Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 155, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
A UNIQUENESS THEOREM ON THE INVERSE PROBLEM FOR THE DIRAC OPERATOR
YU PING WANG, MURAT SAT
Abstract. In this article, we consider an inverse problem for the Dirac op- erator. We show that a particular set of eigenvalues is sufficient to determine the unknown potential functions.
1. Introduction
The inverse spectral problem for a differential operator consists of recovering the operator from its spectral data. In 1929, Ambarzumyan [2] was the first to discuss the following statement
If q ∈ C[0, π] and {n2 : n = 0,1,2, . . .} is the spectral set of the boundary value problem
−y00+q(x)y=λy, x∈[0, π], (1.1) with Neumann boundary conditions
y0(0, λ) =y0(π, λ) = 0, (1.2) thenq(x)≡0 in [0, π].
McLaughlin and Rundell [17] discussed the inverse problem for the Sturm-Liouville equation (1.1) with the boundary conditionsy(0, λ) = 0 andy0(π, λ)+Hky(π, λ) = 0 and showed that the spectral data, for a fixedn(n= 0,1,2, . . .),{λn(q, Hk)}+∞k=1is equivalent to two spectra of boundary value problems with the equation (1.1) and one common boundary condition at x= 0 and two different boundary conditions at x = π. By using McLaughlin and Rundell’s method [17], Koyunbakan [13]
considered a singular Sturm-Liouville problem. Using the spectral data in [17] and Hochstadt and Lieberman’s method, Wang [28] discussed the inverse problem for indefinite Sturm-Liouville operators on the finite interval [a, b]. However, we are motivated by inverse spectral problems for Dirac operators with the above spectral data which are particular set of eigenvalues. As far as we know, inverse spectral problems for Dirac operators have not been considered with the spectral data before.
We consider the system of Dirac operatorsL:=L(p, q, Hk)
Ly =By0+Q(x)y=λy, 0≤x≤π, (1.3)
2010Mathematics Subject Classification. 34A55, 34B24, 34L05, 45C05.
Key words and phrases. Inverse problem; uniqueness theorem; eigenvalue; Dirac operator.
c
2016 Texas State University.
Submitted March 12, 2016. Published June 21, 2016.
1
with the boundary conditions
y1(0, λ) = 0, (1.4)
y2(π, λ) +Hky1(π, λ) = 0, (1.5) where
B=
0 1
−1 0
, Q(x) =
p(x) q(x) q(x) −p(x)
, y(x) = y1(x)
y2(x)
and (p(x), q(x)) are potential functions which are real valued functions in space L2[0, π].
We consider another Dirac operatorLe:=L(e p,eq, He k) which is defined as:
Lye =Bye0+Q(x)e ey=λy,e 0≤x≤π, (1.6) with the boundary conditions
ye1(0, λ) = 0, (1.7)
ye2(π, λ) +Hkye1(π, λ) = 0, (1.8) where
Q(x) =e
p(x)e q(x)e q(x)e −p(x)e
,
Hk ∈R, 0 < H1 < H2 <· · · < Hk < Hk+1 < . . ., the potentials (p(x),e q(x)) aree real valued functions, (p(x),e eq(x))∈L2[0, π] and λis a spectral parameter.
The Dirac equation is a modern presentation of the relativistic quantum me- chanics of electrons intended new mathematical outcomes accessible to a wider audience. It treats in some dept the relativistic invariance of a quantum theory, self-adjointness and spectral theory, qualitative features of relativistic bound and scattering states and the external field problem in quantum electrodynamics, with- out neglecting the interpretational difficulties and limitations of the theory.
Inverse problems for Dirac system were studied by Moses [18], Prats and Toll [24], Verde [27], Gasymov and Levitan [7] and Panakhov [22, 23]. It is well known [8] that two spectra uniquely determine the matrix-valued potential function. In particular, in reference [11], eigenfunction expansions for one dimensional Dirac operators describing the motion of a particle in quantum mechanics are discussed.
