trees and the agmon metric · 2015. 3. 24. · anna maltsev (university of bristol) trees and the...

Trees and the Agmon Metric

Anna Maltsev

University of Bristol

March 23, 2015

Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 1 / 19

Introduction

Main question

How do L2 eigenfunctions decay on a quantum graph?

Prior work:

Liouville and Green for 1D, Agmon: negative energies

Hislop-Post: Worked with “radial trees,” i.e. identical lengths andidentical branching numbers, and a potential added at the vertices asa vertex condition; positive energy localization

Random lengths model

Introduction

Main question

Prior work:

Introduction

Main question

Prior work:

Introduction

Agmon philosophy

For negative eigenvalues (actually E < lim inf V ), use tricky integration byparts to show that ψ ∈ L2 and Hψ = Eψ =⇒ exp(ρ(0, x))ψ ∈ L2, whereρ is a metric. Once L2 exponential decay is established, they can be usedto establish pointwise estimates showing exponential decrease.The method works essentially if x : V (x)− E ≤ 0 is compact, and theexpected metric is essentially an ”action integral” of (V (x)− E )1/2+ overthe minimizing path.

Introduction

Notations:

Our setup:

Quantum graph Γ; edges are segments of the real line

A function space K on Γ so that for φ ∈ K1 φ is twice differentiable on the edges2 φ is continuous at the vertices

3 if v is a vertex, ej are the edges adjacent to v , and φej is φ restrictedto the edge ej , then

∑φ′ej (v) = 0 where the derivative is taken in the

direction away from the vertex v

1 whatever comes in must come out2 Kirchoff condition

4 Notice: if f , g ∈ K then so is fg .Differential operator H = −∆ + V (x) acting on K

Introduction

Notations:

Our setup:

Quantum graph Γ; edges are segments of the real line

A function space K on Γ so that for φ ∈ K1 φ is twice differentiable on the edges2 φ is continuous at the vertices3 if v is a vertex, ej are the edges adjacent to v , and φej is φ restricted

to the edge ej , then∑φ′ej (v) = 0 where the derivative is taken in the

direction away from the vertex v

1 whatever comes in must come out2 Kirchoff condition

4 Notice: if f , g ∈ K then so is fg .Differential operator H = −∆ + V (x) acting on K

Case Studies The Line

Case study: the Line

Definition 1

Let ρE (x , y) =∫ yx

√(V (t)− E )+dt.

Notes:

In multiple dimensions or on graphs with non-trivial topology, takethe minimum over all possible paths from x to y .

Theorem 2 (Agmon)

Let H = −∆ + V with V real and continuous be a closed operatorbounded below with σ(H) ⊂ R. Suppose E is an eigenvalue of H and thatsupp(E − V (x))+ is compact. Suppose ψ ∈ L2 is an eigenfunction of H.Then for any � > 0 there exists a constant c� such that∫

e2(1−�)ρE (x)|ψ(x)|2dx ≤ c�

adapted from Hislop-Sigal book

Case Studies The Line

Case study: the Line

Definition 1

Let ρE (x , y) =∫ yx

√(V (t)− E )+dt.

Notes:

In multiple dimensions or on graphs with non-trivial topology, takethe minimum over all possible paths from x to y .

Theorem 2 (Agmon)

Let H = −∆ + V with V real and continuous be a closed operatorbounded below with σ(H) ⊂ R. Suppose E is an eigenvalue of H and thatsupp(E − V (x))+ is compact. Suppose ψ ∈ L2 is an eigenfunction of H.Then for any � > 0 there exists a constant c� such that∫

e2(1−�)ρE (x)|ψ(x)|2dx ≤ c�

adapted from Hislop-Sigal bookAnna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 5 / 19

Agmon Metric on the Line

Agmon on the Line: Proof ingredient # 1

Lemma 3

For φ ∈ L2 and Fα bounded and satisfying V − E −∣∣∣F ′αFα ∣∣∣2 > δ, we get that

〈Fαφ, (H − E )1

Fαφ〉 ≥ δ‖φ‖2

Proof:〈Fαφ, (H − E )

1

Fαφ

〉=

∫(V − E )|φ|2 + (Fαφ)′

(φ

Fα

)′+ BT

=

∫(V − E )|φ|2 + |φ′|2 −

∣∣∣∣F ′αφFα∣∣∣∣2 ≥ δ‖φ‖2.

