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Inverse Problem for Quantum Graphs:Magnetic Boundary Control

Pavel Kurasov

December 21, 2019

Vienna

Kurasov (Stockholm) Magnetic Boundary Control December 21, 2019, Vienna 1 / 18

Quantum graph

Metric graph

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PPPP

PPP��HHH

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�� pp pp

pp

p pp

Differential expression on the edges

`q,a =

(id

dx+ a(x)

)2

+ q(x)

Matching conditionsVia irreducible unitary matrices Sm associated with each internal vertex Vm

i(Sm − I )~ψm = (Sm + I )∂ ~ψm, m = 1, 2, . . . ,M.


Inverse problem

Task: reconstruct all three members from the family:

the metric graph

the potential(s) q(x) and a(x);

the vertex conditions.

Contact set ∂Γ - a fixed set of vertices containing all degree one vertices.Boundary control - solution to the wave equation subject to control conditionson ∂Γ

(i ddx + a(x)

)2u(x , t) + q(x)u(x , t) = − ∂2

∂t2 u(x , t)

u(x , 0) = ut(x , 0) = 0,

u(·, t)|∂Γ = ~f (t)

Response operatorRT : ~f (t)︸︷︷︸

=~u(·,t)|∂Γ

7→ ∂~u(·, t)|∂Γ.


Response operator and M-function

Response operatorRT : ~f (t)︸︷︷︸

=~u(·,t)|∂Γ

7→ ∂~u(·, t)|∂Γ.

Titchmarsh-Weyl matrix-valued M-function:Consider ψ(x , λ) - solution to the stationary equation(

id

dx+ a(x)

)2

ψ(x , λ) + q(x)ψ(x , λ) = λψ(x , λ)

M(λ) : ~ψ|∂Γ 7→ ∂ ~ψ|∂Γ

Connection

(R~f)

(s) = M(−s2)~f (s)

where · denotes the Laplace transform.


Two explicit formulas

MΓ(λ) = −

( ∞∑n=1

〈ψstn |∂Γ, ·〉`2(∂Γ)ψ

stn |∂Γ

λstn − λ

)−1

,

where λstn and ψstn are the eigenvalues and ortho-normalised eigenfunctions of

Lst.

MΓ(λ)−MΓ(λ′) =∞∑n=1

λ− λ′

(λDn − λ)(λDn − λ′)〈∂ψD

n |∂Γ, ·〉CB∂ψDn |∂Γ,

where λDn and ψDn are the eigenvalues and eigenfunctions of the Dirichlet

Schrodinger operator LD .

These formulas indicate where zeroes and singularities of the M-functionsmay be situated.

Existence of invisible eigenfunctions.


LimitationsThe exact form of the magnetic potential plays no role ⇒ magnetic fluxesthrough cycles (if any)

Φj =

∫Cj

a(y)dy ⇒ fluxes Φj

Vertex conditions can be determined up to phases corresponding to internaledges:

ψ(x) = e iθn ψ(x), x ∈ En. ⇒ phases Θn

One has to require that reflection and transmission from the vertices isnon-trivial.Ex.

x2x1

x3

x4x5x6

x7

x8 S1 =

0 1√

20 1√

21√2

0 1√2

0

0 1√2

0 − 1√2

1√2

0 − 1√2

0

(Sm(∞)

)ij6= 0


Inverse problems for trees

Geometric perturbations

peeling leafs;

trimming brunches;

cutting brunches.

Cleaning procedure

removing potential q on the boundary edgesSchrodinger ⇒ Laplace

All four principles are based on two Lemmas concerning gluing extensions ofsymmetric operators.


Gluing graphs

⇒

Two graphs with 4 and 3 contact vertices are transformed into a graph with 3contact vertices.

One gluing point:

⇒


Gluing extensions of symmetric operatorsLet A1 and A2 be two symmetric operators with the boundary values:

〈A∗j uj , vj〉 − 〈uj ,A∗j vj〉 = 〈~u∂j , ∂~vj〉 − 〈∂~u∂j , ~v∂j 〉.

M-functionsM1,2(λ)~ψ∂ = ∂~u∂ , where A∗j ψj = λψj .

Consider the symmetric extension A = A1 ⊕ A2 determined by the couplingconditions:

~u∂1 |L1 = ~u∂2 |L2 ; ∂~u∂1 |L1 = −∂~u∂2 |L2 ,

where L1, L2 are two identified subspaces of the same dimension.Lemma 1 (following Schur-Frobenius)The M-functions associated with the operators A1,A2, and A are related via

M(λ) =

(M22

1 −M211 (M11

1 + M112 )−1M12

1 −M211 (M11

1 + M112 )−1M12

2

−M212 (M11

1 + M112 )−1M12

1 M222 −M21

2 (M111 + M11

2 )−1M122

).

where M lmj come from the block decomposition of MΓj .