Direct or inverse spectral problem for Dirac and Sturm-Liouville operators were extensively studied in [1, 3, 4, 5, 9, 12, 15, 18, 20, 21, 25, 26, 28]. However, the results on direct or inverse spectral problems of the Dirac operator are less than classical Sturm-Liouville operator, this leads to additional difficulties in connection with inverse spectral problem by the spectral data in [17].
In this article, by using the spectral data in [17] and Hochstadt and Lieberman’s [10] method a uniqueness theorem for Dirac operator on the interval [0, π] will be established, i.e., for a fixed index n (n = 0,±1,±2, . . .), we show that if the spectral set {λn(p, q, Hk)}+∞k=1 for distinctHk can be measured, then the spectral set{λn(p, q, Hk)}+∞k=1 is sufficient to determine the potential functions (p(x), q(x)).
Lemma 1.1([14]). Let the functionϕ(x, λ) =
ϕ1(x, λ) ϕ2(x, λ)
be the solution of (1.3) satisfying the initial condition
ϕ(0, λ) =
ϕ1(0, λ) ϕ2(0, λ)
= 0
−1
= (0,−1)T.
Thenϕ(x, λ) satisfies the integral equations ϕ(x, λ) =
sinλx
−cosλx
+ Z x
0
K(x, t)
sinλt
−cosλt
dt, (1.9)
where kernel K(x, t) is symmetric matrix-valued functions whose entries are con- tinuously differentiable in both of its variables.
Lemma 1.2. The eigenvaluesλn (n6= 0)of the boundary-value problem(1.3)-(1.5) for the coefficientH =Hk in (1.5)are the roots of (1.5)and satisfy the asymptotic formulae:
λn=λ0n+n, (1.10)
where {n} ∈ l2 (l2 consist of sequences {xn} such that P∞
n=1|xn|2 <∞) andλ0n are the zeros of∆0(λ) :=−cosλπ+Hsinλπ, i.e.,
λ0n =n+1
πarctan 1 H. Proof. Letϕ(x, λ) =
ϕ1(x, λ) ϕ2(x, λ)
be the solution of (1.3) satisfying the initial con- dition
ϕ(0, λ) =
ϕ1(0, λ) ϕ2(0, λ)
= 0
−1
,
and ϕ1(0, λ) = 0, ϕ2(π, λ) +Hϕ1(π, λ) = 0. The characteristic function ∆(λ) of the problemLis defined by the following relation:
∆(λ) =ϕ2(π, λ) +Hϕ1(π, λ),
and the zeros of ∆(λ) coincide with the eigenvalues of the problemL.
Using Lemma 1.1, we obtain
∆(λ) =−cosλπ+Hsinλπ+ Z π
0
(K21(π, t) +HK11(π, t)) sinλtdt
− Z π
0
(K22(π, t) +HK12(π, t)) cosλtdt.
Since the eigenvalues are zeros of ∆(λ), we can write the equation
−cosλπ+Hsinλπ+ Z π
0
(K21(π, t) +HK11(π, t)) sinλt dt
− Z π
0
(K22(π, t) +HK12(π, t)) cosλtdt= 0.
Denote
Gn={λ∈C:|λ|=|λ0n|+β, n= 0,±1,±2, . . .}, Gδ={λ:|λ−λ0n| ≥δ, n= 0,±1,±2, . . .}, whereδis sufficiently small number (δβ).
Since|∆0(λ)|> Cδexp(|τ|π) forλ∈Gδ, from [16], lim
|λ|→∞e−|τ|π(∆(λ)−∆0(λ))
= lim
|λ|→∞
e−|τ|π
Z π
0
(K21(π, t) +HK11(π, t)) sinλt dt
−e−|τ|π Z π
0
(K22(π, t) +HK12(π, t)) cosλt dt
= 0
and|∆(λ)−∆0(λ)|< Cδexp(|τ|π) for sufficiently largenandλ∈Gn, we have
|∆0(λ)|>|∆(λ)−∆0(λ)|, whereτ = Imλ.
