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Fractional Laplace operator in the unit ball

Mateusz Kwa±nicki

Wrocªaw University of Science and Technology mateusz.kwasnicki@pwr.edu.pl

Stable processes conference Oaxaca, November 10, 2016

(2)

Outline

(1) Eigenvalues λn of (−∆)α/2 in a ball (2) Eigenvalues µn of (1 − |x|2)α/2+ (−∆)α/2 (3) Detour: Jacobi diusions

(4) Bounds for λn.

Based on joint work with:

• Bartªomiej Dyda (Wrocªaw)

• Alexey Kuznetsov(Toronto)

(3)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Denition

Let Xt denote the isotropic α-stable Lévy process.

Let −L = −(−∆)α/2 be the generator of Xt:

−Lf(x) = lim

t→0+

Exf(Xt) − f(x)

t .

Equivalently:

−Lf(x) = cd,α lim

ε→0+

Z

Rd\Bε

f(y) − f(x)

|y − x|d+α dy.

Remarks:

• We always assume that d = 1, 2, . . . and α ∈ (0, 2).

• Br = B(0, r), B = B(0, 1).

• Above denitions are pointwise; throughout the talk we ignore (important and delicate) questions about domains of unbounded operators.

(4)

Lϕn(x) = λnϕn(x) for x ∈ B,

ϕn(x) = 0 otherwise.

Classical theorem

Solutions ϕn form an orthonormal basis in L2(B), 0 < λ0< λ16 λ2 6 . . .

and ϕ0(x) > 0 for x ∈ B.

Let τ be the time of rst exit from B:

τ =inf{t > 0 : Xt∈ B/ }.

Then:

Ex ϕn(Xt)1{t<τ} = e−λntϕn(x).

(5)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Theorem (consequence of Bochner's relation)

Let V(x) be a solid harmonic polynomial of degree `.

Then:

LV(x) f(|x|) = V(x) g(|x|) in Rd if and only if

Lf(|y|) = g(|y|) in Rd+2`. Remarks:

• True for arbitrary convolution operators L with isotropic kernels.

• Here `solid' = `homogeneous'.

• Examples of V(x): 1, x1, x1x2, x1x2. . . xd, x21− x22.

• Solid harmonic polynomials span L2(∂B).

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We choose V`,j(x) so that

• V`,j(x) is a solid harmonic polynomial of degree `,

• `> 0, j = 1, 2, . . . , Jd,`, where Jd,` = d+2`−2d+`−2 d+`−2`  ,

• V`,j(x) form the basis of L2(∂B). Corollary

Let λradd,n and φradd,n(|x|) be the n-th radial eigenvalue and eigenfunction.

The eigenvalues λn are given by λradd+2`,n, where n, ` > 0.

The corresponding eigenfunctions are V`,j(x) φradd+2`,n(|x|), where j = 1, 2, . . . , Jd,`.

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L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

The eigenvalues λn can thus be arranged in the table:

λradd,0

<

λradd,1

6

λradd,2

6

· · ·

<

λradd+2,0

<

λradd+2,1

6

λradd+2,2

6

· · ·

<

λradd+4,0

<

λradd+4,1

6

λradd+4,2

6

· · ·

<

...

(with `-th row repeated Jd,` times).

We have λ0 = λradd,0. Which one is λ1?

The only possible values are λ1 = λradd,1 and λ1= λradd+2,0.

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The eigenvalues λn can thus be arranged in the table:

λradd,0< < λradd,1 6 λradd,2 6 · · · λradd+2,0< < λradd+2,1 6 λradd+2,2 6 · · · λradd+4,0< < λradd+4,1 6 λradd+4,2 6 · · ·

...

(with `-th row repeated Jd,` times).

We have λ0 = λradd,0. Which one is λ1?

The only possible values are λ1 = λradd,1 and λ1= λradd+2,0.

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L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Conjecture

λradd+2,0< λradd,1

Equivalently: λ1= λradd+2,0, or: ϕ1 is antisymmetric.

ϕ1? not ϕ1?

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Theorem

If d 6 2, or if α = 1 and d 6 9, then indeed λradd+2,0< λradd,1.

Remarks:

• Otherwise this is still an open problem. . .

• . . . strongly supported by numerical bounds.

• Our method: nd two-sided bounds for λradd,n.

