Reconstruction of the Sturm-Liouville operator on a
p-star graph with nodal data
Y. H. Cheng
August 13, 2008
Abstract
In this paper, we deal with the inverse problem of reconstructing the Sturm-Liouville operator defined on a p-star graph. We prove that a dense subset of nodal data can uniquely determine the boundary conditions and the potential functions qi, i = 1, 2, ..., p. We also give reconstruction formulas for them.
1
Introduction
Recently there is a lot of interest in the study of Sturm-Liouville on graphs. On one hand,
the problem is a natural extension of the classical Sturm-Liouville problem on an interval.
On the other hand, it has a number of applications in networks, spider webs, interlocking
springs and even nanostructures. Kuchment called this Sturm-Liouville problem defined on
graphs quantum graphs [4, 5, 6]. In [7], Kuchment and Post studied the spectral properties
of the periodic boundary value problem for the carbon atom in graphite.
In two papers [10, 11], Pivovarchik proved an inverse spectral problem for the p-star
eigenvalues (in fact, one set for each of the p edges and an extra set for the overall eigenvalue
problem) are associated with an overall potential function Q = (q1, q2, . . . , qp). In the course
he gave the asymptotic expansion of eigenvalues and showed that there are p sequences of
eigenvalues which one sequence is simple while the others might not be.
The inverse nodal problem is the problem of understanding the potential function using
the information of the nodal data. The problem is now well studied on the finite interval
(see [12, 8, 9, 1, 2]). The issues of uniqueness, reconstruction and stability are all solved. In
this paper, we plan to study the question of reconstruction for the Sturm-Liouville problem
defined on a p-star graph. Consider the Sturm-Liouville problem on a p-star graph, with
each edge of length 1, defined as the following:
−y00i + qi(x)yi = λyi , i = 1, 2, ..., p , (1)
yi(0, λ) cos αi+ yi0(0, λ) sin αi = 0 , i = 1, 2, ..., p , (2) y1(1, λ) = y2(1, λ) = · · · = yp(1, λ) , (3) p X i=1 y0i(1, λ) = 0 , (4)
where αi ∈ [0, π) and qi are real functions in L2(0, 1), i = 1, 2, ..., p.
We shall reconstruct the potentials using only the nodal data and some constants, namely R1
0 qi (i = 1, . . . , n). Our reconstruction formula is direct and it automatically implies the
uniqueness of this inverse problem. Also we consider boundary conditions other than
Dirich-let ones, including Neumann boundary conditions. We remark that Currier and Watson [3]
also studied the inverse nodal problems on general graphs. They showed that, for qi ∈ L∞,
a set of eigenvalues and nodal positions to reconstruct the potentials qi’s. In this paper, we
shall use only the nodal positions for the reconstruction. The difficulty of the study is to
have a detailed asymptotic expansion of the eigenvalues and nodal points. The situation get
the method of Pivovarchik to deal with these difficulties. Owing to the completeness, we can
only work with p-star graph.
Let xn,ik denote the k-th nodal point of the eigenfunction on the ith edge of the graph, associated with the eigenvalue λn. Define the nodal length l
n,i k by
ln,ik = xn,ik+1− xn,ik .
Obviously, there is a nodal subset, also denoted by xn,ik , whose estimates of nodal length are expressed of order o(n12) (see section 3).
The main theorem is the following.
Theorem 1.1. Let {xn,ik } be a nodal subset of the system (1)-(4) whose nodal length is of the same asymptotic expansion of order o(n12). Then, for |C| 6= 0,
π 2 and π, (i) Either αi = 0, or cot αi = limn→∞nπ2[k −12 − − k−1 2 nπ C − nx n,i k ], if l n,i k = 1 n − C n2π + o(n12) , limn→∞nπ2[k −12 − nxn,ik ], if l n,i k = 1 n+ o( 1 n2) , limn→∞(n − 12)π2[k −12 − (n − 12)x n,i k ], if l n,i k = 1 n−12 + o( 1 n2) , limn→∞(n − 1)π2[k −12 − (n − 1)xn,ik ], if l n,i k = 1 n−1+ o( 1 n2) . (ii) Define Fin(x) = 2n2π2[nln,i j − 1 − Pn−1 k=1(l n,i k − 1 n+ C n2π)], if l n,i k = 1 n− C n2π + o( 1 n2) , 2n2π2[nljn,i− 1 −Pn−1 k=1(l n,i k − 1 n)], if l n,i k = 1 n + o( 1 n2) , 2(n −12)2π2[(n − 1 2)l n,i j − 1 − Pn−1 k=1(l n,i k − 1 n−12)], if l n,i k = 1 n−12 + o( 1 n2) , 2(n − 1)2π2[(n − 1)ln,i j − 1 − Pn−1 k=1(l n,i k − 1 n−1)], if l n,i k = 1 n−1 + o( 1 n2) . Then Fin converges to qi− R1 0 qi pointwisely and in L 1(0, 1)-norm for i = 1, 2, ..., p.
