Recovering Nonlocal Differential Pencils

0.17 MB·English

by  Chuan-Fu Yang

RECOVERING NONLOCAL DIFFERENTIAL PENCILS
CHUAN-FUYANGANDVJACHESLAVYURKO
5
1
0 Abstract. Inverse problems for differential pencils with nonlocal conditions
2
are investigated. Several uniqueness theorems of inverse problems from the

n Weyl-type function and spectra are proved, which are generalizations of the
a well-knownWeylfunctionandBorg’sinverseproblemfortheclassicalSturm-
J Liouvilleoperator.

0
3

] 1. Introduction
P
S Problemswithnonlocalconditionsariseinvariousfieldsofmathematicalphysics
h. [1, 2, 3, 4, 5], biology and biotechnology [6, 7], and in other fields. Nonlocal condi-
t tions come up when value of the function on the boundary is connected to values
a
inside the domain. Recently problems with nonlocal conditions are paid much
m
attention for them in the literature.
[
In this paper we study inverse spectral problems for differential pencils

1 y′′(x)+[λ2−2λp(x)−q(x)]y(x)=0, x∈(0,T), (1)
v
9 and with nonlocal linear conditions
1
T
8 U (y):= y(t)dσ (t)=0, j =1,2. (2)
j j
1
Z0
0
Here p ∈ AC[0,T] (absolutely continuous function) and q ∈ L(0,T) are complex-
.
3 valuedfunctions, σ (t)arecomplex-valuedfunctions ofboundedvariationsandare
j
0
continuous from the right for t>0. There exist finite limits H :=σ (+0)−σ (0).
5 j j j
Linear forms U (y) in (2) can be written as forms
1 j
: T
v
U (y):=H y(0)+ y(t)dσ (t), j =1,2, (3)
i j j j0
X Z0
where σ (t) in (3) are complex-valued functions of bounded variations and are
r j0
a continuous from the right for t≥0, and H 6=0.
1
A complex number λ is called an eigenvalue of the problem (1) and (2) if
0
equation (1) with λ = λ has a nontrivial solution y (x) satisfying conditions (2);
0 0
then y (x) is called the eigenfunction of the problem (1) and (2) corresponding to
0
theeigenvalueλ . Thenumberoflinearlyindependentsolutionsoftheproblem(1)
0
and (2) for a given eigenvalue λ is called the multiplicity of λ .
0 0
Classical inverse problems for Eq.(1) with two-point boundary conditions have
beenstudiedfairlycompletelyinmanyworks(seethemonographs[8,9,10,11,12,
13] and the references therein). The theory of nonlocal inverse spectral problems
now is only at the beginning because of its complexity. Results of the inverse
Date:March9,2015.
2000 Mathematics Subject Classification. 34A55;34L05;47E05.
Key words and phrases. differentialpencils;nonlocalconditions;inversespectralproblems.
1
2 C.F.YANGANDV.YURKO
problem for various nonlocal operators can be found in [14, 15, 16, 17, 18, 19, 20,
21, 22, 23].
In this work by using Yurko’s ideas of the method of spectral mappings [11]
we prove uniqueness theorems for the solution of the inverse spectral problems for
Eq.(1) with nonlocal conditions (2). In Section 2 we formulate our main results
(Theorems 1 and 2). Section 3 introduces some properties of spectral character-
istics. The proofs of Theorems 1 and 2 are given in Section 4. In Section 5 we
providetwocounterexamplesrelatedtothestatementsofthe inverseproblems(see
also[16,24]). InSection6, asa consequenceof Theorem1,we considerthe inverse
problem of recovering the double functions p and q from the given three spectra.
2. main results
Let X (x,λ) and Z (x,λ), k = 1,2, be the solutions of Eq.(1) with the initial
k k
conditions
′ ′
X (0,λ)=X (0,λ)=Z (T,λ)=Z (T,λ)=1,
1 2 1 2
′ ′
X (0,λ)=X (0,λ)=Z (T,λ)=Z (T,λ)=0.
1 2 1 2
Denote by L the boundary value problem (BVP) for Eq.(1) with the conditions
0
U (y)=U (y)=0,
1 2
and ω(λ) := det[U (X )] , and assume that ω(λ) 6≡ 0. The function ω(λ) is
j k j,k=1,2
an entire function of exponential type with order 1, and its zeros Ξ := {ξn}n∈Z
(counting multiplicities) coincide with the eigenvalues of L . The function ω(λ) is
0
called the characteristic function for L .