(11)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Denition

Let P(α,β)n (r)be the Jacobi polynomial and ψradd,n(|x|) = P(nα2,d2−1)(2|x|2− 1),

µradd,n = 2α Γ (α2 + n + 1)Γ (d+α2 + n) n! Γ (d2 + n) . Theorem

L(1 − |x|2)α/2+ ψradd,n(|x|) = µradd,nψradd,n(|x|) for x ∈ B.

Remark: some special cases have been known before.

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Theorem

The eigenvalues of the operator

L(1 − |x|2)α/2+ f(x) are given by µradd+2`,n, where n, ` > 0.

The corresponding eigenfunctions are ψ`,j,n(x) = V`,j(x) P(

α

2,d+2`2 −1)

n (2|x|2− 1), where j = 1, 2, . . . , Jd,`.

These eigenfunctions form an orthogonal basis in weighted L2(B) space with weight (1 − |x|2)α/2dx.

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L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Once it is proved that in B:

L(1 − |x|2)α/2+ f(x)

maps polynomials

of degree n to polynomials of degree n, (F) it follows easily that:

• the eigenfunctions are polynomials;

• they are orthogonal with respect to (1 − |x|2)α/2+ dx;

• they have the form given in the theorem.

(The actual proof follows a completely dierent path).

Open problem

Is there a soft proof of (F)?

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operator eigenfunction eigenvalue

(1) Lf(x) ϕd+2`,j,n λradd+2`,n

(2) L(1 − |x|2)α/2+ f(x)

ψ`,j,n µradd+2`,n (3) (1 −|x|2)α/2+ Lf(x) (1 −|x|2)α/2+ ψ`,j,n µradd+2`,n

These operators are generators of:

(1) XBt, the process Xt killed upon exiting B;

(3) time-changed XBt;

(2) time-changed Doob h-transform of XBt

(corresponding to h(x) = Exτ = cd,α(1 −|x|2)α/2).

(15)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

The two operators on L2(B):

−(1 −|x|2)α/2+ L and (1 −|x|2)∆ − (2 − α)∇

have identical eigenfunctions!

These operators are generators of:

• time-changed XBt;

• d-dimensional Jacobi diusion.

Question

Is time-changed XBt a subordinate Jacobi diusion?

To answer this, one needs to see whether

µd+2`,n= f((2n + α)(2n + d) + (4n + 2 + α)`) for some Bernstein function f.

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200 400 600 800 1

2 3 4 5

d = 1, α = 0.5, ` = 0 (blue) and ` = 1 (yellow) Do these dots lie on a graph of a Bernstein function?

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200 400 600 800 1

2 3 4 5

d = 1, α = 0.5, ` = 0 (blue) and ` = 1 (yellow) Do these dots lie on a graph of a Bernstein function?

Yes!

The corresponding Bernstein function is f(z) = Γ 12(1 + α +p(1 − α)2+ 4z)

Γ 12(1 − α +p(1 − α)2+ 4z)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

(18)

200 400 600 800 1000 1

2 3 4 5

d = 3, α = 0.5, colours correspond to ` = 0, 1, 2, 3, 4 Do these dots lie on a graph of a Bernstein function?

(19)

200 400 600 800 1000 1

2 3 4 5 6

d = 3, α = 0.5, colours correspond to ` = 0, 1, 2, 3, 4, . . . Do these dots lie on a graph

of a Bernstein function?

No!

There is one for each series (xed `), but they are all dierent!

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

(20)

Question

Is time-changed XBt a subordinate Jacobi diusion?

Disappointing theorem

Yes if d = 1. No if d > 2.

Time-changed |XBt|, however, is a subordinate Jacobi diusion in any dimension!

Open problem

Consider time-changed asymmetric 1-dimensional stable process, with clock running at rate (1 + x)ρα(1 − x)^ρα. Is this process a subordinate Jacobi diusion?

(21)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

For x ∈ B we have:

Lϕradd,n(|x|) = λradd,nϕradd,n(|x|) L(1 − |x|2)α/2ψradd,n(|x|) = µradd,nψradd,n(|x|).

Denition

fradd,n(x) = (1 −|x|2)α/2+ ψradd,n(|x|).

We x d and restrict attention to radial functions.

Dropradd, from the notation: µn= µradd,n, fn= fradd,n etc.

Thus, for x ∈ B we have:

(1 −|x|2)α/2Lfn(x) = µnfn(x).