In section 2, we consider the direct problem and derive the eigenvalue asymptotics. The
idea of how to estimate eigenvalues comes from Pivovarchik [10, 11]. In Lemma 2.1, we prove
the counting lemma while the estimates of eigenvalues and the asymptotic expansions are
given in Theorem 2.2 and Theorem 2.3. In section 3, we deal with the inverse nodal problem.
We develop a procedure to reconstruct the potential functions using the nodal points and
obtain our main theorem.
2
Direct Problem
In this section, we deal with the direct problem. The idea of how to estimate eigenvalues
comes from Pivovarchik. One may see also [10, 11]. In Lemma 2.1, we prove the counting
lemma while the sharper estimates of eigenvalues are given in Theorem 2.2.
First, we shall show that the spectrum of the problem (1)-(4) are all real. Denote by K
the operator acting on the Hilbert space H, the direct sum of p copies of L2(0, 1), according
to KY = K y1(x) y2(x) .. . yp(x) ≡ −y00 1(x) + q1(x)y1(x) −y00 2(x) + q2(x)y2(x) .. . −y00 p(x) + qp(x)yp(x) , where y1(x) y2(x) .. . yp(x) ∈ D(K) ≡ y1(x) y2(x) .. . yp(x) : yi ∈ W22(0, 1) , i = 1, 2, ..., p yi(0) cos αi+ yi0(0) sin αi = 0 , i = 1, 2, ..., p , y1(1) = y2(1) = · · · = yp(1) , Pp i=1y 0 i(1) = 0 .
Then the spectrum of λI − K coincides with the spectrum of the problem (1)-(4). It is easy
For every fixed i = 1, 2, ..., p, let ui(x, λ) be the solution of the following initial value problem: −y00 i + qi(x)yi = λyi , yi(0) = sin αi , yi0(0) = − cos αi .
Then ui(x, λ) satisfies the following integral equation
ui(x, λ) = sin αicos( √ λx) − cos α√ i λ sin( √ λx) + Z x 0 sin(√λ(x − t)) √ λ qi(t)ui(t, λ)dt . (5) If yi(x, λ) is an eigenfunction of the system (1)-(4), then there is a nonzero real number
ci such that yi(x, λ) = ciui(x, λ). Moreover, λ is an eigenvalue of (1)-(4) if and only if λ is a
zero of Φ(λ) ≡ u1(1, λ) −u2(1, λ) 0 0 · · · 0 u1(1, λ) 0 −u3(1, λ) 0 · · · 0 · · · · u1(1, λ) 0 · · · −up(1, λ) u01(1, λ) u02(1, λ) · · · u0p(1, λ) , = p X i=1 u0i(1, λ)Y ν6=i uν(1, λ) . (6)
In the following, we denote A1 = Qpi=1sin αi, A2 = Ppi=1cot αi when αi 6= 0 for all
i = 1, 2, ..., p, and A3 =
Pp
j=T +1cot αj when αj 6= 0 for all j = T + 1, T + 2, ..., p where
1 ≤ T ≤ p − 1. Also denote Qji(t) =Pj
ν=iqν(t). Analyzing the function Φ(λ), we can obtain
the asymptotic expansion of the eigenvalue λ.
Lemma 2.1. I. When each αi = 0, there are p sequences of eigenvalues {λn,ν} (ν =
1, . . . , p), with asymptotics
pλn,1= (n −
1
II. When each αi 6= 0, there are p sequences of eigenvalues {λn,ν} (ν = 1, . . . , p), with
asymptotics
pλn,1 = (n − 1)π + o(1) , pλn,ν = (n −
1
2)π + o(1) , ν = 2, 3, ..., p.
