is the size of the population at a given time, r is inherent per-capita growth rate, and K is the carrying capacity.

### Two species

Given two populations, x1 and x2, with logistic dynamics, the Lotka–Volterra formulation adds an additional term to account for the species' interactions. Thus the competitive Lotka–Volterra equations are:

${dx_1 \over dt} = r_1x_1\left(1-\left({x_1+\alpha_{12}x_2 \over K_1}\right) \right)$
${dx_2 \over dt} = r_2x_2\left(1-\left({x_2+\alpha_{21}x_1 \over K_2}\right) \right).$

Here, α12 represents the effect species 2 has on the population of species 1 and α21 represents the effect species 1 has on the population of species 2. These values do not have to be equal. Because this is the competitive version of the model, all interactions must be harmful (competition) and therefore all α-values are positive. Also, note that each species can have its own growth rate and carrying capacity. A complete classification of this dynamics, even for all sign patterns of above coefficients, is available,[1] which is based upon equivalence to the 3-type replicator equation.

### N species

This model can be generalized to any number of species competing against each other. One can think of the populations and growth rates as vectors and the interaction α's as a matrix. Then the equation for any species i becomes

$\frac{dx_i}{dt} = r_i x_i \left(1- \frac{\sum_{j=1}^N \alpha_{ij}x_j}{K_i} \right)$

or, if the carrying capacity is pulled into the interaction matrix (this doesn't actually change the equations, only how the interaction matrix is defined),

$\frac{dx_i}{dt} = r_i x_i \left( 1 - \sum_{j=1}^N \alpha_{ij}x_j \right)$

where N is the total number of interacting species. For simplicity all self-interacting terms αii are often set to 1.

### Possible dynamics

The definition of a competitive Lotka-Volterra system assumes that all values in the interaction matrix are positive or 0 (αij ≥ 0 for all i,j). If it is also assumed that the population of any species will increase in the absence of competition unless the population is already at the carrying capacity (ri > 0 for all i), then some definite statements can be made about the behavior of the system.

1. The populations of all species will be bounded between 0 and 1 at all times (0 ≤ xi ≤ 1, for all i) as long as the populations started out positive.
2. Smale[2] showed that Lotka-Volterra systems that meet the above conditions and have five or more species (N ≥ 5) can exhibit any asymptotic behavior, including a fixed point, a limit cycle, an n-torus, or attractors.
3. Hirsch[3][4][5] proved that all of the dynamics of the attractor occur on a manifold of dimension N-1. This basically says that the attractor cannot have dimension greater than N-1. Why is this important? A limit cycle cannot exist in fewer than two dimensions. An n-torus cannot exist in less than n dimensions, and finally, chaos cannot occur in less than three dimensions. So, Hirsch proved that competitive Lotka–Volterra systems cannot exhibit a limit cycle for N < 3, or any torus or chaos for N < 4. This is still in agreement with Smale that any dynamics can occur for N ≥ 5.
• More specifically, Hirsch showed there is an invariant set C that is homeomorphic to the (N-1)-dimensional simplex
$\Delta_{N-1} = \left \{ x_i : x_i \ge 0, \sum x_i = 1 \right \}$
and is a global attractor of every point excluding the origin. This carrying simplex contains all of the asymptotic dynamics of the system.
4. To create a stable ecosystem the αij matrix must have all positive eigenvalues. For large N systems Lotka-Volterra models are either unstable or have low connectivity. Kondoh [6] and Ackland and Gallagher [7] have independently shown that large, stable Lotka-Volterra systems arise if the elements of αij (i.e. the features of the species) can evolve in accordance with natural selection.

## 4-dimensional example

The competitive Lotka–Volterra system plotted in phase space with the x4 value represented by the color.

A simple 4-Dimensional example of a competitive Lotka–Volterra system has been characterized by Vano et al.[8] Here the growth rates and interaction matrix have been set to

$r_i = \begin{bmatrix} 1 \\ 0.72 \\ 1.53 \\ 1.27 \end{bmatrix} \quad \alpha_{ij} = \begin{bmatrix} 1 & 1.09 & 1.52 & 0 \\ 0 & 1 & 0.44 & 1.36 \\ 2.33 & 0 & 1 & 0.47 \\ 1.21 & 0.51 & 0.35 & 1 \end{bmatrix}.$

This system is chaotic and has a largest Lyapunov exponent of 0.0203. From the theorems by Hirsch, it is one of the lowest-dimensional chaotic competitive Lotka–Volterra systems. The Kaplan–Yorke dimension, a measure of the dimensionality of the attractor, is 2.074. This value is not a whole number, indicative of the fractal structure inherent in a strange attractor. The coexisting equilibrium point, the point at which all derivatives are equal to zero but that is not the origin, can be found by inverting the interaction matrix and multiplying by the unit column vector, and is equal to

$\overline{x} = \left ( \alpha_{ij} \right )^{-1} \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0.3013 \\ 0.4586 \\ 0.1307 \\ 0.3557 \end{bmatrix}.$

Note that there are always 2N equilibrium points, but all others have at least one species' population equal to zero.

The eigenvalues of the system at this point are 0.0414±0.1903i, -0.3342, and -1.0319. This point is unstable due to the positive value of the real part of the complex eigenvalue pair. If the real part were negative, this point would be stable and the orbit would attract asymptotically. The transition between these two states, where the real part of the complex eigenvalue pair is equal to zero, is called a Hopf bifurcation.

A detailed study of the parameter dependence of the dynamics was performed by Roques and Chekroun in.[9] The authors observed that interaction and growth parameters leading respectively to extinction of three species, or coexistence of two, three or four species, are for the most part arranged in large regions with clear boundaries. As predicted by the theory, chaos was also found; taking place however over much smaller islands of the parameter space which makes difficult the identification of their location by a random search algorithm.[8] These regions where chaos occurs are, in the three cases analyzed in,[9] situated at the interface between a non-chaotic four species region and a region where extinction occurs. This implies a high sensitivity of biodiversity with respect to parameter variations in the chaotic regions. Additionally, in regions where extinction occurs which are adjacent to chaotic regions, the computation of local Lyapunov exponents [10] revealed that a possible cause of extinction is the overly strong fluctuations in species abundances induced by local chaos.

## Spatial arrangements

An illustration of spatial structure in nature. The strength of the interaction between bee colonies is a function of their proximity. Colonies A and B interact, as do colonies B and C. A and C do not interact directly, but affect each other through colony B.

### Background

There are many situations where the strength of species' interactions depends on the physical distance of separation. Imagine bee colonies in a field. They will compete for food strongly with the colonies located near to them, weakly with further colonies, and not at all with colonies that are far away. This doesn't mean, however, that those far colonies can be ignored. There is a transitive effect that permeates through the system. If colony A interacts with colony B, and B with C, then C affects A through B. Therefore, if the competitive Lotka–Volterra equations are to be used for modeling such a system, they must incorporate this spatial structure.

### Matrix organization

One possible way to incorporate this spatial structure is to modify the nature of the Lotka–Volterra equations to something like a reaction-diffusion system. It is much easier, however, to keep the format of the equations the same and instead modify the interaction matrix. For simplicity, consider a five species example where all of the species are aligned on a circle, and each interacts only with the two neighbors on either side with strength α-1 and α1 respectively. Thus, species 3 interacts only with species 2 and 4, species 1 interacts only with species 2 and 6, etc. The interaction matrix will now be

$\alpha_{ij} = \begin{bmatrix}1 & \alpha_1 & 0 & 0 & \alpha_{-1} \\ \alpha_{-1} & 1 & \alpha_1 & 0 & 0 \\ 0 & \alpha_{-1} & 1 & \alpha_1 & 0 \\ 0 & 0 & \alpha_{-1} & 1 & \alpha_1 \\ \alpha_1 & 0 & 0 & \alpha_{-1} & 1 \end{bmatrix}.$

If each species is identical in its interactions with neighboring species, then each row of the matrix is just a permutation of the first row. A simple, but non-realistic, example of this type of system has been characterized by Sprott et al.[11] The coexisting equilibrium point for these systems has a very simple form given by the inverse of the sum of the row

$\overline{x}_i = \frac{1}{\sum_{j=1}^N \alpha_{ij}} = \frac{1}{\alpha_{-1} + 1 + \alpha_1}.$

### Lyapunov functions

A Lyapunov function is a function of the system f = f(x) whose existence in a system demonstrates stability. It is often useful to imagine a Lyapunov function as the energy of the system. If the derivative of the function is equal to zero for some orbit not including the equilibrium point, then that orbit is a stable attractor, but it must be either a limit cycle or n-torus - but not a strange attractor (this is because the largest Lyapunov exponent of a limit cycle and n-torus are zero while that of a strange attractor is positive). If the derivative is less than zero everywhere except the equilibrium point, then the equilibrium point is a stable fixed point attractor. When searching a dynamical system for non-fixed point attractors, the existence of a Lyapunov function can help eliminate regions of parameter space where these dynamics are impossible.