0
Denote V (y) := y(j−1)(T), j = 1,2. Consider the BVP L , j = 1,2, for Eq.(1)
j j
withthe conditionsUj(y)=V1(y)=0.TheeigenvaluesetsΛj :={λnj}n∈Z (count-
ing multiplicities) ofthe BVP L coincide with the zerosofthe characteristicfunc-
j
tion ∆ (λ):=det[U (X ),V (X )] .
j j k 1 k k=1,2
Forλ6=λ ,letΦ(x,λ)bethesolutionofEq.(1)undertheconditionsU (Φ)=1
n1 1
and V (Φ) = 0. Denote Weyl-type function M(λ) := U (Φ). It is known [25] that
1 2
for differential pencils with classical two-point separated boundary conditions, the
specificationoftheWeylfunctionuniquelydeterminesthedoublefunctionsp(x)and
q(x). In particular, in [25] it is proved that differential pencils with classical two-
pointseparatedboundary conditionsis uniquely determinedby specifying its Weyl
function, which is equivalent to specification of the spectra of two boundary value
problems with one common boundary condition, and a constructive procedure for
solvingtheinverseproblemisgiven. However,inthecasewithnonlocalconditions,
it is not true; the specification of the Weyl-type function M(λ) does not uniquely
determine the functions p(x) and q(x) (see counterexamples in Section 5). For the
nonlocal conditions the inverse problem is formulated as follows.
T
Throughoutthis paper the functions σ (t) andthe value p(x)dx areknowna
j 0
priori. And condition S: Λ ∩Ξ=∅.
1
R
Inverse problem 1. Given M(λ) and ω(λ), construct the functions p(x) and
q(x).
Let us formulate a uniqueness theorem for Inverse problem 1. For this purpose,
togetherwith(p,q)weconsideranother(p˜,q˜),andweagreethatifacertainsymbol
αdenotesanobjectrelatedto(p,q),thenα˜ willdenoteananalogousobjectrelated
to (p˜,q˜).
RECOVERING NONLOCAL DIFFERENTIAL PENCILS 3
Theorem 1. Let condition S be true. If M(λ)=M˜(λ) and ω(λ)=ω˜(λ), then
a.e.
p(x)=p˜(x) and q(x) = q˜(x) on (0,T).
Thus, under condition S, the specification M(λ) and ω(λ) uniquely determines
the functionpairs(p,q). We note thatif conditionS does nothold,thenthe speci-
fication M(λ) and ω(λ) does not uniquely determine the functions (see Example 1
in Section 5).
Consider the BVP L for Eq.(1) with the conditions U (y) = V (y) = 0. The
11 1 2
eigenvalue set Λ11 := {λ1n1}n∈Z of the BVP L11 coincide with the zeros of the
characteristic function ∆11(λ) := det[U1(Xk),V2(Xk)]k=1,2. Clearly, {λn1}n∈Z ∩
{λ1n1}n∈Z =∅.
Inverse problem 2. Given {λn1,λ1n1}n∈Z, construct p(x) and q(x).
This inverse problem is a generalization of the well-known Borg’s inverse prob-
lem [26] for Sturm-Liouville operatorswith classicaltwo-pointseparatedboundary
conditions.
Theorem2. Ifλ =λ˜ ,λ1 =λ˜1 ,n∈Z,thenp(x)=p˜(x)andq(x)a=.e.q˜(x)
n1 n1 n1 n1
on (0,T).
3. Auxiliary Lemmas
Denote Λ± := {λ : ±Imλ ≥ 0}. It is known (see, for example, [10, 27]) that
there exists a fundamental system of solutions {Y (x,λ)} of Eq.(1) such that
k k=1,2
for λ∈Λ±, |λ|→∞, ν =0,1:
Y(ν)(x,λ)=(iλ)νexp i λx− xp(t)dt (1+O(λ−1)),
1 0 (4)
Y(ν)(x,λ)=(−iλ)νexp −i λx− xp(t)dt (1+O(λ−1)),
2 (cid:0) (cid:0) R 0 (cid:1)(cid:1)
and (cid:0) (cid:0) R (cid:1)(cid:1)
det[Y(ν−1)(x,λ)] =−2iλ(1+O(λ−1)). (5)
k k,ν=1,2
Lemma 1. (See [24]) Let {W (x,λ)} be a fundamental system of solutions
k k=1,2
of Eq.(1), and let Q (y), j =1,2, be linear forms. Then
j
det[Q (W )] =det[Q (X )] det[W(ν−1)(x,λ)] . (6)
j k k,j=1,2 j k k,j=1,2 k k,ν=1,2
It follows from (5)-(6) that
det[Q (Z )] =det[Q (X )] , (7)
j k k,j=1,2 j k k,j=1,2
and
det[Q (Y )] =−2iλ(1+O(λ−1))det[Q (X )] . (8)
j k k,j=1,2 j k k,j=1,2
Introduce the functions
ϕ(x,λ)=−det[X (x,λ),U (X )] , θ(x,λ)=det[X (x,λ),U (X )] ,
k 1 k k=1,2 k 2 k k=1,2
ψ(x,λ)=det[X (x,λ),V (X )] .