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RayleighRitz variational methodgives upper bounds.

The values of

A(n, m) = Z

B

fn(x)Lfm(x) dx, B(n, m) =

Z

B

fn(x) fm(x) dx are given by closed-form expressions.

Fix N and let A, B be N × N matrices with entries A(n, m), B(n, m), respectively.

Theorem

Let λn be the solutions of the eigenvalue problem A~v = λ B~v.

Then λn 6 λn for n = 0, 1, . . . , N − 1.

(23)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Remarks:

• Since fn are orthogonal in weighted L2(B)with weight (1 −|x|2)−α/2dx, in the problem

A~v = λ B~v.

the matrix A is diagonal:

A(n, m) = Z

B

fn(x)Lfm(x) dx

= µm Z

B

fn(x) fm(x) (1 −|x|2)−α/2dx;

the matrix B with entries B(n, m) =R

Bfn(x) fm(x) dx is not diagonal.

• Quality of the bounds improve rapidly as N grows.

• Numerical methods work for relatively large N.

(24)

Aronszajn method of intermediate problemsgives lower bounds.

Two eigenvalue problems in B:

Lf(x) = λ f(x),

Lf(x) = µ (1 − |x|2)−α/2f(x) correspond to Rayleigh quotients:

Q(f) = R

BRf(x)Lf(x) dx

B(f(x))2dx , Q0(f) =

R

Bf(x)Lf(x) dx R

B(f(x))2(1 −|x|2)−α/2dx. Clearly, Q0(f)6 Q(f), and hence µn6 λn.

(25)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

The basic bound µn 6 λn is poor.

Improved bounds come from intermediate problems, corresponding to Reyleigh quotient

QN(f) =

R

Bf(x)Lf(x) dx R

B(f(x))2(1 −|x|2)−α/2dx −R

B(PNf(x))2w(x) dx, where

w(x) = ((1 −|x|2)−α/2− 1)

and PN is the orthogonal projection in weighted L2(B) space with weight w(x) dx onto the linear span of

fn+1(x) − fn(x)

1 − (1 −|x|2)α/2, n = 0, 1, . . . , N − 2.

(26)

Recall that QN(f) =

R

Bf(x)Lf(x) dx R

B(f(x))2(1 −|x|2)−α/2dx −R

B(PNf(x))2w(x) dx. It is rather clear that Q0(f)6 Q1(f)6 . . . → Q(f).

Theorem

The eigenvalues λn corresponding to QN satisfy λn 6 λn.

Surprise: one can actually calculate λn!

(27)

L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Remarks:

• The only non-closed-form expressions here are Z

B

(1 −|x|2)α/2(1 −|x|2n) 1 − (1 −|x|2)α/2 dx.

• The eigenvalues λn of the intermediate problem are equal to either µm or zeros of a polynomial Wn, which is the determinant of an N × N matrix (WeinsteinAronszajn determinant).

• Quality of the bounds improve rapidly as N grows.

• Numerical methods work well for relatively small N;

larger N leads to ill-conditioned matrices.

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We prove the middle inequality in

λradd+2,06 λradd+2,0< λradd,1 6 λradd,1

analytically using N = 2 (that is, 2 × 2 matrices).

Our method could work for d 6 9 and any α ∈ (0, 2).

We managed to work out the technical details only when d6 2 or α = 1.

0.5 1.0 1.5 2.0

0.4 0.5 0.6 0.7 0.8 0.9 1.0

d  1 d  2 d  3 d  4 d  5 d  6

0.5 1.0 1.5 2.0

-0.1 0.1 0.2 0.3 0.4 0.5 0.6

d  7 d  8 d  9 d  10 d  11

radd,1)1/α− (λradd+2,0)1/α

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L = (−∆)α/2 (1 −|x|2)α/2+ L Jacobi Bounds Problems

Open problem 1

Is there a soft proof of the statement:

L(1 − |x|2)α/2+ f(x)

maps polynomials

of degree n to polynomials of degree n? (F) Open problem 2

Consider an asymmetric 1-dimensional stable process, time-changed with clock running at rate (1 − x)α+(1 + x)α. Is it a subordinate Jacobi diusion?

Open problem 3

Explainwhy the spectrum of (1 − |x|2)α/2+ L is so simple.

Open problem 4

Prove that ϕ1 is antisymmetric when d 6 9.

Cytaty

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