The spatial system introduced above has a Lyapunov function that has been explored by Wildenberg et al.[12] If all species are identical in their spatial interactions, then the interaction matrix is circulant. The eigenvalues of a circulant matrix are given by[13]

$\lambda_k = \sum_{j=0}^{N-1} c_j\gamma^{kj}$

for k = 0N − 1 and where $\gamma = e^{i2\pi/N}$ the Nth root of unity. Here cj is the jth value in the first row of the circulant matrix.

The Lyapunov function exists if the real part of the eigenvalues are positive (Re(λk > 0 for k = 0, …, N/2). Consider the system where α-2 = a, α-1 = b, α1 = c, and α2 = d. The Lyapunov function exists if

$\operatorname{Re}(\lambda_k) = \operatorname{Re} \left ( 1+\alpha_{-2}e^{i2 \pi k(N-2)/N}+\alpha_{-1}e^{i2 \pi k(N-1)/N}+\alpha_1e^{i2 \pi k/N}+\alpha_2e^{i4 \pi k/N} \right )$ $= 1+(\alpha_{-2}+\alpha_2)\cos \left ( \frac{i4 \pi k}{N} \right ) + (\alpha_{-1}+\alpha_1)\cos \left ( \frac{i2 \pi k}{N} \right ) > 0$

for k = 0, … ,N − 1. Now, instead of having to integrate the system over thousands of time steps to see if any dynamics other than a fixed point attractor exist, one need only determine if the Lyapunov function exists (note: the absence of the Lyapunov function doesn't guarantee a limit cycle, torus, or chaos).

Example: Let α−2 = 0.451, α−1 = 0.5, and α2 = 0.237. If α1 = 0.5 then all eigenvalues are negative and the only attractor is a fixed point. If α1 = 0.852 then the real part of one of the complex eigenvalue pair becomes positive and there is a strange attractor. The disappearance of this Lyapunov function coincides with a Hopf bifurcation.

### Line systems and eigenvalues

The eigenvalues of a circle, short line, and long line plotted in the complex plane

It is also possible to arrange the species into a line.[12] The interaction matrix for this system is very similar to that of a circle except the interaction terms in the lower left and upper right of the matrix are deleted (those that describe the interactions between species 1 and N, etc.).

$\alpha_{ij} = \begin{bmatrix}1 & \alpha_1 & 0 & 0 & 0 \\ \alpha_{-1} & 1 & \alpha_1 & 0 & 0 \\ 0 & \alpha_{-1} & 1 & \alpha_1 & 0 \\ 0 & 0 & \alpha_{-1} & 1 & \alpha_1 \\ 0 & 0 & 0 & \alpha_{-1} & 1 \end{bmatrix}$

This change eliminates the Lyapunov function described above for the system on a circle, but most likely there are other Lyapunov functions that have not been discovered.

The eigenvalues of the circle system plotted in the complex plane form a trefoil shape. The eigenvalues from a short line form a sideways Y, but those of a long line begin to resemble the trefoil shape of the circle. This could be due to the fact that a long line is indistinguishable from a circle to those species far from the ends.

## Notes

1. ^ I. Bomze, Lotka–Volterra equation and replicator dynamics: a two-dimensional classification. Biological Cybernetics 48, 201–211 (1983); I. Bomze, Lotka–Volterra equation and replicator dynamics: new issues in classification. Biological Cybernetics 72, 447–453 (1995).
2. ^ S. Smale, J. Math. Biol. 3, 5., 1976
3. ^ M. Hirsch, SIAM J. Math. Anal. 16, 423., 1985
4. ^ M. Hirsch, Nonlinearity 1, 51., 1988
5. ^ M. Hirsch, SIAM J. Math. Anal. 21, 1225., 1990
6. ^ Kondoh M Science 299, 1388 (2003)
7. ^ Ackland GJ and Gallagher ID Phys. Rev. Lett. 93, 158701 (2004)
8. ^ a b Vano, J.A., Wildenberg, J.C., Anderson, M.B., Noel, J.K., Sprott, J.C. Chaos in low-dimensional Lotka--Volterra models of competition. Nonlinearity, 19, 2391-- 2404
9. ^ a b L.Roques, and M.D. Chekroun (2011): Probing chaos and biodiversity in a simple competition model, Ecological Complexity, 8 (1), 98--104.
10. ^ Nese, J.M., 1989. Quantifying local predictability in phase space. Physica D 35,237--250.
11. ^ Sprott, J.C., Wildenberg, J.C., Azizi, Y. A simple spatiotemporal chaotic Lotka–Volterra model. Chaos, Solitons & Fractals 26, 1035., 2005
12. ^ a b Wildenberg, J.C., Vano, J.A., Sprott, J.C. Complex spatiotemporal dynamics in Lotka–Volterra ring systems. Ecological Complexity 3, 140. 2006
13. ^ Hofbauer, J., Sigmund, K., 1988. The Theory of Evolution and Dynamical Systems. Cambridge University Press, Cambridge, U.K, p. 352.


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# How does changing the angular frequency ω of the generator affect the current flowing in the circuit and the power transferred to the electric motor?

When an electric motor is used with an electric generator, the situation can be viewed as an RLC circuit where the generator is an AC power source. The source produces a sinusoidal AC voltage with an value of V and an angular frequency ω. The work done by the motor is reflected in the resistance R of the circuit. How does changing the angular frequency ω of the generator affect the current flowing in the circuit and the power transferred to the electric motor?

Let's define inductive reactance $$\chi_L=\omega L$$ and capacitive reactance $$\chi_c=\frac{1}{\omega C}$$. Then, using resistance $$R$$, the impedance of the circuit is $$Z=\sqrt{R^2+(\chi_L-\chi_C)^2}$$ The rms current is $$I=\frac{V}{Z}$$ where $$V$$ is rms voltage.
The phase between the current and voltage is given by $$\tan\delta=\frac{\chi_L-\chi_C}{R}$$ Then, the power is simply $$P=IV\cos\delta$$ Inserting the initial variables, $$P=\frac{V^2}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}\cdot\sqrt{\frac{\omega L-\frac{1}{\omega C}-R}{R}}$$ The power is the highest when $$\delta=0$$ which happens when $$\chi_L=\chi_C$$.


text:
# Entries explained

Dictionary entries typically consist of a headword (in bold), a part of speech, and a definition.

In the above entry for bakoa, the part of speech is given as "a." (adjective). Byington is extremely thorough in listing alternative translations of each Choctaw word. Three example sentences with translations are given above. At the end is a cross-reference to an alternative spelling.

In the following entry, the abbreviations "sing." (for singular) and "pl." (for plural) help describe usage.

The cross-reference in the above entry to akama is to a related verb form. Byington often refers to examples in the Bible to support specific translations. These citations are sometimes to older versions of the Choctaw Bible, however.


text:
?

Free Version
Difficult

# Location of Maximum for Function Involving Power of Natural Log

APCALC-EJKJVL

A relative maximum of the function $g(x)= \cfrac {\ln^3 {x}}{x}$ occurs when $x =$

A

1

B

3

C

$e^3$

D

$\ln{3}$


text:
FAQ/wald - CBU statistics Wiki
location: FAQ / wald

What is the Wald statistic?

The wald statistic, outputted in logistic and Cox model regressions, is simply the square of the regression coefficient, B, divided by its variance. This is also calculated as the t-statistic squared.

If the regression coefficient is zero the wald statistic should, for large samples, be asymptotically equal to a chi-square statistics on one df.

Univariate Wald = $$\frac{\mbox{ B }}{\mbox{variance (B)}}$$

There are multivariate Wald statistics which may be used to compare more than one regression coefficient to zero simultaneously which are approximately chi-squared on p-1 df when testing if p regression estimates are equal to zero.

This uses the Covariance matrix of ther regression estimates (COV). The univariate Wald is a special case.

Multivariate Wald = $$B^T$$ COV B

None: FAQ/wald (last edited 2013-03-08 10:17:15 by localhost)


text:
# How do I force two arrays to hold the same horizontal space?

In a previous topic (Why do horizontal matrices look different than multistory matrices?), it was asked why a vector holds a different horizontal space from what a matrix holds. One solution devised is to use bvector environment. The matrices in the question are for state-space representation of physical systems dynamics. Thus, they are of interest to me.

When I illustrate or try to explain concepts in a clear fashion, I usually use matrices with rules to divide the matrices as follows

As you can see above, the space held by the vector is slightly larger than the one held by the matrix. Further, the space held internally (i.e., between the brackets [ and ]) are slightly larger with the matrix. This leads to visually distracting equations. Thus, in case that I use array to draw matrices, how can I force the two matrices to

1. hold the same total space
2. hold the same space between the brackets surrounding them

Also, the vector x in the last equation does not align with the one above it unless I add a phantom content as shown in the empty box. Can I get the two vectors to align without inserting the phantom content?