k 1 k k=1,2
Then
U (ϕ)=0, U (ϕ)=ω(λ), V (ϕ)=∆ (λ), V (ϕ)=∆ (λ),
1 2 1 1 2 11
U (θ)=ω(λ), U (θ)=0, V (θ)=−∆ (λ),
1 2 1 2
U (ψ)=∆ (λ), V (ψ)=0, V (ψ)=−1.
j j 1 2
Moreover,by calculation, Eqs.(6)-(7) yield that
det[θ(ν−1)(x,λ),ϕ(ν−1)(x,λ)] =ω(λ),
ν=1,2 (9)
det[ψ(ν−1)(x,λ),ϕ(ν−1)(x,λ)] =∆ (λ),
ν=1,2 1
4 C.F.YANGANDV.YURKO
∆ (λ)=−U (Z ), ∆ (λ)=−U (Z ), ∆ (λ)=U (Z ). (10)
1 1 2 2 2 2 11 1 1
Note that the functions Φ,ψ,ϕ and θ are all the solutions of Eq.(1) with some
conditions, comparing boundary conditions on Φ,ψ,ϕ and θ, we arrive at
ψ(x,λ)
Φ(x,λ)= , (11)
∆ (λ)
1
1 ∆ (λ)
2
Φ(x,λ)= θ(x,λ)+ ϕ(x,λ) . (12)
ω(λ) ∆ (λ)
1
(cid:16) (cid:17)
Hence,
∆ (λ)
2
M(λ):=U (Φ)= , (13)
2
∆ (λ)
1
det[Φ(ν−1)(x,λ),ϕ(ν−1)(x,λ)] =1. (14)
ν=1,2
Let v (x,λ) and v (x,λ) be the solutions of Eq.(1) with the conditions
1 2
′ ′
v (T,λ)=v (T,λ)=1, v (T,λ)=0, U (v )=0.
1 2 1 1 2
Obviously,
v (x,λ)=Z (x,λ), v (x,λ)=Z (x,λ)+N(λ)Z (x,λ),
1 1 2 2 1
det[v(ν−1)(x,λ)] =1, (15)
k k,ν=1,2
where
∆ (λ) U (Z )
1 1 2
N(λ)= =− . (16)
∆ (λ) U (Z )
11 1 1
Denote
a
Ua(y):= y(t)dσ (t), a∈(0,T].
1 1
Z0
Clearly, U =UT, and if σ (t)≡C (constant) for t≥a, then U =Ua.
1 1 1 1 1
For sufficiently small δ >0, we denote
Π :={λ: argλ∈[δ,π−δ]}, G :={λ: |λ−λ |≥δ, ∀n∈Z},
δ δ n1
and
G′ :={λ: |λ−λ1 |≥δ, ∀n∈Z},
δ n1
where λ ∈Λ and λ1 ∈Λ .
n1 1 n1 11
Lemma 2. For λ∈Π , |λ|→∞, we have
δ
(iλ)ν x
Φ(ν)(x,λ)= exp i λx− p(t)dt (1+o(1)), x∈[0,T), (17)
H
1 (cid:18) (cid:18) Z0 (cid:19)(cid:19)
(iλ)ν T−x
v(ν)(x,λ)= exp −i λ(T −x)− p(t)dt (1+O(λ−1)), x∈[0,T),
1 2
Z0 !!
(18)
∆ (λ)=−H1 exp −i λT − T p(t)dt (1+o(1)),
1 2iλ 0
(19)
∆11(λ)= H21 exp (cid:16)−i (cid:16)λT − 0RT p(t)dt (cid:17)(cid:17)(1+o(1)).
(cid:16) (cid:16) R (cid:17)(cid:17)
RECOVERING NONLOCAL DIFFERENTIAL PENCILS 5
Let σ (t)≡C (constant) for t≥a (i.e. U =Ua). Then for λ∈Π , |λ|→∞,
1 1 1 δ
H x
ϕ(ν)(x,λ)= 1 (−iλ)ν−1exp −i λx− p(t)dt
2
(cid:18) (cid:18) Z0 (cid:19)(cid:19)
×[1+o(1)+O(exp(iλ(2x−a)))], x∈(0,T], (20)
T−x
v(ν)(x,λ)=(−iλ)ν−1exp i λ(T −x)− p(t)dt
2
Z0 !!
