Inverse scattering: Mathematical properties of phase shifts · Farkas Miklós Seminar, Budapest, 2017.10.19. Properties of phase shifts Miklós Horváth Introduction Uniqueness and

Properties ofphase shifts

MiklósHorváth

Introduction

Uniquenessand stability

Distribution ofthe phaseshifts

Inverse scattering: Mathematical properties ofphase shifts

Miklós Horváth

Farkas Miklós Seminar, Budapest, 2017.10.19.


MiklósHorváth

Introduction



The main topics of this talk

1 Introduction

2 Uniqueness and stability

3 Distribution of the phase shifts


MiklósHorváth

Introduction



How to detect atomic forces?

A beam of particles is shot to a material.

After interacting with atomic forces, the particles scatter.

The number of particles scattered to different directions iscounted by detectors.

The potential energy of the atomic force space can berecovered from these scattering data.


MiklósHorváth

Introduction



The wave function

Schrödinger equation:

−∆u(x) + V (x)u(x) = k2u(x), x ∈ R3,

V (x) is the potential energy of the atomic force space, k isthe energy of the incoming particles.For V = 0 u(x) = eikr/r and u(x) = e−ikr/r , r = |x | aresolutions. The wave function U(x , t) = <(u(x)e−iωt )corresponds to outgoing, resp. incoming spherical wave.The Sommerfeld radiation condition

limr→∞

r(∂u∂r− iku

)= 0

excludes incoming spherical waves.


MiklósHorváth

Introduction



The wave function

For decaying potential V there is a unique solution of−∆u + Vu = k2u of the form

u = ui + us,

where

ui(x) = eikx ·α incoming wave from direction α

and the scattered wave us satisfies the Sommerfeldcondition

limr→∞

r(∂us

∂r− ikus

)= 0.


MiklósHorváth

Introduction



Scattering amplitude

The unique solution satisfies

us(x) =eikr

rA(x̂ , α, k) + O

(1r2

)r = |x | → ∞, x̂ = x/r .

Here A(x̂ , α, k) is the scattering amplitude.|A| can be measured: the number per unit time of particlesscattered to the solid angle dx̂ is proportional to |A|2dx̂ .A can be reconstructed from |A|.


MiklósHorváth

Introduction



Spherical symmetry

Let the potential be spherically symmetrical:

V (x) = q(r), r = |x |.

Suppose∞∫

0

r |q(r)|dr <∞.

"Singularity less than r−2 near 0, decay more than r−2 at∞."Separation of variables:

u(x) =yn(r)

rY m

n (θ, ϕ), r = |x |, n ≥ 0, −n ≤ m ≤ n.

In variable r we get the radial Schrödinger equations

−yn”(r)+

(q(r) +

n(n + 1)

r2

)yn(r) = k2yn(r), r ≥ 0, n = 0,1, . . . .


MiklósHorváth

Introduction



Asymptotics of the solutions

−yn”(r)+

(q(r) +

n(n + 1)

r2

)yn(r) = k2yn(r), r ≥ 0, n = 0,1, . . . .

If r → 0+, −yn” + n(n+1)r2 yn � 0, hence yn � crn+1 + dr−n.

Only the first one is allowed, otherwise u is singular.If r →∞, −yn” � k2yn, hence yn � const · sin(kr + const).There is a unique solution of the form

yn(r) =

{cnrn+1(1 + o(1)), r → 0+, cn > 0,sin(kr − nπ/2 + δn) + o(1), r →∞.

The constants δn are the phase shifts.

q = 0⇒ δn = 0 (and yn(r) =√πkr/2Jn+1/2(kr)).


MiklósHorváth

Introduction



Scattering amplitude in case of spherical symmetry

The wave function and the scattering amplitude expressedby phase shifts:

u(x) =∞∑

n=0

ineiδn2n + 1

kyn(r)

rPn(x̂ · α),

A(x̂ , α, k) =1k

∞∑n=0

(2n + 1)eiδn sin δnPn(x̂ · α),

wherePn(t) =

12nn!

[(t2 − 1)n

](n)are the Legendre polynomials.

The scattering amplitude depends only on the angle ofincident and scattered directions.


MiklósHorváth

Introduction



Uniqueness

There are slowly decaying transparent potentials for whichall δn = 0 (R. Newton 1961).

For exponentially decaying potentials the δn uniquelydetermine the potential q (Novikov 1988).

For general nonexponentially decaying potentialsuniqueness is still open.