\documentclass[10pt,norsk, fleqn]{article}
\usepackage[x11names]{xcolor}
\usepackage{amsmath}
\usepackage{calc}
\usepackage{eqparbox}
\usepackage{nccmath}

\begin{document}

\newcommand{\ssarray}{
\begin{array}{cc !{\vrule width 0.1mm}c}
2 &1 &1
\\
3 &-2 &-2
\\
\hline
1 &1 &2
\end{array}
}

\begin{fleqn}
\begin{aligned} \dot{\vec{x}}(t) &= \textbf{A}\vec{x}(t) + \textbf{B}\vec{u}(t)\\ \vec{y}(t) &= \textbf{C}\vec{x}(t) + \textbf{D}\vec{u}(t) \end{aligned} \medspace \to \medspace
\end{fleqn}
\begin{fleqn}
\begin{alignat*}{2}
\dot{\textbf{x}} = &\adjustbox{cframe=green 0.1mm}{$\left[\ssarray\right]$} &\vec{x} + \left[
\begin{array}{c}
2
\\
1
\\
\hline
2
\end{array}
\right] \vec{u}
\\
$\left[ \begin{array}{cc !{\vrule width 0.1mm}c} 2 &-2 &-2 \end{array} \right]$
}
&\vec{x} \adjustbox{phantom, cframe=green 0.1mm}{$\medspace + \left[ \begin{array}{c} 2 \\ 1 \\ \hline 2 \end{array} \right] \vec{u}$}
\end{alignat*}
\end{fleqn}

\end{document}

• one way is to draw them with matrix library from the tikz package. – Zarko Mar 28 at 8:28
• There is also \mathmakebox from the mathtools package. – John Kormylo Mar 28 at 15:54

Here is a solution using \mathmakebox.

Not sure why you want all the vertical space in the second equation.

\documentclass[10pt,norsk, fleqn]{article}
\usepackage[x11names]{xcolor}
\sbox0{$% measure inside width \begin{array}{cc|c} 2 &1 &1 \\ 3 &-2 &-2 \\ \hline 1 &1 &2 \end{array}$}%
\sbox1{$\left[\usebox0\right]$}% measure outside width
\sbox2{$\left[\begin{array}{c} 2\\1\\ \hline 2 \end{array}\right]$}% measure height
\begin{aligned} \vec{x}\ '(t) &= \textbf{A}\vec{x}(t) + \textbf{B}\vec{u}(t)\\ \vec{y}(t) &= \textbf{C}\vec{x}(t) + \textbf{D}\vec{u}(t) \end{aligned}\ \ \ \to\ \ \ \begin{aligned} \dot{\textbf{x}} & = \usebox1 \vec{x} + \usebox2 \vec{u}\\ \vec{y} & = \mathmakebox[\wd1][c]{\left[ \mathmakebox[\wd0][c]{\begin{array}{cc|c}0&-1&-1\end{array}} \right]} \vec{x} \raisebox{0pt}[\ht2][\dp2]{}% or \vphantom{usebox2} \end{aligned}


text:
Using EKL to Publish Elements at the 3DPart

Using EKL to Publish Elements at the 3DPart

To publish elements at the 3DPart level we must get the VPMReference of the 3DPart, this is achieved by using the GETPLMOwner and AggregatingReference EKL Method’s. Once we have the VPMReference we can call the CreatePublication method, the first argument is the instance but in our case it’s allowed to be ‘NULL’, the second and third arguments are the Element to be published and the name of the publication.

Let iAxisSystem1 , iAxisSystem2 ( AxisSystem )
iAxisSystem1 = Geometrical Set.1\FST_AxisSystem_1
iAxisSystem2 = Geometrical Set.1\FST_AXisSystem_2

Let ioPart ( VPMReference )
Set ioPart  = ( GetPLMOwner( iAxisSystem1 ) : VPMRepReference ).AggregatingReference : VPMReference

ioPart->CreatePublication( NULL , iAxisSystem1, iAxisSystem1.Name )
ioPart->CreatePublication( NULL , iAxisSystem2, iAxisSystem2.Name )


text:
# GATE2005-4

2.3k views

Which one of the following are essential features of an object-oriented programming language?

1. Abstraction and encapsulation

2. Strictly-typedness

3. Type-safe property coupled with sub-type rule

4. Polymorphism in the presence of inheritance

1. I and II only
2. I and IV only
3. I, II and IV only
4. I, III and IV only

B

There have been several attempts at formalizing the concepts used in object-oriented programming. The following concepts and constructs have been used as interpretations of OOP concepts:

ans b)
ans should be B

object-oriented programming language is object based programming language +abstration+inheritance.

for any OOP language ABSTRACTION and INHERITANCE are compulsory.

Type safety is the extent to which a programming language discourages or prevents type errors. A type error is erroneous or undesirable program behaviour caused by a discrepancy between differing data types for the program's constants, variables, and methods (functions), e.g., treating an integer (int) as a floating-point number (float).

https://stackoverflow.com/questions/805168/what-is-a-strictly-typed-language

Hence, II and III are clearly not essentials.

I & IV on the other hand form an OOPs language.

1 vote

Ans: B it completes the three pillars of oops

## Related questions

1
3.3k views
A common property of logic programming languages and functional languages is: both are procedural languages both are based on $\lambda$-calculus both are declarative both use Horn-clauses
double foo(int n) { int i; double sum; if(n == 0) { return 1.0; } else { sum = 0.0; for(i = 0; i < n; i++) { sum += foo(i); } return sum; } } Suppose we modify the above function $foo()$ ... modification the time complexity for function $foo()$ is significantly reduced. The space complexity of the modified function would be: $O(1)$ $O(n)$ $O(n^2)$ $n!$


text:
# Why does different wavelengths of light travel at different speeds in an optical fibre core?

I've recently read about material dispersion in an optical fibre which is caused when light that enters have several different wavelengths. What I don't understand is why would wavelength affect the speed of light in the core in anyway?

This is because the speed of light inside the fibre depends on the index of refraction $$n$$: $$v=\frac{c}{n}\, .$$ In turn, $$n$$ depends on the wavelength. It is possible to compute this dependence based on the response of the material to the electric field of the wave, which leads to the permittivity $$\epsilon$$ to depend on the frequency $$\omega$$ of the wave, and thus on $$\lambda$$. It is simplest to observe this experimentally.

• Thanks this one makes sense! – KSP May 11 at 12:56

In multimode fibre there are a number of different modes each of which has a different $$\omega(k)$$ relation: Typcally we will have something like $$\omega(k)=\left(\frac{c}{n_{\rm glass}}\right)\sqrt{k^2+\Omega_{mn}^2}$$ in which the frequency has a lower cutoff $$\Omega_{mn}$$ below which thelight cannot propagate in the fibre. The numbers $$m$$, $$n$$ label the different modes which look like standing waves in the transverse direction and propagating waves in the direction along the fibre.

This is quite different from dispersion in bulk glass where it is the refractive index $$n_{\rm glass}$$ which varies with frequency. The different modes can be loosely though of as the light bouncing (by total internal refrcation) off the cladding of the fibre at different angles. If the light bounces at a lorger angle to the directon of propagation. it spends moretime going sideways and less time going in the direction along the fibre --- think of a sailboat tacking at an angle to the direction in which it wants to go -- and so makes its way along the fibre at a slower rate.

The wikipedia article https://en.wikipedia.org/wiki/Optical_fiber#Multi-mode_fiber
has some illustrations of this.

• Thank you for your comment. This whole topic has become a reminder for me that just because something seems to explain and makes sense, doesn't make it right. – Peter May 12 at 13:16

When light enters a medium such as glass or water it slows down. If it enters at an angle the slowing causes the path of the light to bend, which is why a lens works. The colour (equivalent to wavelength) affects how much it slows down, and therefore how much it bends, which is why a prism produces colours and why there are colours in a rainbow.

The reason for this slowing has to do with the way the light waves (which are actually variations in the electric and magnetic fields) are affected by the (electrically charged) electrons in the medium as the waves pass through the atoms. Different wavelengths of light will have slightly different interactions with the electrons.

To complicate things, the speed of the light is only indirectly related to the time signals take to travel through an optic fibre. When light within a piece of glass strikes the inside surface at certain angles it may be unable to leave the glass, instead being totally internally reflected in the glass. Optic fibres take advantage of this phenomenon to stop the light from escaping the fibre (as it otherwise would). The angles involved depend on the colour/wavelength of the light. If you change the angles you change the distance the light must travel, and so you change the speed of the signal in the fibre

• Thanks, your explanation makes sense! – KSP May 11 at 12:56
• Apologies to KSP and any others who may have been misled by my original answer. I believe this is now more accurate, and I hope it is clear. – Peter May 11 at 13:27


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# Hardy spaces for Bessel-Schr\"odinger operators