Lemma 2Assume that dim L1 ≡ dim L2 = 1, then any two out of three M-functions in theabove formula determine uniquely the third one.


Three inverse problems

I Reconstruction of the metric graphGlobally: Every metric tree is uniquely reconstructed from the travellingtimes between the boundary verticesLocally: Travelling times associated with a bunch determine the bunch

II Reconstruction of the potentialBoundary Control: (the diagonal of) the response operator RT determinesthe potential on the boundary edges everywhere at the distance d < T/2from the boundary

III Reconstruction of the vertex conditionsThe response operator for an equilateral bunch with zero potentialdetermines the vertex conditions up to the phase θn associated with theroot edge.


Three inverse problems

III Reconstruction of the vertex conditionsThe response operator for an equilateral bunch with zero potentialdetermines the vertex conditions up to the phase θn associated with theroot edge.The kernel of the response operator is of the form:

r(t) = −δ′(t)+2PSv(∞)Pδ′(t−2`)+4P(I−P−1)A(I−P−1)Pδ(t)+smooth

whereI Sv(∞) - the limit of the vertex scattering matrix;I P - the projector on the d − 1 components associated with the bunch;I A - the matrix in the quadratic form parameterisation of vertex conditions;I P−1 - the eigenprojector for S corresponding to e.v. −1.

Sv(∞) = I− 2P−1


Fusion1 Selecting a bunch in the tree.

Using global or local procedure (Subproblem I)2 Reconstruction of the potential on the bunch.

Boundary Control determines potential on the boundary edges from thebunch (Subproblem II).

3 Cleaning of the bunch - removing potential.Based on gluing Lemmas.

4 Trimming of the bunch.Based on gluing Lemmas.

5 Recovering vertex conditions at the root.The response operator for equilateral bunch with zero potential determinesthe vertex condition at the root of the bunch (Subproblem III).

6 Cutting away bunches - pruning of the tree.Based on gluing Lemmas.

Repeating this procedure sufficiently many times we solve the inverse problemfor trees, of course up to the phases θn at internal edges.Assumptions:

q ∈ L1(Γ),Sv(∞) does not have zero entries.Kurasov (Stockholm) Magnetic Boundary Control December 21, 2019, Vienna 12 / 18

Inverse problems for graphs with cycles

Magnetic potential is equivalent to change of vertex conditions ⇒ onlystandard vertex conditions (continuity + Kirchhoff).

The response operator and M-matrices are considered as functions of theset of fluxes Φj , fixed each time.Magnetic boundary control

M(λ,Φj) 7→ q(x)

Key feature:existence of invisible eigenfunctionsEvery such eigenfunction produces ambiguity in the solution of the inverseproblem.


Loop graph

V1x1x2 x3

MΓloop`1

(λ,Φ) = M11 + 2 cos ΦM12 + M22

=2 cos Φ− TrT (k)

t12(k),

where T (k) = {tij(k)} is the transfer matrix for [x1, x2].

NB! The M-function depends on cos Φ, i.e. Φ and −Φ lead to the same MΓloop`1


Solution using the transfer matrix

Solving inverse problem:

t12(k) =1

M12(λ);

u+(k) :=1

2

(t11(k) + t22(k)

)=

1

2

TrM(λ)

M12(λ).

t12(kj) = 0 - the spectrum of the Dirichlet-Diricihlet operator on [x1, x2].At these points we know:

t11(kj)t22(kj) = 1 detT (k) = 1;

t11(kj) + t22(kj) = u+(kj)− known function.

⇒ t11(kj), t22(kj) are determined up to a sign

tii (k) are analytic functions of exponential type and are uniquely determined bytheir values at kj , provided

∫q(x)dx = 0.


Solution using the M-function

M12(λ) =1

4

(MΓloop

`1

(λ, 0)−MΓloop`1

(λ, π)),

Tr M(λ) =1

2

(MΓloop

`1

(λ, 0) + MΓloop`1

(λ, π)).

Consider λDj = k2j and check for which (complex) Φ the M-function does not

have a singularity at this point.

MΓ(λ,Φ)) = MΓ(λ′,Φ)+∞∑n=1

λ− λ′

(λDn − λ)(λDn − λ′)〈∂ψD

n (Φ)|∂Γ, ·〉`2(∂Γ)∂ψDn (Φ)|∂Γ,

Find Φj for which the M-function is not singular at λDj . NB! The phase Φj isdetermined up to a sign, since the formula for M(λ,Φ) contains cos Φ.Invisible eigenfunction{

ψ(x1) = ψ(x2) = 0,

ψ′(x1)− e iΦjψ′(x2) = 0.⇒ ψ′(x1)

ψ′(x2)= e iΦj


Cycle opening procedure

Note that the last formula does not require that M is an M-function for aninterval - we did not use properties of the transfer matrix.

⇒ ⇒

⇒

Each cycle requires a sequence of signs + a real constant.


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