MiklósHorváth

Introduction



Uniqueness from partial data

For compactly supported potentials a zero densitysubsequence is enough:

Theorem (Ramm 1999.)

If q(r) = 0 for r > a and rq(r) ∈ L2(0,a), then∑n∈L,n 6=0

1n

=∞⇒ {δn : n ∈ L} determines q


MiklósHorváth

Introduction



Uniqueness from partial data

Theorem (H. 2006.)

a. L1 instead of L2.b. Let for 0<σ < 1B(σ) = {q : q(r) = 0 for r > a, r1−σq(r) ∈ L1(0,a)}. If∑

n∈L,n 6=0

1n<∞

then ∀ q ∈ B(σ) ∃q∗ 6= q ∈ B(σ) for whichδn(q) = δn(q∗) ∀n ∈ L.

Müntz condition:∑n∈L,n 6=0

1n

=∞⇔ {e−nx : n ∈ L} closed in L1(0,∞).


MiklósHorváth

Introduction



Stability

Logarithm of the input error in the output: weak stability

Theorem (H. with M. Kiss, 2009)

Suppose that k = 1, supp(q) ⊂ [0,a], |rq(r)| ≤ D,∫ a0 |d(r2q(r))| ≤ D and similarly for q∗. Now if

| sin(δ∗n − δn)| < ε(ae

2n

)2n+2, 0 ≤ n ≤ N

then

‖r2(q∗(r)− q(r))‖L2(0,a) ≤ c(a,D)(

N−1/2 + log(1/ε)−1/2).


MiklósHorváth

Introduction



Decay rate of the phase shifts

Let k = 1.

Theorem (Ramm 1998.)

Let rq(r) ∈ L2(0,a), q = 0 for r > a and q has constant signon (a− δ, a) for some δ > 0. Then

limn→∞

n|δn|1/(2n) =ae2.

So the support of the potential can be simply recoveredfrom phase shifts.


MiklósHorváth

Introduction



Decay rate of the phase shifts

Theorem (H. 2010)

Let rq(r), rq∗(r) ∈ L1(0,∞).a. If q = q∗ a.e. on (a,∞), then for all sufficiently large n

|δn − δ∗n| ≤ c(ae

2n

)2n+2.

b. Conversely, if q and q∗ have compact support and

δn − δ∗n = O((a1e

2n

)2n), n→∞

holds for all a1 > a, then q = q∗ a.e. on (a,∞).


MiklósHorváth

Introduction



Growth rate of the scattering amplitude

The analogous result for the scattering amplitude

A(x̂ , α, 1) = F (t) =∞∑

n=0

(2n + 1)eiδn sin δnPn(t), t = x̂ · α :

Theorem (H. 2010)

Let rq(r), rq∗(r) ∈ L1(0,∞) and a > 0. Now if q = q∗ a.e.on (a,∞) then F (t)− F ∗(t) is entire and

|F (t)− F ∗(t)| ≤ c(1 + |t |)ea√

2|t |, t ∈ C.

Conversely, if q and q∗ have compact support, F (t)− F ∗(t)is entire and

F (t)− F ∗(t) = O(ea1√

2|t |), t ∈ C ∀a1 > a

then q = q∗ a.e. on (a,∞).


MiklósHorváth

Introduction



Variational formula for the phase shifts

Ho and Rabitz 1988, J. Chem. Phys.:

−δ̇n =

∫ ∞0

q̇y2n .

So if q increases, the phase shifts decrease. The precisedomain of δn for compactly supported potentials:

Theorem (H. 2007)

Let q(r) = 0 a.e for r > a and rq(r) ∈ L1(0,a). Then

δn−1(χ(0,a)) < δn(q) <∞, n ≥ 1

and neither bounds can be improved. In particular−a < δ0 <∞, −a + arctan a < δ1 <∞,π2 − a + arctan a2−3

3a < δ2 <∞ etc.


MiklósHorváth

Introduction



Inequalities between phase shifts

µn = n + 1/2− aJ ′n+1/2(a)− tan δnY ′n+1/2(a)

Jn+1/2(a)− tan δnYn+1/2(a).

∆0µn = µn, ∆k+1µn = ∆kµn −∆kµn+1.

Theorem (H. 2009)

If rq(r) ∈ L1(0,a), supp q ⊂ [0,a], q ≤ 1 a.e and∫ a0 r(1− q(r)) dr < 2n0 + 1 then the backward differences ofµn satisfy

∆kµn ≥ 0 ∀n ≥ n0, ∀k ≥ 0.