Dropped the term |φ′|2, since positive. Integrated by parts, since φ ∈ L2and Fα is bounded. Note δ is independent of the upper bound on Fα.

Agmon on the Line: Proof ingredient # 1

Lemma 3

For φ ∈ L2 and Fα bounded and satisfying V − E −∣∣∣F ′αFα ∣∣∣2 > δ, we get that

〈Fαφ, (H − E )1

Fαφ〉 ≥ δ‖φ‖2

Proof:〈Fαφ, (H − E )

1

Fαφ

〉=

∫(V − E )|φ|2 + (Fαφ)′

(φ

Fα

)′+ BT

=

∫(V − E )|φ|2 + |φ′|2 −

∣∣∣∣F ′αφFα∣∣∣∣2 ≥ δ‖φ‖2.

Dropped the term |φ′|2, since positive. Integrated by parts, since φ ∈ L2and Fα is bounded. Note δ is independent of the upper bound on Fα.

Proof Ingredient # 2

Lemma 4

Let η be a smoothed characteristic function of {V − E > δ} such that η′has compact support and ψ be an L2 solution of (H − E )ψ = 0. Then〈

F 2αηψ, (H − E )ηψ〉

=〈η′(ηF 2α

)′ψ,ψ

〉≤ C‖ψ‖2. (3.1)

For the equality can use that (H − E )ψ = 0 and integrate by partsThe inequality follows because η′ has compact support

Can construct such η because we assume E to be an eigenvaluebelow the spectrum, equivalent to V − E ≤ δ compact. Here η = 1on the exterior set and 0 on the compact set

Here we have taken φ from before to be ηFαψ, so thatδ‖ηFαψ‖2 ≤ C‖ψ‖2.

Lemma 4

=〈η′(ηF 2α

)′ψ,ψ

〉≤ C‖ψ‖2. (3.1)

Lemma 4

=〈η′(ηF 2α

)′ψ,ψ

〉≤ C‖ψ‖2. (3.1)

Lemma 4

=〈η′(ηF 2α

)′ψ,ψ

〉≤ C‖ψ‖2. (3.1)

Proof ingredient # 3

We want to take F as large as possible, but still satisfying the constraint

V − E −∣∣∣F ′F ∣∣∣2 > δ, so we construct it by

F (x) = e(1−δ)∫ x

0 (V (t)−E)+ = e(1−δ)ρE (x)

In above lemmas, we use Fα instead where for α > 0

Fα :=F

1 + αF< Cα

We observe that

(i) Fα(x) < F (x),

(ii) |F ′α(x)| < |F ′(x)|,

(iii)∣∣∣F ′α(x)Fα(x) ∣∣∣ < ∣∣∣F ′(x)F (x) ∣∣∣.

Then can take α→ 0, since the constants throughout do not depend on α.

Proof ingredient # 3

We want to take F as large as possible, but still satisfying the constraint

V − E −∣∣∣F ′F ∣∣∣2 > δ, so we construct it by

F (x) = e(1−δ)∫ x

0 (V (t)−E)+ = e(1−δ)ρE (x)

In above lemmas, we use Fα instead where for α > 0

Fα :=F

1 + αF< Cα

We observe that

(i) Fα(x) < F (x),

(ii) |F ′α(x)| < |F ′(x)|,

(iii)∣∣∣F ′α(x)Fα(x) ∣∣∣ < ∣∣∣F ′(x)F (x) ∣∣∣.

Then can take α→ 0, since the constants throughout do not depend on α.Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 8 / 19

For a quantum tree:

Same argument carries over to rooted trees with few modifications

Integration by parts still works because of the Kirchoff condition

Can show that if ψ ∈ L2(Γ) then e(1−�)ρE (x)ψ ∈ L2, whereρ =

∫ x0 (V (t)− E )

1/2+ and the integral is taken over the unique path

from the root to x

By a standard method can go from L2 estimates to pointwise.

Are we done?

Agmon Metric on the Line The Regular Tree

Case study: the regular tree

Regular tree: starts at a root, each edge has length L and at eachvertex splits into b edges, e.g. the Bethe Lattice.