## Abstract

Consider the Bessel operator with a potential on L^2((0,infty), x^a dx), namely Lf(x) = -f''(x) - a/x f'(x) + V(x)f(x). We assume that a>0 and V\in L^1_{loc}((0,infty), x^a dx) is a non-negative function. By definition, a function f\in L^1((0,infty), x^a dx) belongs to the Hardy space H^1(L) if sup_{t>0} |e^{-tL} f| \in L^1((0,infty), x^a dx). Under certain assumptions on V we characterize the space H^1(L) in terms of atomic decompositions of local type. In the second part we prove that this characterization can be applied to L for a \in (0,1) with no additional assumptions on the potential V.
arXiv:1603.07685v1 [math.CA] 23 Mar 2016
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS
EDYTA KANIA AND MARCIN PREISNER
Abstract. Consider the Bessel operator with a potential on L2((0,), xαdx), namely
Lf(x) = f′′(x)α
xf(x) + V(x)f(x).
We assume that α > 0and VL1
loc((0,), xαdx)is a non-negative function. By deﬁnition,
a function fL1((0,), xαdx)belongs to the Hardy space H1(L)if
sup
t>0
etLf
L1((0,), xαdx).
Under certain assumptions on Vwe characterize the space H1(L)in terms of atomic decompositions
of local type. In the second part we prove that this characterization can be applied to Lfor α(0,1)
with no additional assumptions on the potential V.
1. Introduction
1.1. Background. The Schrödinger operator on Rdis given by
e
Lf=f+V · f,
where is the Laplace operator and Vis a function called a potential. If we assume that V ∈ L1
loc(Rd)
and V 0then one can ﬁnd a densely deﬁned, self-adjoint operator Lon L2(Rd), that corresponds to
e
L. It is well known that Lgenerates the semigroup of contractions Kt= exp(tL)and Ktadmits an
integral kernel Kt(x, y)such that
Ktf(x) = ZRdKt(x, y)f(y)dy,
0≤Kt(x, y)(4πt)d/2exp |xy|2
4t.
There have been wide studies on harmonic analysis related to Schrödinger operators and, more
generally, operators with Gaussian bounds. We refer the reader to [1], [2], [3], [5], [7], [8], [9], [10], [11],
[12], [14], [15], [16], [17], [18], [20], [21], [25], and references therein. In particular, the Hardy spaces
(1.1) H1(L) = (fL1(Rd) : kfkH1(L):=
sup
t>0|Ktf|
L1(Rd)
<)
related to Lwere intensively studied. At this point let us mention that the classical Hardy space
H1(∆) has many equivalent deﬁnitions, e.g. in terms of: various maximal functions, singular integrals,
square functions, etc. A particulary useful result is the atomic decomposition theorem (see [6], [22]):
a function fH1(∆) can be decomposed as f(x) = Pkλkak(x), where Pk|λk| ≃ kfkH1(∆) and
akare classical atoms, that is, there exist balls Bksuch that:
supp akBk,
kakk≤ |Bk|1,
Za(x)dx = 0.
In other words atoms satisfy some localization, size, and cancellation conditions.
2010 Mathematics Subject Classiﬁcation. 42B30, 42B25, 35J10 (primary), 47D03, 43A85 (secondary).
Key words and phrases. Hardy space, Schrödinger operator, Bessel operator, maximal function, atomic decomposition.
The second author was supported by Polish funds for sciences grant DEC-2012/05/B/ST1/00672 from Narodowe
Centrum Nauki.
1
2 EDYTA KANIA AND MARCIN PREISNER
Let us mention that Hofmann et al. [21] have found general results (for Vsatisfying 0≤ V L1
loc(Rd))
saying that H1(L)given above is equal to the Hardy spaces via: square functions, atomic or molecular
decompositions. However, atoms used in [21] are given in terms of L, to be more precise: a function
aL2(Rd)is an atom if there exist a ball Band bDom(L), such that: a=Lb,supp bBand b, Lb
satisfy some size condition.
An another approach, started by Dziubański and Zienkiewicz in the 90’s, was to ﬁnd atomic spaces
with simple geometric conditions that characterize H1(L). It appeared that this cannot be done in full
generality, and the properties of atoms depend strictly on the potential Vand the dimension d. For
example, if V C
c(Rd)and V 6≡ 0then atoms have modiﬁed cancellation condition Ra(x)ω(x)dx = 0,
where ωis such that 0< C1ω(x)C. For this result and generalizations see [15], [17], [24]. Other
results, see [14], lead to Hardy spaces with local atoms. It was ﬁrst observed by Goldberg [19] that
if we take supremum for 0< t τ2in (1.1), then one obtains atomic space with classical atoms
complemented with the atoms of the form |B|1B(x), where the ball Bhas radius τ. In [14] the
authors assume that for 0 V L1
loc(Rd)there exists a family of cubes Q={Qk:kN}such that
kQk=Rd,|QkQj|= 0 for k6=j, d(Qk)d(Qj)if Q∗∗∗
kQ∗∗∗
j6=.
Here d(Q)is the diameter of Qand Qis a cube that has the same center as Qbut with slightly enlarged
diameter. The atomic space H1
at(Q)is built on classical atoms and atoms of the form |Qk|1Qk(x).
The main result of [14] states that under two additional assumptions on V, Q,Kt(see [14, p.41], con-
ditions: (D),(K)) we have that H1(L) = H1
at(Q), see [14, Thm. 2.2]. In other words the atoms for
H1(L)are either classical atoms or local atoms related to some Qk∈ Q.
Among examples of atoms for which one can ﬁnd a family Qsuch that the assumptions of [14, Thm.
2.2] are satisﬁed, there are potentials Vin the Reverse Hölder class in dimension d3. For more
examples see [14]. Later, Czaja and Zienkiewicz [7, Thm. 2.4] proved that in dimension one for any
0≤ V L1
loc(R)there is a family of intervals such that [14, Thm. 2.2] gives local atomic decompositions
for H1(L).
A question that we are concerned with is: what happens if we replace by the Bessel operator
Bf(x) = f′′(x)α/x f (x)on L2((0,), xαdx)? It is known that if α+ 1 Nthen Bcorresponds
to on radial functions on Rdwith d=α+ 1, however Bexists and generates a semigroup for all
α > 1, which can be considered as the Laplace operator on spaces with non-integer dimensions.
In this paper we prove results similar to [14] and [7] for the Bessel operator with a potential.
1.2. Deﬁnitions. For α > 0let (X, ρ, µ)be a metric-measure space, where X= (0,),ρ(x, y) =
|xy|and (x) = xαdx. Denote B(x, r) = {yX:ρ(x, y)< r}and observe that Xis a space of
homogeneous type in the sense of Coifman-Weiss [6], i.e. the doubling condition holds
µ(B(x, 2r)) (B(x, r)),
where Cdoes not depend on xXand r > 0.