×[1+o(1)+O(exp(iλ(2x−a)))], x∈[0,T). (21)
Proof. The function Φ(x,λ) can be expressed as
Φ(x,λ)=A (λ)Y (x,λ)+A (λ)Y (x,λ), (22)
1 1 2 2
together with U (Φ)=1 and V (Φ)=0, which yields that
1 1
A (λ)U (Y )+A (λ)U (Y )=1, A (λ)V (Y )+A (λ)V (Y )=0. (23)
1 1 1 2 1 2 1 1 1 2 1 2
Using (4), one gets that for λ∈Π , |λ|→∞:
δ
U (Y )=H (1+o(1)), U (Y )=O(exp(−iλT)), (24)
1 1 1 1 2
V (Y )=exp i λT − T p(t)dt (1+O(λ−1)),
1 1 0
(25)
V1(Y2)=exp(cid:16)−(cid:16)i λT −R 0T p(t)d(cid:17)t(cid:17) (1+O(λ−1)).
Solving linear algebraic system(cid:16) (2(cid:16)3) by uRsing (24)-(cid:17)((cid:17)25), we obtain
T
A (λ)=H−1(1+o(1)), A (λ)=O exp 2i λT − p(t)dt .
1 1 2
Z0 !!!
Substituting these relationsinto(22), wehaveproved(17). Formulas(18)-(21)can
be proved similarly, and are omitted. ✷
By the well-known method (see, for example, [10]) the following estimates hold
for x∈(0,T), λ∈Λ+ :
T−x
v(ν)(x,λ)=O λνexp −i λ(T −x)− p(t)dt , (26)
1
Z0 !!!
x
Φ(ν)(x,λ)=O λνexp i λx− p(t)dt , ρ∈G . (27)
δ
(cid:18) (cid:18) (cid:18) Z0 (cid:19)(cid:19)(cid:19)
Moreover, if σ (t) ≡ C (constant) for t ≥ a (i.e. U = Ua), then for x ≥ a/2, λ ∈
1 1 1
Λ+:
x
ϕ(ν)(x,λ)=O λν−1exp −i λx− p(t)dt , (28)
(cid:18) (cid:18) (cid:18) Z0 (cid:19)(cid:19)(cid:19)
T−x
v(ν)(x,λ)=O λν−1exp i λ(T −x)− p(t)dt , ρ∈G′. (29)
2 δ
Z0 !!!
6 C.F.YANGANDV.YURKO
4. Proofs of Theorems
Proof of Theorem 2 We know that the characteristic function ∆ (λ) of the
1
BVPL isanentirefunctionoforderonewithrespecttoλ. Followingthetheoryof
1
Hadamard’s factorization(see [28]), ∆ (λ) can be expressed as an infinite product
1
as
∆1(λ)=c1ea1λ 1− λ eλλn1+12(λλn1)2+···+p1(λλn1)p,
λ
n∈Z(cid:18) n1(cid:19)
Y
whereλ aretheeigenvaluesofthe problemL , pis thegenusof∆ (λ), c anda
n1 1 1 1 1
are constants. Since for the order ρ of ∆ (λ), p ≤ ρ ≤ p+1 (see [28]), and ∆ (λ)
1 1
is an entire function of exponential type with order 1, we find that the genus of
∆ (λ) is 0 or 1 (that is, p=0∨1). Thus ∆ (λ) can be rewritten by
1 1
∆1(λ)=c1ea1λ 1− λ eλλn1p.
λ
n∈Z(cid:18) n1(cid:19)
Y
Since ∆ (λ) and ∆˜ (λ) are both entire functions of order one with respect to λ,
1 1
and λ = λ˜ for all n ∈ Z, by the Hadamard’s factorization theorem, we may
n1 n1
suppose (the case when ∆ (0)=0 requires minor modifications)
1
∆1(λ)=c1ea1λ 1− λ eλnλ1p
λ
n∈Z(cid:18) n1(cid:19)
Y
and
∆˜1(λ)=c˜1ea˜1λ 1− λ eλλn1p˜,
λ
n∈Z(cid:18) n1(cid:19)
Y
forsome constantsc ,c˜ ,a ,a˜ andp,p˜, whichcanbe determinedfromthe asymp-
1 1 1 1
totics. From this we get for all λ∈C
∆˜1(λ) = c˜1eh(a˜1−a1)+(p˜−p)Pn∈Z λn11iλ.