Using the Rouch´e’s theorem, we conclude that, for sufficiently largen, the func- tions ∆0(λ) and ∆0(λ) +{∆(λ)−∆0(λ)}= ∆(λ) have the same number of zeros inside the contourGn, namely 2n+ 1 zerosλ−n, . . . , λ0, . . . , λn. Thus the eigenval- uesλn are of the formλn =λ0n+n, where
n→∞lim n= 0.
Substitutingλ0n+nforλnin the last equality and using the fact that ∆0(λ0n+n) =
∆00(λ0n)[1 +o(1)]n. We conclude that n∈l2. The proof is complete.
2. Main results and proofs
Lemma 2.1. Letσ(Lkj) :={λn(p, q, Hkj)}(j = 1,2)be the spectrum of the bound- ary value problem (1.3)-(1.5)for the coefficient Hkj. IfHk1 6=Hk2, then
σ(Lk1)∩σ(Lk2) =∅, (2.1)
wherekj∈N,∅ denotes an empty set.
Lemma 2.2. Letλn(p, q, Hk)be then-th eigenvalue of the boundary-value problem (1.3)-(1.5). Then the spectral set {λn(p, q, Hk)}+∞k=1 is a bounded infinite set.
The above lemema plays an important role in the proof of next theorem.
Theorem 2.3. Letλn(p, q, Hk)be then-th eigenvalue of the boundary-value prob- lem (1.3)-(1.5) andλn(p,eq, He k)be then-th eigenvalue of the boundary-value prob- lem (1.6)-(1.8), for a fixed indexn(n∈Z). If
λn(p, q, Hk) =λn(ep,eq, Hk) for allk∈N, then
(p(x), q(x)) = (p(x),e eq(x)) a.e. on[0, π].
Proof of Lemma 2.1. Suppose that the conclusion is false. Denote λnj(Hkj) = λnj(p, q, Hkj), j = 1,2. Then there exists λn1(Hk1) = λn2(Hk2) ∈ R, where λnj(Hkj) ∈ σ(Lkj) nj ∈ Z. Let ϕj(x, λnj(Hkj)) be the solution of (1.3)-(1.5) corresponding to the eigenvalueλnj(Hkj) and that it satisfies the initial conditions ϕj,1(0, λnj(Hkj)) = 0 whereϕj = (ϕj,1, ϕj,2)T. We get
Bϕ01(x, λn1(Hk1)) +Q(x)ϕ1(x, λn1(Hk1)) =λn1(Hk1)ϕ1(x, λn1(Hk1)), (2.2) and
Bϕ02(x, λn2(Hk2)) +Q(x)ϕ2(x, λn2(Hk2)) =λn2(Hk2)ϕ2(x, λn2(Hk2)). (2.3) If we multiply (2.2) by ϕ2(x, λn2(Hk2)), and (2.3) byϕ1(x, λn1(Hk1)) respectively (in the sense of scalar product i.e.
h(ϕ1,1, ϕ1,2)T,(ϕ2,1, ϕ2,2)Ti=ϕ1,1ϕ2,1+ϕ1,2ϕ2,2
and subtract from each other and integrate from 0 toπ, we obtain
ϕ2,2(x, λn2(Hk2))ϕ1,1(x, λn1(Hk1))−ϕ2,1(x, λn2(Hk2))ϕ1,2(x, λn1(Hk1))|πx=0= 0.
(2.4)
Using the initial conditions, we obtain
ϕ2,2(π, λn2(Hk2))ϕ1,1(π, λn1(Hk1))−ϕ2,1(π, λn2(Hk2))ϕ1,2(π, λn1(Hk1)) = 0.