If equality occurs for some k an n then q = 1 a.e. on [0,a].


MiklósHorváth

Introduction




In particular, if∫ a

0 r(1− q(r)) dr < 1 and q = 0 for r > a andq ≤ 1 then

a2

1− a cot(δ1 + a)≥ 2 + a cot(δ0 + a)

with equality if and only if q = 1 a.e. on (0,a).


MiklósHorváth

Introduction




Theorem (H. with O. Safar, 2010)

If suppq ⊂ [b,a] with some values 0 < b < a < π/2 andq ≤ 1 then

(1/a− cot a)2

(1− b cot b)2 δ0 ≤ δ1

≤ b2

9 max(1, (1/a + tan a)2(1/a− cot a)2)δ0.


MiklósHorváth

Introduction



δ0 versus δ1

Regge 1959:δ1 < δ0 +

π

2.

Theorem (H. and O. Sáfár 2017.)

Let q ≥ 0, rq(r) ∈ L1(0, π/2), q = 0 for r > π/2. Then

δ1 ≤ δ0 − arctan(δ0).

Stronger than δ1 < δ0 + π/2. It turned out that

−a + arctan(a) < δ1 ≤ δ0 − arctan(δ0) if q = 0 for r > a

(recall that −a < δ0 and t 7→ t − arctan(t) is increasing).Key points of the proof:


MiklósHorváth

Introduction



y1/y0 is increasing

Lemma

y1/y0 is strictly increasing on r ∈ (0, π).

[y1

y0

(y1

y0

)′y2

0

]′= y2

0

[(y1

y0

)′]2

+y1

y0(y ′1y0 − y ′0y1)′

= y20

[(y1

y0

)′]2

+2r2 y2

1 > 0

⇒ y1y0

(y1y0

)′y2

0 increasing. It tends to 0 near r = 0+, hencey1y0

(y1y0

)′y2

0 > 0 on (0, π) and then(

y1y0

)′> 0.


MiklósHorváth

Introduction



Upper estimate for y1/y0

For 0 < b ≤ π/2 and M > 0 let q = Mχ(0,b) be abox-potential.

Lemma

If q = Mχ(0,b) then

y21 (b)

y20 (b)

≤ b2

1 + b2 .

Technical proof, e.g.

coth xx− 1

x2 <1

1 +√

2 + x2 coth2 x, x > 0.


MiklósHorváth

Introduction



Optimal potential must be box-potential

AM = {q : 0 ≤ q ≤ M, rq(r) ∈ L1(0, π/2), q = 0 for r > π/2}.

Compact set in the weighted L1 space. Fix 0 ≤ K ≤ π2/41+π2/4 .

Suppose q ∈ AM maximizes δ1 − K δ0. Let

y21

y20

{< K , if r < r0

> K , if r > r0.

(δ1 − K δ0). =

∫ π/2

0q̇(Ky2

0 − y21 )⇒ q = Mχ(0,r0).

If K = r∗2

1+r∗2 then r∗2

1+r∗2 = K =y2

1 (r0)

y20 (r0)

≤ r20

1+r20⇒ r∗ ≤ r0.


MiklósHorváth

Introduction



What if M →∞?

If M →∞ then r0 → r∗, δ0(Mχ(0,r0))→ −r∗,δ1(Mχ(0,r0))→ −r∗ + arctan r∗, hence

δ1(q)− K δ0(q) ≤ δ1(Mχ(0,r0))−r∗2

1 + r∗2δ0(Mχ(0,r0))

≤ −r∗ + arctan r∗ − r∗2

1 + r∗2(−r∗)

⇒ δ1(q) ≤ inf0<r∗<π/2

(r∗2

1 + r∗2δ0(q)− r∗ + arctan r∗ +

r∗3

1 + r∗2

).

Minimum attained at r∗ = −δ0(q), hence

δ1(q) ≤ δ0(q)− arctan δ0(q).


MiklósHorváth

Introduction



Some open problems

Extensions to nonpositive q, to q nonzero beyond π/2Lower estimate of δ1 − K δ0

Estimates of any expression of the δn

Inner characterization of the sequences of phase shifts.Recall that the whole sequence can be identified from somesubsequences of zero density. Loeffel 1968. gave a highlysophisticated description for compactly supported potentials.

Unique recovery of q from the phase shifts fornonexponentially decaying potentials

Documents

Inverse scattering: Mathematical properties of phase shifts · Farkas Miklós Seminar, Budapest, 2017.10.19. Properties of phase shifts Miklós Horváth Introduction Uniqueness and