The classical Bessel operator is given by
Bf(x) = f′′(x)α
xf(x).
Slightly abusing notation, we shall also write Bfor the densely deﬁned, self-adjoint operator on L2(X, µ)
that corresponds to the diﬀerential operator above, see Subsection 2.1 for the semigroup generated by B.
In this paper we consider the Bessel-Schrödinger operator L,
(1.2) Lf=Bf+V·f,
where VL1
loc(X, µ), V 0. To be more precise, denote hf , giµ=Rfg dµ and deﬁne a quadratic
form
Q(f, g) = hf, giµ+hV f, V giµ,
with the domain
Dom(Q) = nfL2(X, µ) : f,V f L2(X , µ)o.
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 3
The quadratic form Qis positive and closed. Therefore, it deﬁnes a self-adjoint operator Lwith the
domain
Dom(L) = fDom(Q) : hL2(X, µ)gDom(Q)Q(f, g) = hh, giµ.
For f , h as above we put Lf:= h. Let Kt= exp(tL)be the semigroup generated by L. Denote by
Bsthe Bessel process on (X, µ). By using the Feynman-Kac formula,
Ktf(x) = Exexp Zt
0
V(Bs)dsf(Bt),
one gets that Kthas an integral kernel Kt(x, y)and
(1.3) 0Kt(x, y)Pt(x, y),
where Pt(x, y)is the kernel related to Pt= exp (tB), see Subsection 2.1.
We deﬁne the Hardy space H1(L)by means of the maximal operator associated with Kt, namely
(1.4) H1(L) = (fL1(X, µ) : kfkH1(L)=
sup
t>0|Ktf|
L1(X, µ)
<).
The goal of this paper is to give an atomic characterizations of local type for H1(L). Let |I|be the
diameter of I.
Deﬁnition 1.5. Let Ibe a collection of intervals that are closed with respect to the topology on (0,).
We call a family Ia proper section of Xif:
(a) for I, J ∈ I,I6=J, the intersection IJis either the empty set or a singleton,
(b) X=SI∈I I,
(c) there exists a constant C0>0such that for I, J ∈ I,IJ6=we have
C1
0|I| ≤ |J| ≤ C0|I|.
Denote τB(c, r) := B(c, τ r). For an interval I=B(x, r)(if I= (0,2A)we take I=B(A, A)), let
cI := B(x, cr). For a family Ias in Deﬁnition 1.5 we set I:= βI for some ﬁxed β > 1, such that
I∗∗∗ J∗∗∗ 6=if and only if IJ6=.
We say that a function a:XCis an (I, µ)-atom if:
(i)there exist I∈ I and an interval JI∗∗,such that: supp(a)J, kakµ(J)1,Za dµ = 0,
or
(ii)there exists I∈ I, such that a(x) = µ(I)1I(x).
The atoms as in (ii)are called local atoms.
The atomic Hardy space H1
at(I, µ)associated with the collection Iis deﬁned in the following way.
We say that f∈ H1
at(I, µ)if
(1.6) f(x) = X
n
λnan(x),
where λnC,anare (I, µ)-atoms, and Pn|λn|<. Set
(1.7) kfkH1
at(I):= inf X
n|λn|,
where the inﬁmum is taken over all possible representations of fas in (1.6).
For a collection Ias above and V0,VL1
loc(X, µ)we consider the following two conditions:
- there exist constans C, ε > 0such that
(D) sup
yI∗∗ ZX
K2k|I|2(x, y)(x)Ck1εfor I∈ I, k N,
- there exist constans C, δ > 0such that
(K) Z2t
0ZX
Ps(x, y)I∗∗∗ (y)V(y)(y)ds Ct
|I|2δ
for xX, I ∈ I, t ≤ |I|2.
4 EDYTA KANIA AND MARCIN PREISNER
1.3. Statement of results. Our ﬁrst main result is the following, cf. [14, Thm. 2.2]
Theorem 1.8. Assume that a proper section Iand 0VL1
loc(X, µ)are given, so that (D)and
(K)hold. Then H1(L) = H1
at(I, µ)and there exists a constant C > 0, such that
C1kfkH1
at(I)≤ kfkH1(L)CkfkH1
at(I).
In the second part we give an important application of Theorem 1.8. Let us restrict ourselves to
α(0,1). We prove that for any 0VL1
loc(X, µ)we can ﬁnd a family I(V)such that the
assumptions of Theorem 1.8 hold, cf. [7, Thm. 2.4] for the case α= 0. To be more precise, let Dbe a
family of dyadic intervals on (0,), that is D={[k2n,(k+ 1)2n] : kN∪ {0}, n Z}. Consider the
family I(V)that consists of maximal dyadic closed intervals Ithat satisfy
(S) |2I|2
µ(2I)Z2I
V(y)(y)1.
In Section 4we prove that I(V)is a well deﬁned proper section. The second main result is the
following.
Theorem 1.9. Let α(0,1) and 0VL1
loc(X, µ). Then the family I(V)satisﬁes the assumptions
of Theorem 1.8.
Corollary 1.10. Let α(0,1) and 0VL1
loc(X, µ). Then there is C > 0, such that
C1kfkH1
at(I(V))≤ kfkH1(L)CkfkH1
at(I(V)).
The paper is organized as follows. In Section 2we study the atomic Hardy spaces related to Band
its local versions. This is used in a proof of Theorem 1.8, which is provided in Section 3. Finally, in
Section 4a proof of Theorem 1.9 is given.
2. Hardy spaces for the Bessel operator
2.1. Global Hardy space for B.In this section we consider the case α > 0. Let Pt= exp(tB)be
the Bessel semigroup given by
Ptf(x) = ZX
Pt(x, y)f(y)(y),
Pt(x, y) = (2t)1exp x2+y2
4tIα1
2xy
2t(xy)α1
2,
where Iα(x) = P
m=0 1
m!Γ(m+α+1) x
22m+αis the modiﬁed Bessel function of the ﬁrst kind. It is clear
that Pt(x, y) = Pt(y, x)and, since B(0,)(x) = 0, we have that
(2.1) ZX
Pt(x, y)(x) = 1.
Let us recall that
(2.2) µ(B(x, t)) t(x+t)α.
It is known that the kernel Pt(x, y)satisﬁes the two-side Gaussian estimates (see, e.g. [13, Lem. 4.2]),
(2.3) C1µ(B(x, t))1exp |xy|2
c1t!Pt(x, y)Cµ(B(x, t))1exp |xy|2
c2t!,
while the derivative satisﬁes
(2.4)
∂x Pt(x, y)Ct1/2µ(B(x, t))1exp |xy|2
ct !.
Let H1(B)be the Hardy space related to Pt, i.e. the space deﬁned as in (1.4) with Land Ktreplaced
by Band Pt, respectively. We call a function aan µ-atom if
(iii)there exists an interval JX, such that: supp(a)J, kakµ(J)1,Za dµ = 0.
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 5
The atomic Hardy space H1
at(µ)is deﬁned as in (1.6) and (1.7) with anbeing µ-atoms.
Theorem 2.5. [4, Thm. 1.7] Let α > 0. There is C > 0such that
C1kfkH1
at(µ)≤ kfkH1(B)CkfkH1
at(µ).
2.2. Local Hardy space for B.For τ > 0we deﬁne h1
τ(B),the local Hardy space related to B, as
the set of L1(X, µ)functions for which the norm
kfkh1
τ(B)=
sup
tτ2|Ptf|
L1(X, µ)
is ﬁnite.
Let Iτbe a proper section of Xthat consists of closed intervals of length τ.
Theorem 2.6. (a)There exists C > 0such that for τ > 0and an (Iτ, µ)-atom awe have
kakh1
τ(B)C.
(b)Let Ibe an interval such that supp(f)Iand fh1
|I|(B). Then
f=
X
n=0
λnan,
X
n=0 |λn| ≤ Ckfkh1
|I|(B),
where a0(x) = µ(I)1I(x)and anare µ-atoms supported in I∗∗ for n1.
Let us remark that another characterization of h1
τ(B), by mean of a local Riesz transform, was given
in [23, Thm. 2.11]. Theorem 2.6 will be used to prove Theorem 1.8.