(2.5) On the other hand, note the equality
ϕ2,2(π, λn2(Hk2))ϕ1,1(π, λn1(Hk1))−ϕ2,1(π, λn2(Hk2))ϕ1,2(π, λn1(Hk1))
=ϕ1,1(π, λn1(Hk1))[ϕ2,2(π, λn2(Hk2)) +Hk2ϕ2,1(π, λn2(Hk2))]
−ϕ2,1(π, λn2(Hk2))[ϕ1,2(π, λn1(Hk1)) +Hk1ϕ1,1(π, λn1(Hk1))]
+ (Hk1−Hk2)ϕ1,1(π, λn1(Hk1))ϕ2,1(π, λn2(Hk2))
= (Hk1−Hk2)ϕ1,1(π, λn1(Hk1))ϕ2,1(π, λn2(Hk2)).
(2.6)
SinceHk1−Hk2 6= 0, ifϕ1,1(π, λn1(Hk1))ϕ2,1(π, λn2(Hk2)) = 0, then
ϕ1,1(π, λn1(Hk1)) = 0 orϕ2,1(π, λn2(Hk2)) = 0. (2.7) This and (1.5) yield
ϕ1,1(π, λn1(Hk1)) =ϕ1,2(π, λn1(Hk1)) = 0, (2.8) or
ϕ2,1(π, λn2(Hk2)) =ϕ2,2(π, λn2(Hk2)) = 0. (2.9) Then (2.8) and (2.9) yield
ϕ1(x, λn1(Hk1))≡0 or ϕ2(x, λn2(Hk2))≡0 on [0, π], (2.10) where ϕ1 = (ϕ1,1, ϕ1,2)T and ϕ2 = (ϕ2,1, ϕ2,2)T. This is impossible. Thus, we obtain
ϕ2,2(π, λn2(Hk2))ϕ1,1(π, λn1(Hk1))−ϕ2,1(π, λn2(Hk2))ϕ1,2(π, λn1(Hk1))6= 0.
(2.11) It is obvious that the contradiction between (2.5) and (2.11) implies that (2.1)
holds. Hence the proof is complete.
Proof of Lemma 2.2. We prove the lemma by two steps. For the problemL1 :=
L1(q), µn is the n-th eiegenvalue with boundary conditions ϕ1(0) = ϕ1(π) = 0, for the problem L2:=L2(q, h),λn is the n-th eigenvalue problem (1.3)-(1.5) with H =Hk.
Step 1. We show that (see [6])
µn< λn≤µn+1. (2.12)
From the Green identity, we have
[ϕ2(x, λ)ϕ1(x, µ)−ϕ1(x, λ)ϕ2(x, µ)]
x=π x=0
= (µ−λ) Z π
0
[ϕ1(x, µ)ϕ1(x, λ) +ϕ2(x, µ)ϕ2(x, λ)]dx.
Hence, we have (µ−λ)
Z π
0
[ϕ1(x, µ)ϕ1(x, λ) +ϕ2(x, µ)ϕ2(x, λ)]dx
= [ϕ2(π, λ)ϕ1(π, µ)−ϕ1(π, λ)ϕ2(π, µ)] =d(µ)∆(λ)−d(λ)∆(µ), whered(µ) =ϕ1(π, µ), ∆(λ) =ϕ2(π, λ) +Hϕ1(π, λ).
Whenλ→µ, we obtain Z π
0
[ϕ21(x, µ) +ϕ22(x, µ)]dx=d·(µ)∆(µ)−d(µ)∆·(µ), with ∆·(µ) =dµd∆(µ) andd·(µ) = dµdd(µ). In particular, this yields
αn=−∆·(µn)d(µn), 1
d2(µ) Z π
0
[ϕ21(x, µ) +ϕ22(x, µ)]dx=− d dµ(∆(µ)
d(µ)), for−∞< µ <∞andd(µ)6= 0, whereαn are norming constants.
Thus the function ∆(µ)d(µ) is monotonically decreasing onR− {λn:n∈Z}with
µ→λlimn
∆(µ)
d(µ) =±∞.