III. When αi = 0, i = 1, 2, ..., T and αi 6= 0, i = T + 1, T + 2, ..., p where 1 ≤ T ≤ p − 1,
there are p sequences of eigenvalues {λn,ν} (ν = 1, . . . , p), with asymptotics
pλn,ν = nπ + (−1)νarcsin s T p + o(1) , ν = 1, 2 , pλn,ν = nπ + o(1) , ν = 3, 4, .., T + 1 , pλn,ν = (n − 1 2)π + o(1) , ν = T + 2, T + 3, ..., p.
Proof. Assume αi 6= 0 for all i = 1, 2, ..., p, and
√
λ ∈ C. According to (5) and (6), it is easy to show that
Φ(λ) = −√λpA1sin
√
λ cosp−1√λ + O(ep|Im
√ λ|) .
(7)
Denote by s2
n = λn the zeros of Φ(λ). If there is a subsequence {snk}
∞
k=1 such that
limk→∞Imsnk = ∞, then (7) implies
Φ(λnk) = −snkpA1sin snkcos p−1s nk+ O(e p|Imsnk|) , = −snkpA1 e−i(p−2)snk − e−ipsnk 2i 1 + e2isnk 2 p−1 + O(ep|Imsnk|) .
But this makes a contradiction to Φ(λnk) = 0. This means that Imsn is bounded above.
Similarly, it also can be shown that Imsn is bounded below. Hence, there exists M > 0 such
that |Imsn| < M .
To count the eigenvalues, we compare Φ(λ) with √λpA1sin
√
λ cosp−1√λ. By (7), when
λ is sufficiently large and |Im√λ| < M , there is a constant C > 0 such that Φ(λ) + √ λpA1sin √ λ cosp−1√λ < C .
On the other hand, the zeros of √s sin√s cosp−1√s, denoted by ρ2n,i, are
ρn,1 = (n − 1)π ,
ρn,ν = (n −
1
2)π , ν = 2, 3, ..., p .
For any > 0, let Dn,i be the disk of the radius at the centre ρn,i satisfying Dn,1∩ Dn,2= ∅
for all n. Note that Dn,2 = Dn,3 = · · · = Dn,p for each n. Then there is a constant d > 0
such that pA sin √ λ cosp−1√λ > d ,
for all λ ∈ {λ : |Im√λ| < M }\(∪n,iDn,i). Furthermore, for λ ∈ {λ : |Im
√
λ| < M, |√λ| >
C
d}\(∪n,iDn,i), we can obtain
Φ(λ) + √ λpA1sin √ λ cosp−1√λ < C < | √ λ|d < √ λpA1sin √ λ cosp−1√λ .
Since can be arbitrary small, we can obtain II according to Rouche’s Theorem.
The part I for αi = 0 for all i = 1, 2, ..., p is similar to II. For III, (5) can be rewritten as
Φ(λ) = B (−√λ)T −1 sin T −1√λ cosp−T −1√λp sin2√λ − T+ O(e p|Im√λ| |√λ|T ) , (8) where B = Qp
i=T +1sin αi. The zeros of p sin2
√
s − T sinT −1√s cosp−T −1√s, denoted by
κ2n,ν, are κn,ν = nπ + (−1)νarcsin s T p , ν = 1, 2 , κn,ν = nπ , ν = 3, 4, .., T + 1 , κn,ν = (n − 1 2)π , ν = T + 2, T + 3, ..., p.
Theorem 2.2. I. When each αi = 0, there are p sequences of eigenvalues {λn,ν} (ν = 1, . . . , p), with asymptotics pλn,1 = (n − 1 2)π + 1 2p(n − 12)π Z 1 0 (1 − cos((2n − 1)πt))Qp1(t)dt + O( 1 n2) , pλn,ν = nπ + Λn,ν nπ + O( 1 n2) , ν = 2, 3, ..., p.
where Λn,ν is the root of the polynomial equation of degree (p − 1) p X i=1 " Y ν6=i (Λ − 1 2 Z 1 0 (1 − cos(2(n − 1)πt))qν(t) dt) # = 0 , (9)
II. When each αi 6= 0, there are also p sequences of eigenvalues {λn,ν} (ν = 1, . . . , p), with
asymptotics pλn,1 = (n − 1)π + 1 (n − 1)pπ(−A2+ 1 2 Z 1 0 (1 + cos(2(n − 1)t))Qp1(t)dt) + O( 1 n2) , pλn,ν = (n − 1 2)π − Λn,ν (n − 12)π + O( 1 n2) , ν = 2, 3, ..., p.