Fix E < 0. Consider H = −∆, i.e. V = 0 and look for a L2 solution.

The problem: Does there exist an L2 solution and does it decay faster

than e−√|E |x , which would be the corresponding decay on the line.

Case study: the regular tree

Regular tree: starts at a root, each edge has length L and at eachvertex splits into b edges, e.g. the Bethe Lattice.

Fix E < 0. Consider H = −∆, i.e. V = 0 and look for a L2 solution.The problem: Does there exist an L2 solution and does it decay faster

than e−√|E |x , which would be the corresponding decay on the line.

Construction for the Regular Tree

Transfer matrix: T =

(cosh kL sinh kL1b sinh kL

1bcosh kL

)Both eigenvalues λ1, λ2 are real Since detT = 1/b andTrT =

(1 + 1b

)cosh kL > 2/

√b

Since all transfer matrices are equal by construction, the eigenvectorcorresponding to λ1 will give us an initial condition.

The eigenvalue λ1 is given by

λ1 =

(1

2+

1

2b

)cosh kL−

√((1

2+

1

2b

)cosh kL

)2− 1/b

=1(

b2 +

12

)cosh kL +

√((b2 +

12

)cosh kL

)2 − b ≤1

b cosh kL

(3.2)

Construction for the Regular Tree

Transfer matrix: T =

(cosh kL sinh kL1b sinh kL

1bcosh kL

)Both eigenvalues λ1, λ2 are real Since detT = 1/b andTrT =

(1 + 1b

)cosh kL > 2/

√b

Since all transfer matrices are equal by construction, the eigenvectorcorresponding to λ1 will give us an initial condition.

The eigenvalue λ1 is given by

λ1 =

(1

2+

1

2b

)cosh kL−

√((1

2+

1

2b

)cosh kL

)2− 1/b

=1(

b2 +

12

)cosh kL +

√((b2 +

12

)cosh kL

)2 − b ≤1

b cosh kL

(3.2)

the above solution is in L2 for the tree since∫Γ|φ|2 = C

∑n

bnλ2n1 .

If λ1 < α/√b for α < 1 then the above sum converges.

The factor of 1bn makes the pointwise decay faster than the case ofthe line, where the decay is just e−kx , but this comparison is “unfair.”

If a solution is to be in L2 for a tree the 1/√b factor is required for

convergence. However, we have a factor of 1/b instead, which meansthat even if we consider partial integrals, the decay on the tree will befaster than on the line.

Does this hold more generally?

∑n

bnλ2n1 .

∑n

bnλ2n1 .

Our new result

The Agmon Metric for a General Tree

We label the maximum out-degree on the tree to be dmax we will labeleach vertex by a dmaxary string s. We also denote the edge whichterminates at vertex vs by es .

Theorem 5

Suppose ψ ∈ L2 is a solution of H. Suppose at vertex vs , psj is thefraction of the derivative continuing down branch esj . For the path P fromthe root to x , let

ρ(x) =

∫P

(V (t)− E )+ + 2∑v∈P

δV (t) log(1/pv ).

Then if ψ ∈ L2, then e(1−�)ρψ is square integrable on the path P.

Note: e(1−�)ρψ is not in L2(Γ) but sufficient for pointwise bounds.

Our new result

Key change

For ingredient # 1 inequality V − E −∣∣∣F ′F ∣∣∣2 > δ has to be satisfied.

We sacrifice continuity.

Get boundary terms in integration by parts. Recall that φ = Fαηψ sofor an edge e starting at v1 and ending at v2:∫

eFφ

d2

dx2

(1

Fφ

)=

∫e

d

dx

(F 2ηψ

d

dx(ηψ)

)−∫e

d

dx(Fφ)

d

dx

(1

Fφ

)= F 2ηψ

d

dx(ηψ)

∣∣∣∣v2v1

−∫e

d

dx(Fφ)

d

dx

(1

Fφ

)Want to take F 2 so that its discontinuity cancels with thediscontinuity of ddxψ, so we can increase F

2 by a factor of 1/p.