Consequently from the asymptotic behavior ofλn andµn, we prove (2.12).
Step 2. We show that the following formula holds,
λn(H0)<· · ·< λn(Hk+1)< λn(Hk)< . . . . (2.13) Let ϕ(x, λn(H)) be the solution of the boundary value problem (1.3)-(1.5) cor- responding to the eigenvalue λn(H) and that it satisfies the initial conditions ϕ1(0, λn(H)) = 0,ϕ2(0, λn(H)) =−1 andϕ1(0, λn(H + ∆H)) = 0,ϕ2(0, λn(H+
∆H)) =−1. We have
Bϕ0(x, λn(H)) +Q(x)ϕ(x, λn(H)) =λn(H)ϕ(x, λn(H)), (2.14) Bϕ0(x, λn(H+ ∆H)) +Q(x)ϕ(x, λn(H+ ∆H))
=λn(H+ ∆H)ϕ(x, λn(H+ ∆H)), (2.15) where ∆H is the increment of H. Multiplying (2.14) by ϕ(x, λn(H + ∆H)), and multiplying (2.15) byϕ(x, λn(H)) and subtracting from each other and integrating from 0 toπ, we obtain, from the initial conditions at zero,
∆λn(H) Z π
0
hϕ1(x, λn(H))ϕ1(x, λn(H+ ∆H)) +ϕ2(x, λn(H))ϕ2(x, λn(H+ ∆H))i
dx
= ∆Hϕ1(π, λn(H))ϕ1(π, λn(H+ ∆H)),
(2.16)
where ∆λn(H) =λn(H+ ∆H)−λn(H).
It is well known thatϕ(x, λn(H)) andλn(H) are real and continuous with respect toH. Letting ∆H→0, we have
∂λn(H)
∂H = ϕ21(π, λn(H)) Rπ
0[ϕ21(x, λn(H)) +ϕ22(x, λn(H))]dx >0. (2.17) This implies that (2.13) holds. Therefore, from Step 1 and Step 2 we have that the spectral set{λn(p, q, Hk)}∞k=1 is a bounded infinite set. The proof is complete Finally, using Lemma 2.2, the properties of entire functions and the result of [29], we have shown that Theorem 2.3 holds.
Proof of Theorem 2.3. By Lemma 1.1 the solutions to Equation (1.3) satisfying ϕ(0, λ) = (0,−1)T, and solutions of (1.6) satisfying ϕ(0, λ) = (0,e −1)T can be respectively expressed in the integral forms:
ϕ(x, λ) =
sinλx
−cosλx
+ Z x
0
K(x, t)
sinλt
−cosλt
dt, (2.18)
ϕ(x, λ) =e
sinλx
−cosλx
+ Z x
0
K(x, t)e
sinλt
−cosλt
dt, (2.19)
where kernels K(x, t) and K(x, t) are symmetric matrix-valued functions whosee entries are continuously differentiable in both of its variables.
If we multiply (1.3) byϕ(x, λ) and (1.6) bye ϕ(x, λ) respectively (in the sense of scalar product inR2) and subtract from each other, then we obtain
d
dx{ϕ1(x, λ)ϕe2(x, λ)−ϕe1(x, λ)ϕ2(x, λ)}=h[Q(x)−Q(x)]ϕ(x, λ),e ϕ(x, λ)i.e (2.20) Integrating the last equality from 0 toπwith respect to the variable x, we give
{ϕ1(x, λ)ϕe2(x, λ)−ϕe1(x, λ)ϕ2(x, λ)}
π x=0=
Z π
0
h[Q(x)−Q(x)]ϕ(x, λ),e ϕ(x, λ)idx.e (2.21) Becauseϕ(x, λ) andϕ(x, λ) satisfy the same initial conditions, it follows thate
ϕ1(0, λ)ϕe2(0, λ)−ϕe1(0, λ)ϕ2(0, λ) = 0. (2.22) Define
P(x) =Q(x)−Q(x),e p1(x) =p(x)−p(x),e q1(x) =q(x)−q(x),e (2.23) and
H(λ) :=
Z π
0
hP(x)ϕ(x, λ),ϕ(x, λ)idx.e (2.24) Considering the properties ofϕ(x, λ) andϕ(x, λ), we conclude thate H(λ) is an entire function inλ. Because the first term of (2.21) forλ=λn(p, q, Hk) andx=π is zero, then
H(λn(p, q, Hk)) = 0.