where Λn,ν is the root of the polynomial equation of degree (p − 1) p X i=1 " Y ν6=i (Λ − cot αν + 1 2 Z 1 0 (1 + cos((2n − 1)πt))qν(t) dt) # = 0 , (10)
III. When αi = 0, i = 1, 2, ..., T and αi 6= 0, i = T + 1, T + 2, ..., p where 1 ≤ T ≤ p − 1,
there are p sequences of eigenvalues {λn,ν} (ν = 1, . . . , p), with asymptotics
pλn,ν = nπ + (−1)νarcsin s T p + δn+ o( 1 n) , ν = 1, 2 , pλn,ν = nπ + Λ0n,ν nπ + O( 1 n2) , ν = 3, 4, ..., T + 1, pλn,ν = (n − 1 2)π − Λ1 n,ν (n − 12)π + O( 1 n2) , ν = T + 2, T + 3, ..., p,
where δn = − 1 4pnπ 2T A3− (p − T ) Z 1 0 QT1(t)dt − T Z 1 0 QpT +1(t)dt + o(1 n) ,
Λ0n,ν is the root of the polynomial equation of degree (T − 1)
T X i=1 " T Y ν6=i,ν=1 (Λ − 1 2 Z 1 0 (1 − cos(2nπt))qν(t) # = 0 , (11) and Λ1
n,ν is the root of the polynomial equation of degree (p − T − 1) p X i=T +1 " p Y ν6=i,ν=T +1 (Λ − cot αν + 1 2 Z 1 0 (1 + cos((2n − 1)πt))qν(t) dt) # = 0 . (12)
Proof. The proofs of I, II and III are similar. We shall prove III here. For αi = 0, i =
1, 2, ..., T and αi 6= 0, i = T + 1, T + 2, ..., p, let λ be an eigenvalue and sufficiently large.
According to Lemma 2.1, we can sort the eigenvalues by the their estimates. For √λn = 1
n+ o(1) and
√
λn = n−11 2
+ o(1), the argument are similar and then we skip the second one
here.
(i) √λn = nπ + o(1). In this case, (6) is equal to T X i=1 − cos√λ + O(√1 λ) T Y ν6=i,ν=1 G2ν(λ) p Y ν=T +1 G1ν(λ) + p X i=T +1 −√λ sin αisin √ λ + O(1) T Y ν=1 G2ν(λ) p Y ν6=i,ν=T +1 G1ν(λ) = 0 , where G1ν(λ) = sin ανcos √ λ + O(√1 λ) , G2ν(λ) = −sin √ λ √ λ + cos√λ 2λ Z 1 0 (1 − cos(2√λt))qν(t)dt −sin √ λ 2λ Z 1 sin(2 √ λt)qν(t)dt + O( 1 λ3/2) .
This implies T X i=1 T Y ν6=i,ν=1 √ λ tan√λ − 1 2 Z 1 0 (1 − cos(2√λt))qν(t)dt −tan √ λ 2 Z 1 0 sin(2√λt)qν(t)dt # + O(√1 λ) = 0 .
Observing above equation, we can obtain√λ tan√λ = O(1) and hence
T X i=1 T Y ν6=i,ν=1 √ λ tan √ λ − 1 2 Z 1 0 (1 − cos(2 √ λt))qν(t)dt + O(√1 λ) = 0 .