Proof. (a)Obviously, if ais µ-atom, then the statement follows from Theorem 2.5. Assume then that
a(x) = µ(I)1I(x)and |I|=τ. It is well known that (2.3) implies the boundedness of the maximal
operator supt>0|Ptf|on L2(X, µ). Using this fact and the Schwarz inequality,
sup
tτ2|Pta|
L1(I∗∗)Cµ(I)1/2
sup
t>0|Pta|
L2(X,µ)(I)1/2kakL2(X,µ)C.
Denote by cIthe center of Iand notice that |xy| ≃ |xcI|when yIand x(I∗∗)c. By (2.3),
sup
tτ2|Pta|
L1((I∗∗)c)CZ(I∗∗ )c
sup
tτ2ZI
µ(B(x, t))1exp |xcI|2
c3tµ(I)1(y)(x)
CZ(I∗∗)c
sup
tτ2
t1/2exp |xcI|2
c3tdx
CZ(I∗∗)c|I|1exp |xcI|2
c4|I|2dx C.
(b)Deﬁne λ0:= Rfand g(x) = f(x)λ0µ(I)1I(x). Notice that
|λ0| ≤ kfkL1(X,µ)≤ kfkh1
|I|(B).
Therefeore, kgkL1(X,µ)2kfkL1(X,µ). Our goal is to prove that
(2.7) kgkH1(B)Ckfkh1
|I|(B).
For t≤ |I|2using (a)we obtain
sup
t≤|I|2|Ptg|
L1(X,µ)≤ kfkh1
|I|(B)+|λ0|
µ(I)1I
h1
|I|(B)Ckfkh1
|I|(B)
6 EDYTA KANIA AND MARCIN PREISNER
As for t≥ |I|2notice that Rgdµ = 0. Therefore, using (2.4),
sup
t≥|I|2|Ptg|
L1((I∗∗)c)Z(I∗∗ )c
sup
t≥|I|2ZI
(Pt(x, y)Pt(x, cI)) g(y)(y)(x)
Z(I∗∗)cZI
sup
t≥|I|2
|ycI|
tµ(B(x, t))1exp |xcI|2
c2t|g(y)|(y)(x)
CkgkL1(X,µ)|I|Z(I∗∗)c
sup
t≥|I|2
t1exp |xcI|2
c2tdx
CkfkL1(X,µ)|I|Z(I∗∗)c|xcI|2dx Ckfkh1
|I|(B).
Likewise,
sup
t≥|I|2|Ptg|
L1(I∗∗)CZI∗∗ ZI
sup
t≥|I|2
µ(B(x, t))1|g(y)|(y)(x)
CkgkL1(X,µ)ZI∗∗
sup
t≥|I|2
t1/2dx Ckfkh1
|I|(B).
From (2.7) we have that gH1(B), so using Theorem 2.5 we obtain λkand µ-atoms aksuch that
g=P
k=1 λkak. Consequently,
f=
X
k=0
λkak,
X
k=0 |λk| ≤ Ckfkh1
|I|(B),
where a0(x) = µ(I)1I(x).
The only problem we have to deal with is that akare not necessarily supported in I∗∗ . Let ψbe
a function such that ψ1on I,ψ0on (I∗∗ )c, and kψkC|I|1. To complete the proof we
will show, that for every µ-atom akthere exist a sequence aj
k, such that each aj
kis either µ-atom or
aj
k=µ(I)1I. Moreover, ak=Pjλj
kaj
kand Pj|λj
k|< C, with Cthat not depend on k. Fix a=ak
and an interval JXsuch that: supp(a)J,kakµ(J)1. Obviously supp(ψa)I∗∗ J,
and if JI, then ψa =ais an µ-atom. Furthermore, if J(I∗∗ )c, then ψa = 0, so it suﬃces
to consider the case that J(I)cI∗∗ 6=. Observe that |J| ≤ |I|and let NNbe such that
1/2N+1|I| ≤ |J| ≤ 1/2N|I|.
Deﬁne λ:= Rψa dµ and notice that
|λ|=ZJ
ψ(x)a(x)(x)=ZJ
a(x)(ψ(x)ψ(cJ)) (x)
µ(J)1ZJ|xcJ||ψ(ξ)|(x)
C|J||I|1C2N.
Let us choose intervals Ij, such that J=: I0I1⊂ ··· ⊂ INI∗∗, where |Ij+1 |/|Ij|= 2 and
|IN| ≃ |I|. Then
ψa =ψa λµ(I0)1I0+
N
X
j=1
λµ(Ij1)1Ij1µ(Ij)1Ij
+λ(µ(IN)1INµ(I)1I) + λµ(I)1I=
N+2
X
j=0
bj.
Observe that:
(1) for j= 0, ..., N we have: supp(bj)Ij,Rbj= 0, and
kbjkC|λ|µ(Ij)1C2Nµ(Ij)1,
(2) supp(bN+1)I∗∗ ,kbN+1kC|λ|µ(I)1and RbN+1 = 0.
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 7
We conclude that bjare multiples of (I|I|, µ)-atoms and
N+2
X
j=0
bj
H1
at(I|I|)
C
N+2
X
j=0
2NC.
Corollary 2.8. There exists a constant C > 0such that for τ > 0we have
C1kfkH1
at(Iτ)≤ kfkh1
τ(B)CkfkH1
at(Iτ).
The right inequality in Corolarry 2.8 follows easily from Theorem 2.6(a). For the left inequality one
uses Theorem 2.6(b)with a suitable partition of unity and methods as in Lemmas 3.2 and 3.5 below.
We omit the details.
3. Proof of Theorem 1.8
3.1. Auxiliary estimates. For a proper section Ilet {φI}I∈I be a partition of unity associated with
I, that is a family of Cfunctions on X, such that supp(φI)I,0φI1,kφ
Ik≤ |I|1and
PI∈I φI(x) = 1 for all xX.
The perturbation formula states that
(3.1) Pt(x, y)Kt(x, y) = Zt
0ZX
Pts(x, z)V(z)Ks(z, y)(z)ds.
To prove Theorem 1.8 we closely follow the proof of [14, Thm. 2.2]. However, in our weighted space,
some technical diﬃculties appear. Therefore we present all the details for convenience of the reader.
Lemma 3.2. For I∈ I and fL1(X, µ),
sup
t4|I|2|Pt(φIf)|
L1((I∗∗)c, µ)CkφIfkL1(X, µ).
Proof. Denote by cIthe center of the interval I∈ I. For x(I∗∗)cand yIwe have |xy| ≃
|xcI|. Notice that (2.2) implies
(3.3) µ(B(x, t))1xαt1/2.
Using (2.3) the left-hand side is bounded by
Z(I∗∗)c
sup
t4|I|2ZI
µ(B(x, t))1e|xy|2
c2t|φI(y)f(y)|(y)(x)
CkφIfkL1(X,µ)Z(I∗∗)c
sup
t4|I|2
xαt1
2e|xcI|2
c3t(x)
CkφIfkL1(X,µ)Z(I∗∗)c|I|1e|xcI|2
c4|I|2dx
CkφIfkL1(X, µ).
Corollary 3.4. For I∈ I and fL1(X, µ),
kφIfkh1
|I|(B)
sup
t≤|I|2|Pt(φIf)|
L1(I∗∗, µ)
.
Denote
e
fI:= X
J:IJ6=
φJf, ¯
fI:= fe
fI=X
J:IJ=
φJf.
8 EDYTA KANIA AND MARCIN PREISNER
Lemma 3.5. For I∈ I and fL1(X, µ),
sup
t≤|I|2KtφIe
fIφIKte
fI
L1(I∗∗, µ)CX
J:IJ6=kφJfkL1(X, µ)
Proof. Denote e
I=J:IJ6=J. Note that for xI∗∗ and ye
I, there is |xy| ≤ C|I|and
(3.6) sup
t≤|I|2
t1
2e|xy|2
c2tC|xy|1.
Using (1.3), (2.3), (3.3), and (3.6),
sup
t≤|I|2KtφIe
fIφIKte
fI
L1(I∗∗, µ)ZI∗∗
sup
t≤|I|2Ze
I|φI(y)φI(x)|Kt(x, y)e
fI(y)(y)(x)
CZI∗∗ Ze
I|xy|kφ
Iksup
t≤|I|2
t1
2e|xy|2
c2te
fI(y)(y)dx
CZI∗∗ |I|1dx
e
fI
L1(X,µ).
Lemma 3.7. Assume that Vand Iare given, so that (D)holds. Then
X
I∈I
sup
t>0Kt¯
fI
L1(I∗∗∗, µ)CkfkL1(X, µ).
Proof. Denote sm= 2m|I|2and let m2. By the semigroup property, (1.3), (2.3),
sup
smtsm+1
Kt(|φJf|)
L1(X, µ)
ZX
sup
smtsm+1 ZXZX
Ptsm1(x, z)Ksm1(z, y)|φJ(y)f(y)|(z)(y)(x)
CZX|φJ(y)f(y)|ZX
Ksm1(z, y)ZX
s1/2
mexp |xz|2
c3smdx dµ(z)(y)
C(m1)1ε· kφJfkL1(X, µ).
In the last inequality we have applied (D). Using the above estimate and Lemma 3.2,
X
I∈I
sup
t>0Kt X
J:IJ=
φJf!
L1(I∗∗∗, µ)
X
J∈I X
I:JI=
sup
t>0
Kt(|φJf|)
L1(I∗∗∗, µ)CX
J∈I
sup
t>0
Kt(|φJf|)
L1((J∗∗)c, µ)
CX
J∈I
sup
t4|J|2
Kt(|φJf|)
L1((J∗∗)c, µ)
+CX
J∈I
X
m=2
sup
smtsm+1
Kt(|φJf|)
L1((J∗∗)c, µ)
CX
J∈I kφJfkL1(X, µ)+CX
J∈I
X
m=2
m1ε· kφJfkL1(X, µ)CkfkL1(X, µ).
Lemma 3.8. Z
0ZX
V(z)Ks(|f|)(z)(z)ds CkfkL1(X, µ)
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 9
Proof. By integrating (3.1) and using (2.1) we obtain
Zt
0ZX
V(z)Ks(z, y)(z)ds C.
The proof is ﬁnished by setting t→ ∞.
Lemma 3.9. Assume that Vand Iare given, so that (K)holds. Then for I∈ I,
sup
t≤|I|2|(PtKt)(φIf)|
L1(X, µ)CkφIfkL1(X, µ).
Proof. By (1.3) and Lemma 3.2,
sup
t≤|I|2|(PtKt)(φIf)|
L1((I∗∗)c, µ)CkφIfkL1(X, µ).
Consider the integral on I∗∗ . From (3.1),
(PtKt)(φIf)(x) = Zt
0
PtsV′′Ks(φIf)(x)ds +Zt
0
PtsVKs(φIf)(x)ds
where V=I∗∗∗ V+(I∗∗∗)cV=V+V′′ . Repeating the argument from the proof of Lemma 3.2 and
using Lemma 3.8 we obtain
sup
t≤|I|2Zt
0
PtsV′′Ks(φIf)ds
L1(I∗∗
,µ)CkφIfkL1(X, µ).