Technicality: we have to work with Fα so the discontinuity will be inFα then take limα→∞

Our new result

Key change

eFφ

d2

dx2

(1

Fφ

)=

∫e

d

dx

(F 2ηψ

d

dx(ηψ)

)−∫e

d

dx(Fφ)

d

dx

(1

Fφ

)= F 2ηψ

d

dx(ηψ)

∣∣∣∣v2v1

−∫e

d

dx(Fφ)

d

dx

(1

Fφ

Our new result

Key change

eFφ

d2

dx2

(1

Fφ

)=

∫e

d

dx

(F 2ηψ

d

dx(ηψ)

)−∫e

d

dx(Fφ)

d

dx

(1

Fφ

)= F 2ηψ

d

dx(ηψ)

∣∣∣∣v2v1

−∫e

d

dx(Fφ)

d

dx

(1

Fφ

Our new result

Case study: the harmonic millipede

We consider a millipede with segments of length L, at each vertex vk ofwhich there dangles an infinitely long leg ek . There is one leg at eachvertex The energy parameter is E = −1, no potential on the main path, soon each bodily edge, ψ′′ = ψ. On the dangling legs the potential isV (x) = (x + 2)2 − 2, where x is the distance from vk , and we see thatV − E > 0 on the legs. The L2 solutions on the legs solveψ′′ = (V (x)− (−1))ψ, which easily gives cst.e−(x+2)2/2. We differentiateand find that ψ′ek (vk) = −2ψ(vk). The transfer matrix for connectingsolutions from one main-body segment to the next is(

cosh L sinh Lsinh L− 2 cosh L cosh L− 2 sinh L

).

The eigenvalues of this are

eL

2

1±√

1−(

2

eL

)2 ≈ e±Lwhen L is large. The solution decays one-dimensionally on the millipede’sbody and superexponentially on the oscillating legs.

Our new result

Notice:

1 Not L2 if any p = 0, so 1/p’s are ok

2 This requires a lot of information on ψ

Consider instead an averaged function Ψ.

Can use a similar Agmon argument to show that FΨ ∈ L2

Do not need information on the p’s, but only get information

Only get information on ψ in an averaged sense

Our new result

Details

Consider a tree with equal edge lengths and equal branching numbersat each generation. Let ψ ∈ L2 be an eigenfunction.

Let Ψ(x) =∑

y :dist(0,y)=x wyψ(x)

wy =∏

v 1/bv where the product is taken over all vertices v on thepath from 0 to y and bv is the out-degree.

Notice:∑

y :dist(0,y)=x wy = 1, so indeed an average

Ψ is continuous everywhere, and Ψ′(v−) = bΨ′(v+) at the vertices

Notice: the p’s are gone!

‖Ψ(x)‖2L2 ≤∫∞

0 (∑

y |wy |2)(∑

y |ψ(y)|2) ≤ ‖ψ(x)‖2L2At each vertex can increase F 2 by a factor of bv , so

e(1−�)∫

(V−E)1/2+ +12

∑n log(bn)δvn (t)dtΨ is also in L2.

Our new result

Details

Consider a tree with equal edge lengths and equal branching numbersat each generation. Let ψ ∈ L2 be an eigenfunction.Let Ψ(x) =

∑y :dist(0,y)=x wyψ(x)

wy =∏

Notice:∑

‖Ψ(x)‖2L2 ≤∫∞

0 (∑

y |wy |2)(∑

e(1−�)∫

(V−E)1/2+ +12

Our new result

Details

wy =∏

Notice:∑

‖Ψ(x)‖2L2 ≤∫∞

0 (∑

y |wy |2)(∑

e(1−�)∫

(V−E)1/2+ +12

Our new result

Details

wy =∏

Notice:∑

‖Ψ(x)‖2L2 ≤∫∞

0 (∑

y |wy |2)(∑

e(1−�)∫

(V−E)1/2+ +12

Work in Progress

Questions to be settled:

1 Is the L2 eigenfunction unique? Analogues of the the limit point -limit circle dichotomy.

2 What if we add leaves and finite subtrees to our regular infinite tree?“Obvious” that this will only improve the decay.

3 What happens in case of a more general graph? What if there arecycles?

Thanks

Thank you!

IntroductionCase StudiesThe Line

Agmon Metric on the LineThe Regular Tree

Our new resultWork in ProgressThanks

trees and the agmon metric · 2015. 3. 24. · anna maltsev (university of bristol) trees and the...

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