Denote by Λn,ν, ν = 3, 4, ..., T + 1, the root of the (T − 1) degree polynomial equation
of T X i=1 T Y ν6=i,ν=1 Λ − 1 2 Z 1 0 (1 − cos(2nπt))qν(t)dt = 0 . Then pλn,νtanpλn,ν = Λn,ν + O(√1 λn,ν
). Combining with Lemma 2.1, we have
pλn,ν = nπ + Λn,ν pλn,ν + O( 1 n2) , ν = 3, 4, ..., T − 1 . (ii) √λn = nπ + (−1)νarcsin q T
p + o(1). In this case, (6) implies T X i=1 − cos√λ − cos √ λ 2√λ R1 0 sin(2 √ λt)qidt − sin √ λ 2√λ R1 0(1 − cos(2 √ λt))qi(t)dt +O(1 λ) T Y ν=1,ν6=i −sin√√λ λ G3 ν(λ) p Y ν=T +1 sin ανcos √ λ G4 ν(λ) + p X i=T +1 −√λ sin αisin √ λ − cos αicos √ λ +sin αi 2 cos √ λR01(1 + cos(2√λt))qi(t)dt +sin αi 2 sin √ λR1 0 sin(2 √ λt)qi(t)dt + O(√1λ) YT ν=1 −sin√√λ λ G3 ν(λ) p Y ν=T +1,ν6=i sin ανcos √ λ G4ν(λ) = 0 ,
where G3ν(λ) = 1 − cot √ λ 2√λ Z 1 0 (1 − cos(2√λt))qν(t)dt + 1 2√λ Z 1 0 sin(2√λt)qν(t)dt + O( 1 λ) , G4ν(λ) = 1 − cot α√ ν λ tan √ λ +tan √ λ 2√λ Z 1 0 (1 + cos(2√λt))qν(t)dt − 1 2√λ Z 1 0 sin(2√λt)qν(t)dt + O( 1 λ) . This implies T X i=1 p X j=T +1 Hi,j(λ) T Y ν=1,ν6=i G3ν(λ) p Y ν=T +1,ν6=j G4ν(λ) = 0 , where Hi,j(λ) =p sin2√λ − T 1 + 1 2√λ R1 0 sin(2 √ λt)qi(t)dt − 2√1λ R1 0 sin(2 √ λt)qj(t)dt +p sin √ λ cos√λ 2√λ 2 cot αj− R1 0(1 − cos(2 √ λt))qi(t)dt − R1 0(1 + cos(2 √ λt))qj(t)dt , or equivalently T X i=1 p X j=T +1 Hi,j(λ) G3 i(λ)G4j(λ) = 0 . Furthermore p sin2√λ − T = −sin √ λ cos√λ 2√λ 2T A3− (p − T ) Z 1 0 (1 − cos(2 √ λt))QT1(t)dt −T Z 1 0 (1 + cos(2√λt))QpT +1(t)dt + O(1 λ) .
Now, let √λn = nπ + arcsin
q
T
p + δn where δn= o(1). Then
p sin2pλn− T = 2pT (p − T ) δn+ o(δn) , = −sin √ λncos √ λn 2√λn 2T A3− (p − T ) Z 1 0 QT1(t)dt − T Z 1 0 QpT +1(t)dt + o(1 n) , = −pT (p − T ) 2pnπ 2T A3− (p − T ) Z 1 0 QT1(t)dt − T Z 1 0 QpT +1(t)dt + o(1 n) , This implies δn= − 1 4pnπ 2T A3− (p − T ) Z 1 0 QT1(t)dt − T Z 1 0 QpT +1(t)dt + o(1 n) .
On each i-th edge of the graph, the problem can be reduced to a scalar Sturm-Liouville
system. Hence we can refer to the papers [1, 8] to obtain the asymptotic expansion of the
nodal points, with sn,ν =pλn,ν,:
I. If αi = 0, then as n → ∞, xn,ik,ν = kπ sn,ν + 1 2s2 n,ν Z xn,ik,ν 0 (1 − cos(2sn,νt))qi(t)dt + o( 1 s3 n,ν ) . II. If αi 6= 0, then as n → ∞, xn,ik,ν = (k − 1 2)π sn,ν − 1 s2 n,ν cot αi+ 1 2s2 n,ν Z xn,ik,ν 0 (1 + cos(2sn,νt))qi(t)dt + o( 1 s3 n,ν ) .
and hence, in both case, we have
lk,νn,i = π sn,ν + 1 2s2 n,ν Z xn,ik+1,ν xn,ik,ν (1 + γicos(2sn,νt))qi(t)dt + o( 1 s3 n,ν ) , where γi = 1 if α
i > 0, γi = −1 if αi = 0. Combining the asymptotic expansion of eigenvalues
Theorem 2.3. The asymptotic expansions of the nodal points are given by
I. Suppose that αi = 0 for all i = 1, 2, ..., p. Then
xn,ik,1 = k n − 12 + 1 2(n −12)2π2 Z xn,ik,1 0 (1 − cos(2(n −1 2)πt))qi(t)dt −k R1 0 Pp i=1qi(t)dt 2p(n − 12)3π2 + o( 1 n3) , xn,ik,ν = k n + 1 2n2π2 Z xn,ik,ν 0 (1 − cos(2nπt))qi(t)dt − kΛn,ν n3π2 + o( 1 n3) ,
where ν = 2, 3, ..., p, and Λn,ν is given by (9).