To estimate the integral that contains Vwrite
Zt
0
PtsVKs(φjf))(x)dsZt/2
0
PtsVPs(|φjf|)(x)ds +Zt
t/2
PtsVPs(|φjf|)(x)ds
=A1,t(x) + A2,t (x).
Let tm= 2m|I|2. Similarly as in proof of the Lemma 3.7 we obtain
sup
t≤|I|2
A1,t
L1(X, µ)
X
m=0 ZX
sup
tm+1ttmZt/2
0ZX
Pts(x, y)V(y)Ps(|φIf|)(y)(y)ds dµ(x)
C
X
m=0 ZXZtm
0
V(y)Ps(|φIf|)(y)ZX
t1
2
me|xy|2
c3tmdx ds dµ(y)
C
X
m=0 Ztm
0ZX
I∗∗∗ (y)V(y)Ps(|φIf|)(y)ds dµ(y)
C
X
m=0 2m|I|2
|I|2δ
kφIfkL1(X, µ).
In the last inequality we have used (K). To estimate A2,t we proceed similarly noticing that for
t[tm+1, tm]and s[t/2, t]we have s[tm+2, tm]. The details are left to the reader.
3.2. Proof of Theorem 1.8. First inequality. Observe that φIf=φIe
fIand
Kt(φIf) = KtφIe
fIφIKte
fIφIKt¯
fI+φIKt(f).
From Lemmas 3.5 and 3.7 we deduce that
10 EDYTA KANIA AND MARCIN PREISNER
X
I∈I
sup
t≤|I|2|Kt(φIf)|
L1(I∗∗, µ)X
I∈I
sup
t≤|I|2KtφIe
fIφIKte
fI
L1(I∗∗, µ)
+X
I∈I
sup
t≤|I|2φIKt¯
fI
L1(I∗∗, µ)
+X
I∈I
sup
t≤|I|2|φIKtf|
L1(I∗∗, µ)
CkfkL1(X, µ)+kfkH1(L)CkfkH1(L).
The above estimate, together with Corollary 3.4 and Lemma 3.9, lead to
X
I∈I kφIfkh1
|I|(B)CX
I∈I
sup
t≤|I|2|(PtKt)(φIf)|
L1(I∗∗, µ)
+CX
I∈I
sup
t≤|I|2|Kt(φIf)|
L1(I∗∗, µ)
CkfkH1(L).
Now we use Theorem 2.6(b)for each φIfgetting (λI
n)nand (I, µ)-atoms (aI
n)n, so that
φIf(x) = X
n
λI
naI
n(x),and X
nλI
nCkφIfkh1
|I|(B).
Summing up for all I∈ I we ﬁnish the ﬁrst part of the proof.
Second inequality. Let abe an (I, µ)-atom, such that supp(a)I∗∗ . There exists an integer
m0, independent of I, such that
inf |I|2:JI6=2m|I|2.
Denote tn= 2n|I|2. Observe that
sup
ttm|Kt(a)|
L1(X, µ)
sup
ttm|(KtPt)(a)|
L1(X, µ)
+
sup
ttm|Pt(a)|
L1(X, µ)≤ kakL1(X, µ).
In the last inequality we applied Lemma 3.9 and Theorem 2.6(b), since ais also an (I|I|, µ)-atom.
It suﬃces to estimate
supt>tm|Kta|
L1(X, µ). This is done by using (1.3) and (K). Indeed, using
similar methods as in the proof of Lemma 3.2,
sup
t>tm|Kta|
L1(X, µ)X
nm
sup
tnttn1|Kta|
L1(X, µ)
X
nm
sup
tn+1t3tn+1
KtKtn+1 a
L1(X, µ)
CX
nm+1 ZXZX
K2n|I|2(x, y)|a(y)|(y)(x)
CkakL1(X, µ)C,
since the operator suptn+1 t3tn+1 Ktis bounded on L1(X, µ), see (2.3).
4. Proof of Theorem 1.9.
In the whole section we assume that α(0,1). Recall that for 0a < b,
(4.1) µ((a, b)) = b1+αa1+α
1 + α(bα+1 2ab
(ba)aα2ab.
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 11
By (4.1) and the Mean-Value Theorem we easily get
(4.2) |I|2
µ(I)b1αa1α.
Lemma 4.3. For α(0,1) and IJ(0,),
|I|2
µ(I)ZI
V(y)(y)|J|2
µ(J)ZJ
V(y)(y).
Proof. Obviously, since V0, it is enough to prove that
(4.4) |I|2
µ(I)|J|2
µ(J).
Let abcdand I= (b, c)(a, d) = J. Denote Γ(x, y) = (yx)2/(yα+1 xα+1 ). Now, (4.4) is
equivalent to Γ(b, c)Γ(a, d). This is done in two steps.
Step 1: Γ(b, c)Γ(a, c). Denote a=sc and b=tc, where 0< s t < 1. It is enough to prove
(1 t)2
1tα+1 (1 s)2
1sα+1 .
By a simple calculus argument, the function F1(x) = (1 x)2/(1 xα+1 )is monotonically decreasing
for x(0,1).
Step 2: Γ(a, c)Γ(a, d). Similarly, let c=ta,d=sa,1< ts. The function F2(x) =
(x1)2/(xα+1 1) is monotonically increasing in (1,), thus
(t1)2
(t)α+1 1(s1)2
(s)α+1 1.
Proposition 4.5. Let VL1
loc(X, µ),V0. Then the family I(V)of maximal dyadic intervals
satisfying (S)is a well-deﬁned proper section (see Deﬁnition 1.5).
Proof. For a closed dyadic interval Iconsider F(I) = |2I|2µ(2I)1R2IV dµ and denote by Idthe
smallest dyadic interval containing I. Notice that 2I2Idand, by Lemma 4.3, we have F(I)F(Id).
Also, for an increasing sequence of dyadic intervals InIn+1 we have limn→∞ F(In) = , see (4.2).
This justiﬁes the choice of I(V)as maximal dyadic intervals such that (S) holds. What is left to
prove is that I(V)is a proper section, namely we need to show that for I , J ∈ I(V),IJ6=we have
|I| ≃ |J|.
By contradiction, suppose that there exist Ik,Jksuch that IkJk6=and |Ik|/|Jk| → ∞. We can
assume that 2Jd
k2Ikfor all k. Denote, ak=|Ik|2µ(Ik)1|Jk|2µ(Jk). By the choice of I,
1|2Ik|2
µ(2Ik)Z2Ik
V(y)(y)|2Ik|2µ(2Jd
k)
µ(2Ik)2Jd
k22Jd
k2
µ(2Jd
k)Z2Jd
k
V(y)(y)C1ak.
The proof will be ﬁnished when we show that ak→ ∞. This follows from (4.1) by considering several
cases. Let a, b, c be such that 0a < b < c.
Case 1: Jk= [a, b],Ik= [b, c].
Subcase 1: 4a2bc. Using (4.1) we have
ak(c/b)2(b/c)1+α= (c/b)1α(|Ik|/|Jk|)1α.
Subcase 2: 4a2bc. This subcase can hold only for ﬁnite k.
Subcase 3: 4a2bc. Using (4.1) we have
akc2
(ba)2
(ba)aα
cα+1 =aαc
cα(ba)c1α
(ba)1α(|Ik|/|Jk|)1α.
12 EDYTA KANIA AND MARCIN PREISNER
Subcase 4: 4a2bc. Using (4.1) we have
akcb
ba2(ba)aα
(cb)bα≃ |Ik|/|Jk|.
Case 2: Jk= [b, c],Ik= [a, b]. Then
akC|Ik|
|Jk|2|Jk|bα
|Ik|bα≃ |Ik|/|Jk|.
Recall that hf, g iµ=RXfg dµ, so that h−Bφ, ψ iµ=hφ, Bψiµfor appropriate ψ , φ. For y > 0the
distributional equation
Bφ=δy
has the solution given by φy(x) = 1
2(1α)x1αy1α. We shall use φyto construct superharmonic
functions that will be crucial in the proof of (D).
Lemma 4.6. Let α(0,1),0VL1
loc(X, µ), and I(V)is as in Proposition 4.5. Then
ZX
K2n|I|2(x, y)(x)C21α
2n
for yI∗∗,I∈ I(V), and n0.
Proof. Let Ibe a dyadic interval such that
|2I|2
µ(2I)Z2I
V dµ 1,|2Id|2
µ(2Id)Z2Id
V dµ > 1.
By continuity argument there exists Jsuch that I∗∗ 2IJ2Idand |J|2
µ(J)RJV dµ = 1. Let
J= (a, b)and observe that |J| ≃ |I|. Deﬁne
φI(x) = 1 + 1
2(1 α)ZJ
V(y)x1αy1α(y).
Fix zI∗∗. By (4.2) and the doubling condition,
(4.7) φI(z)1 + Csup
y,yJ|y1α(y)1α|ZJ
V dµ C.
Also, we claim that for xX,
(4.8) φI(x)1 + µ(J)x1αz1α
|J|2.
Indeed, if xis such that |x1αz1α| ≤ C|J|2µ(J)1, with Clarge enough, this follows exactly as in
(4.7). In the opposite case |x1αz1α| ≥ C|J|2µ(J)1, we have |x1αz1α| ≃ |x1αy1α|for
yJand the claim follows.
Now we proceed to a crucial argument that uses superharmonicity. Observe that formal calculation
gives
LφI(x) = BφI(x)V(x)φI(x) = V(x)( J(x)φI(x)) 0
and, consequently,
∂t KtφI(z) = ZX
Kt(z, x)(LφI(x)) (x)0.
(4.9) KtφI(z)φI(z), t > 0.
However, φIis not in Dom(L)(or even in L2(X, µ)), thus we provide a detailed proof of (4.9) in
Appendix.
Denote
θ(t) = ZX
Kt(z, x)(x).
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 13
Our goal is prove that
(4.10) θ(2n|I|2)c021α
2n.