From Lemma 2.2, we see that the spectral set {λn(p, q, Hk)}+∞k=1 is a bounded infinite set. Hence, there exists λn0(p, q)∈R, such that λn0(p, q) is a finite accu- mulation point of the spectrum set {λn(p, q, Hk)}+∞k=1. It is well known that the set of zeros of every entire function which is not identically zero hasn’t any finite accumulation point. Therefore
H(λ) = 0, ∀λ∈C. We can show from (2.24) that
H(λ) = Z π
0
p1(x)n
−cos 2λx+ Z x
0
R1(x, t) exp(2iλt)dt
− Z x
0
R2(x, t) exp(−2iλt)dto dx+
Z π
0
q1(x)n
−sin 2λx +
Z x
0
R3(x, t) exp(2iλt)dt+ Z x
0
R4(x, t) exp(−2iλt)dto dx= 0
(2.25)
whereRl(x, t),l= 1,4 are piecewise-continuously differentiable on 0≤t≤x≤π.
Moreover, by using Euler’s formula, (2.25) can be written as Z π
0
f1(x)n
exp(2iλx) + Z x
0
S11(x, t) exp(2iλt)dt +
Z x
0
S12(x, t) exp(−2iλt)dto dx+
Z π
0
f2(x){exp(−2iλx) +
Z x
0
S21(x, t) exp(2iλt)dt+ Z x
0
S22(x, t) exp(−2iλt)dt}dx= 0,
(2.26)
where
f1(x) =−1
2i(q1(x) +ip1(x)), f2(x) = 1
2i(q1(x)−ip1(x)), i=√
−1, (2.27) and the matrix S(x, t) = (Sij(x, t)), i, j = 1,2 with entries being piecewise-con- tinuously differentiable on 0 ≤t ≤ x≤π. By changing the order of integration, (2.26) can be written as
Z π
0
exp(2iλt)h f1(t) +
Z π
t
(f1(x)S11(x, t) +f2(x)S21(x, t))dx]dt +
Z π
0
exp(−2iλt)h f2(t) +
Z π
t
(f1(x)S12(x, t) +f2(x)S22(x, t))dxi dt= 0,
or Z π
0
e0(λt), f(t) + Z π
t
S(x, t)f(x)dx
dt= 0. (2.28)
Here e0(λt) = (exp(2iλt),exp(−2iλt))T andf(x) = (f1(x), f2(x))T. Thus from the completeness of the functionse0(λt), it follows that
f(t) + Z π
t
S(x, t)f(x)dx= 0, for 0< t < π.
But this equation is a homogeneous Volterra integral equation and has only the zero solution. Thus we havef(x) = (f1(x),f2(x))T = 0 on the interval [0, π]. From (2.27)), it holds
q1(x) +ip1(x) = 0 =q1(x)−ip1(x), i.e. q1(x) = 0 and p1(x) = 0. From (2.23) we obtain
(p(x), q(x)) = (p(x),e q(x)),e
a.e. on [0, π]. This result completes the proof.
Acknowledgements. The authors would like to thank the anonymous referees for their careful reading and valuable comments in improving the original manuscript.
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Yu Ping Wang
Department of Applied Mathematics, Nanjing Forestry University, Nanjing, 210037, Jiang Su, China
E-mail address:[email protected]
Murat Sat
Department of Mathematics, Faculty of Science and Art, Erzincan University, Erzin- can, 24100, Turkey
E-mail address:murat [email protected]