II. Suppose that αi 6= 0 for all i = 1, 2, ..., p. Then
xn,ik,1 = k − 1 2 n − 1 − 1 (n − 1)2π2 cot αi− 1 2 Z xn,ik,1 0 (1 + cos(2(n − 1)πt))qi(t)dt ! + (k − 1 2) p(n − 1)3π2 p X i=1 cot αi− 1 2 Z 1 0 qi(t)dt + o( 1 n3) , xn,ik,ν = k − 1 2 n − 12 − 1 (n − 12)2π2 cot αi− 1 2 Z xn,ik,ν 0 (1 + cos(2(n − 1 2)πt))qi(t)dt ! +(k − 1 2)Λn,ν (n − 12)3π2 + o( 1 n3) ,
where ν = 2, 3, ..., p and Λn,ν is given by (10).
(I) for i = 1, 2, ..., T , we have xn,ik,ν = k n − k n2π (−1) ν arcsin s T p + δn ! + 1 2n2π2 Z xn,ik,ν 0 (1 − cos(2snt))qi(t)dt − (−1)νarcsinqT p n3π3 Z xn,ik,ν 0 (1 − cos(2snt))qi(t)dt + o( 1 n3) , ν = 1, 2 , xn,ik,ν = k n + 1 2n2π2 Z xn,ik,ν 0 (1 − cos(2nπt))qi(t)dt − kΛ0 n,ν n3π2 + o( 1 n3) , ν = 3, 4, ..., T + 1 , xn,ik,ν = k n − 12 + 1 2(n −12)2π2 Z xn,ik,ν 0 (1 − cos(2(n − 1 2)πt))qi(t)dt + kΛ 1 n,ν (n − 12)3π2 + o( 1 n3) , ν = T + 2, T + 3, ..., p .
(II) for i = T + 1, T + 2, ..., p, we have
xn,ik,ν = k − 1 2 n − k −12 n2π (−1) νarcsin s T p + δn ! − 1 n2π2 cot αi− 1 2 Z xn,ik,ν 0 (1 + cos(2snt))qi(t)dt ! + 2(−1)νarcsinqTp n3π3 cot αi− 1 2 Z xn,ik,ν 0 (1 + cos(2snt))qi(t)dt ! +o( 1 n3) , ν = 1, 2 , xn,ik,ν = k − 1 2 n − 1 n2π2 cot αi− 1 2 Z xn,ik,ν 0 (1 + cos(2nπt))qi(t)dt ! −(k − 1 2)Λ 0 n,ν n3π2 + o( 1 n3) , ν = 3, 4, ..., T + 1 , xn,ik,ν = k − 1 2 n − 12 − 1 (n − 12)2π2 cot αi− 1 2 Z xn,ik,ν 0 (1 + cos(2(n −1 2)πt))qi(t)dt ! +(k − 1 2)Λ 1 n,ν (n − 12)3π2 + o( 1 n3) , ν = T + 2, T + 3, ..., p . In both cases, Λ0
3
Inverse Problem
The following lemma can be referred to [8] and hence we skip the proof here.
Lemma 3.1. Let {x(n)k } be the nodal set corresponding to the eigenvalue s2
n= λn. Then Z x(n)k+1 x(n)k cos(2snt)q(t)dt = o( 1 n) , Z x(n)k+1 x(n)k q(t)dt = O(1 n) .