This will follow by induction argument. By (1.3) and (2.3),
(4.11) K2t(z, x) = ZX
Kt(z, y)Kt(y, x)(y)Cθ(t)µ(B(x, t))1
and
θ(2t) = Z|x1αz1α|<R
K2t(z, x)(x) + Z|x1αz1α|>R
K2t(z, x)(x) = A1+A2.
By (4.8), (4.9), and (4.7) we have
A2|J|2
(J)Z|x1αz1α|>R
K2t(z, x)µ(J)x1αz1α
|J|2(x)
CR1|J|2µ(J)1ZK2t(z, x)φI(x)(x)
CR1|J|2µ(J)1.
To estmate A1we use (4.11) and (2.2),
A1(t)Z|x1αz1α|<R
µ(B(x, t))1(x)
(t)t1/2Z|x1αz1α|<R
dx.
Case A: z1α2R, then
Z|x1αz1α|<R
dx =1
1α(z1α+R)1
1α(z1αR)1
1αCRzα.
In this case
(4.12) θ(2t)c1θ(t)t1/2Rzα+R1|J|2µ(J)1.
Case B: z1α<2R,
Z|x1αz1α|<R
dx C(z1α+R)1
1αCR 1
1α.
In this case
(4.13) θ(2t)c2θ(t)t1/2R1
1α+R1|J|2µ(J)1.
Now we proceed to the proof of (4.10). The ﬁrst step, θ(|I|2)C, follows simply by (1.3). Assume
that (4.10) holds for some n. Consider the following cases:
Case 1: ρ(0, I )2|I|. In this case µ(I)≃ |I|zα.
Subcase 1.1: θ(2n|I|2)2n/2|I|2z2. Observe that
R1:= 21θ(2n|I|2)1/2(2n|I|2)1/4|I|1/2zα< z1α/2.
Putting R=R1and t= 2n|I|2into (4.12) and using the induction hypothesis,
θ(2n+1|I|2)c3θ(2n|I|2)1/2(2n|I|2)1/4|I|1/2
c3c1/2
02n1α
42n
4c02(n+1) 1α
2.
The last inequality holds if we choose c0such that c0c2
321α.
Subcase 1.2: θ(2n|I|2)2n/2|I|2z2. One easily checks that
R2:= 21θ(2n|I|2)1(2n|I|2)1/2|I|zα1α
2α> z1α/2.
14 EDYTA KANIA AND MARCIN PREISNER
Putting R=R2and t= 2n|I|2into (4.13) and using the induction hypothesis,
θ(2n+1|I|2)c4θ(2n|I|2)(2n|I|2)1/2|I|1
1αzα
1α1α
2α
c4θ(2n|I|2)(2n|I|2)1/2|I|1α
2α
c4c0(2n)1α
2+1
21α
2αc02(n+1) 1α
2.
In the last inequality we choose c0such that c0c2α
42(1α)(2α)/2. Notice that we have used z≥ |I|,
which follows from ρ(0, I)2|I|.
Case 2: ρ(0, I )2|I|. In this case µ(I)≃ |I|α+1 . Notice that z4|I|.
Subcase 2.1 θ(2n|I|2)2n/2|I|2αzα2. Observe that
R3:= 21θ(2n|I|2)1/2(2n|I|2)1/4|I|1α
2zα/2< z1α/2.
Putting R=R3and t= 2n|I|2into (4.12) and using the induction hypothesis,
θ(2n+1|I|2)c5θ(2n|I|2)1/22n/4zα/2|I|α/2
c5c1/2
02n1α
42n
4zα/2|I|α/2
c5c1/2
02α2n1α
42n
4
c02(n+1) 1α
2.
The last inequality holds if we choose c0such that c0c2
521+α.
Subcase 2.2 θ(2n|I|2)<2n/2|I|2αzα2. One easily checks that
R4:= 21θ(2n|I|2)1(2n|I|2)1/2|I|1α1α
2α> z1α/2.
Putting R4and t= 2n|I|2into (4.13) we obtain
θ(2n+1|I|2)c6θ(2n|I|2)(2n|I|2)1/2|I|1α
2α
c6c0(2n)1α
2+1
21α
2αc02(n+1) 1α
2,
similarly as in subcase 1.2.
Lemma 4.14. Let α(0,1),0VL1
loc(X, µ), and I(V)is as in Proposition 4.5. Then the pair
(V, I(V)) satisﬁes (K).
Proof. Case 1: ρ(0, I )2|I|. In this case µ(I)≃ |I|1+α. Using (2.3) and (2.2),
Z2t
0ZI∗∗∗
Ps(x, y)V(y)(y)ds CZ2t
0
s1+α
2ds ·ZI∗∗∗
V dµ C t 1α
2µ(I)
|I|2Ct
|I|21α
2
.
Case 2: ρ(0, I)>2|I|. In this case µ(I)≃ |I|cα
I, where cIdenotes the center of I. Using (2.3) and
the doubling condition,
Z2t
0ZI∗∗∗
Ps(x, y)V(y)(y)ds CZ2t
0ZI∗∗∗
µ(B(y, s))1V(y)(y)ds
CZ2t
0
s1
2cα
Ids ·ZI∗∗∗
V(y)(y)
Ct 1
2µ(I)
|I|2cα
ICt
|I|21
2
.
Combining Lemmas 4.6 and 4.14 we obtain Theorem 1.9.
HARDY SPACES FOR BESSEL-SCHRÖDINGER OPERATORS 15
Appendix
The goal of this Appendix is to give a precise proof of the formula (4.9). Recall that J= (a, b). By
the deﬁnition of φIwe have
(4.15)
φI(x) = 1 + 1
2(1 α)
RJV(y)y1α(y)x1αRJV(y)(y), x < a
Rx
aV(y)(x1αy1α)(y)Rb
xV(y)(y1αx1α)(y), x (a, b)
x1αRJV(y)(y)RJV(y)y1α(y), x > b.
Recall that RJV=µ(J)|J|2. The formula (4.15) easily implies that
(4.16) |φ
I(x)| ≤ (J)|J|2xα
for all xX.
Lemma 4.17. Let ψDom(Q),ψ0,ηC
c(X, µ),η1on supp(ψ). Then φIηDom(Q)and
Q(ψ, φIη) = ZX
ψ(x)V(x)(φI(x)J(x)) (x)0.
Proof. Observe that using (4.15) we obtain that
φ′′
I(x)α
xφ
I(x) = J(x)V(x).
Since supp(η)is compact, using facts (4.8), (4.16) we deduce that φIηDom(Q). Since η1on
supp (ψ),ZX
ψ(x)(φI(x)η(x))(x) = ZX
ψ(x)φ
I(x)(x)
=ZX
ψ(x)φ′′
I(x)α
xφ
I(x)(x)
=ZX
ψ(x)J(x)V(x)(x).
The lemma follows, since φI1.
Recall that zJis ﬁxed and denote ϑ(u) = KuφI(z).Our goal is to prove that ϑ(t+s)ϑ(t)for
t, s > 0.
Denote k(x) := Kt(x, z) = Kt/2(Kt/2(·, z ))(x). Since the semigroup Ktis analytic and Kt/2(·, z )
L2(X, µ)(see (1.3) and (2.3)), we have kDom(L)Dom(Q). Let ηnX,ηnC
c(X, µ),
supp(ηn)(0, n + 1],ηn1on (0, n], and kη
nk+kη′′
nkC.
First, observe that
(Ku(k)ηn)(φIηn+1)= (Ku(k)ηn)φ
I=Ku(k)φ
Iηn+Ku(k)φ
Iη
n.
Using this,
Q(Ku(k), φIηn) = ZKu(k)(φIηn)+ZV Ku(k)φIηn
=ZKu(k)φ
Iηn+ZKu(k)φIη
n+ZV Ku(k)φIηn
=Z(Ku(k)ηn)(φIηn+1)ZKu(k)φ
Iη
n+ZKu(k)φIη
n
+ZV(Ku(k)ηn)(φIηn+1)
=Q(Ku(k)ηn, φIηn+1)2ZKu(k)φ
Iη
nZKu(k)φIη′′
nZKu(k)φIη
n
α
x
=Q(Ku(k)ηn, φIηn+1)2B1B2B3.
(4.18)
16 EDYTA KANIA AND MARCIN PREISNER
Proposition 4.19. The function ϑis non-increasing.
Proof. Using (4.18),
ϑ(t+s)ϑ(t) = ZX
(Ks(k)k) (x)φI(x)(x)
= lim
n→∞ ZX
(Ks(k)k)(x)φI(x)ηn(x)(x)
= lim
n→∞ ZXZs
0
(L)Ku(k)(x)duφI(x)ηn(x)(x)
=lim
n→∞ Zs
0ZX
LKu(k)(x)φI(x)ηn(x)(x)du
=lim
n→∞ Zs
0
Q(Ku(k), φIηn)du
=lim
n→∞ Zs
0
(Q(Ku(k)ηn, φIηn+1)2B1B2B3)du.
Having in mind Lemma 4.17 it is enough to show that Rs
0Bidu 0as n→ ∞ for i= 1,2,3.
This follows from Lebesgue’s Dominated Convergence Theorem and the estimates we have already
established. For example, for B2observe that |φI(x)η′′
n(x)| ≤ Cµ(J)|J|2|x|1α[n,n+1](x)for nN
with Nlarge enough. Then the majorant is
Zs
0ZX
sup
nN|Ku(k)(x)φI(x)η′′
n(x)|(x)du CZs
0ZX
|x|1α
µ(B(x, t+u)) exp |xz|2
c2(t+u)xαdx du
CZs
0ZX
|x|1α
(t+u)1/2x
x+t+uα
exp |xz|2
c2(t+u)dx du
Ct1/2sZX|x|1αexp |xz|2
c2(t+s)dx
C(I, t, s).
The integrals with B3and B1goes similarly. For the latter one we use (4.16).
Acknowledgments: The authors would like to thank Jacek Dziubański for his helpful remarks.
References
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Edyta Kania
Instytut Matematyczny, Uniwersytet Wrocławski
pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland
Marcin Preisner
Instytut Matematyczny, Uniwersytet Wrocławski
pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland
... We are interested in proving the lower Gaussian estimates, but this can be done only for some potentials V . For other potentials Hardy spaces may have a local character, (see e.g., [17]). Let ...
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... This can be done by a standard procedure, for details see e.g. [24,Thm. 2.2(b)]. ...
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