By Theorem 2.3 and above lemma, we can obtain the asymptotic expansion of nodal
lengths as follows: I. For αi = 0, i = 1, 2, ..., p, ln,ik,1 = 1 n − 12 + 1 2(n − 12)2π2 Z xn,ik+1,1 xn,ik,1 qi(t)dt − R1 0 Q p 1(t)dt 2p(n − 12)3π2 + o( 1 n3) , lk,νn,i = 1 n + 1 2n2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt − Λn,ν n3π2 + o( 1 n3) , where ν = 2, 3, ..., p. II. For αi 6= 0, i = 1, 2, ..., p, ln,ik,1 = 1 n − 1+ 1 2(n − 1)2π2 Z xn,ik+1,1 xn,ik,1 qi(t)dt + 2A2− R1 0 Q p 1(t)dt 2p(n − 1)3π2 + o( 1 n3) , ln,ik,ν = 1 n − 12 + 1 2(n − 12)2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt + Λn,ν (n − 12)3π2 + o( 1 n3) , where ν = 2, 3, ..., p.
(I) for i = 1, 2, ..., T , we have ln,ik,ν = 1 n − 1 n2π (−1) νarcsin s T p + δn ! + 1 2n2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt + o( 1 n3) , ln,ik,ν = 1 n + 1 2n2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt − Λ0 n,ν n3π2 + o( 1 n3) , ln,ik,ν = 1 n − 12 + 1 2(n − 12)2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt + Λ1n,ν (n − 12)3π2 + o( 1 n3)
(II) for i = T + 1, T + 2, ..., p, we have
ln,ik,ν = 1 n − 1 n2π (−1) νarcsin s T p + δn ! + 1 2n2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt + o( 1 n3) , ln,ik,ν = 1 n + 1 2n2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt − Λ0n,ν n3π2 + o( 1 n3) , ln,ik,ν = 1 n − 12 + 1 2(n − 12)2π2 Z xn,ik+1,ν xn,ik,ν qi(t)dt + Λ1 n,ν (n − 12)3π2 + o( 1 n3)
Now given a nodal subset, also denoted by {xn,ik }, of i-th branch of the p-star graph whose nodal length is of the same asymptotic expansion of order o(n12), we shall first
distin-guish {ln,ik } into which one of asymptotic expansion given as above, and then build up the reconstruction formulas. We derive the following procedure:
1. If {lkn,i} = 1 n− C n2π+o( 1 n2) for some |C| 6= 0, π
2, π, go to procedure 2, else go to procedure
3. 2. Define Θni = −n2π2 xn,ik −k − 1 2 n + k − 12 n2π C , Fin = 2n2π2 nln,ik − 1 + C nπ − 2n2π2 n−1 X k=1 ln,ik − 1 n + C n2π .
3. If ln,ik = 1n+ o(n12), define Θni = −n2π2 xn,ik −k − 1 2 n , Fin = 2n2π2 nln,ik − 1 − 2n2π2 n−1 X k=1 lkn,i− 1 n , else if lkn,i= 1 n−12 + o( 1 n2), define Θni = −(n − 1 2) 2π2 xn,ik − k − 1 2 n − 12 , Fin = 2(n − 1 2) 2π2 (n − 1 2)l n,i k − 1 − 2(n − 1 2) 2π2 n−1 X k=1 lkn,i− 1 n − 12 ,
else if lkn,i= n−11 + o(n12), define
Θni = −(n − 1)2π2 xn,ik − k − 1 2 n − 1 , Fin = 2(n − 1)2π2 (n − 1)lkn,i− 1 − 2(n − 1)2π2 n−1 X k=1 ln,ik − 1 n − 1 . Since 1 n − 1 = 1 n − −π n2π + 1 n2(n − 1) , 1 n − 12 = 1 n − −1 2π n2π + 1 4 n2(n − 1 2) ,
and 1 ≤ T ≤ p − 1, we can distinguish ln,ik into n1 − C n2π + o(
1
n2), where |C| 6= 0,
π 2, π, or
not. Hence, procedure 1 makes sense. In procedure 2 and 3, we cancel the noise from nodal
data first and reconstruct q −R01q using corrected nodal data. Since each nodal data only determine one reconstruction formula and the reconstruction formula only depends on nodal
data, the uniqueness holds obviously.
According to above procedure and the result of [1, 8], we have either αi = 0 or
cot αi = lim n→∞Θ
n i ,
and Fin converges to qi −
R1
0 q(t)dt pointwisely and in L
1. The proof of Theorem 1.1 is
complete.
ACKNOWLEDGEMENT
The author would like to thank Prof. Chun-Kong Law for his valuable suggestions and
the helpful comments.
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Y. H. Cheng
Mathematics Division,
National Center for Theoretical Sciences,
HsinChu, Taiwan, R.O.C.