∆ (λ) c
1 1
The expression (19) implies that
H T
1
∆ (λ)=− exp −i λT − p(t)dt (1+o(1))
1
2iλ
Z0 !!
and
H T
∆˜ (λ)=− 1 exp −i λT − p˜(t)dt (1+o(1)).
1
2iλ
Z0 !!
T T
We get that, using the assumption that p(t)dt= p˜(t)dt,
0 0
∆˜1(λ =1+o(1)≡ c˜1eRh(a˜1−a1)+(p˜−Rp)Pn∈Z λn11iλ,
∆ (λ) c
1 1
which yields that
1
(a˜ −a )+(p˜−p) =0, c˜ =c .
1 1 1 1
λ
n∈Z n1
X
Consequently, ∆ (λ) ≡ ∆˜ (λ). Analogously, from λ1 = λ˜1 for all n ∈ Z we get
1 1 n1 n1
∆ (λ)≡∆˜ (λ). By virtue of (16), this yields
11 11
N(λ)≡N˜(λ). (30)
RECOVERING NONLOCAL DIFFERENTIAL PENCILS 7
Define the functions
P (x,λ):=v (x,λ)v˜′(x,λ)−v˜′(x,λ)v (x,λ),
1 1 2 1 2 (31)
P (x,λ):=v (x,λ)v˜ (x,λ)−v˜ (x,λ)v (x,λ).
2 2 1 2 1
Using (15) and (30), one gets
P (x,λ)=(Z (x,λ)Z˜′(x,λ)−Z˜′(x,λ)Z (x,λ))+(N˜(λ)−N(λ))Z (x,λ)Z˜′(x,λ)
1 1 2 1 2 1 1
=Z (x,λ)Z˜′(x,λ)−Z˜′(x,λ)Z (x,λ),
1 2 1 2
P (x,λ)=Z (x,λ)Z˜ (x,λ)−Z˜ (x,λ)Z (x,λ)+(N(λ)−N˜(λ))Z (x,λ)Z˜ (x,λ)
2 2 1 2 1 1 1
=Z (x,λ)Z˜ (x,λ)−Z˜ (x,λ)Z (x,λ).
2 1 2 1
Thus, for each fixed x ∈ (0,T), the functions P (x,λ), k = 1,2, are entire in λ.
k
On the other hand, taking (18) and (21) into account we calculate for each fixed
x≥T/2 and k =1,2:
P (x,λ)−δ Ω (x)=o(1), |ρ|→∞, ρ∈Π ,
k 1k 1 δ
where δ is the Kroneckersymbol and
1k
T−x T−x
exp[i (p(t)−p˜(t))dt]+exp[−i (p(t)−p˜(t))dt]
Ω (x)= 0 0 .
1
2
R R
Also, applying (26) and (29), we get for k =1,2,
′
P (x,λ)=O(1), |ρ|→∞, ρ∈G .
k δ
UsingthemaximummodulusprincipleandLiouville’stheoremforentirefunctions,
we conclude that
P (x,λ)≡Ω (x), P (x,λ)≡0, x≥T/2.
1 1 2
From (31) it yields
v (x,λ)=Ω (x)v˜ (x,λ), v (x,λ)=Ω (x)v˜ (x,λ).
1 1 1 2 1 2
Again, using the asymptotic expression (18) for v (x,λ) and v˜ (x,λ), we have
1 1
T−x
exp −i(λ(T −x)− p(t)dt) [1+o(1)]
Z0 !
T−x
=Ω (x)exp −i(λ(T −x)− p˜(t)dt) [1+o(1)],
1
Z0 !
which leads to
T−x
exp i (p(t)−p˜(t))dt =Ω (x).
1
Z0 !
This deduces for x≥T/2,
T−x
(p(t)−p˜(t))dt=0,
Z0
which yields
T
p(x)=p˜(x) for x∈ 0, .
2
(cid:20) (cid:21)
T−x
At this case we have (p(t)−p˜(t))dt = 0 for x ≥ T/2. Thus Ω (x) = 1 for
0 1
x≥T/2.
R
8 C.F.YANGANDV.YURKO
Together with (31) this yields that for x≥T/2,
v (x,λ)=v˜ (x,λ), Z (x,λ)=Z˜ (x,λ), p(x)=p˜(x), q(x)a=.e.q˜(x). (32)
k k k k
Next let us now consider the BVPs La and La for Eq.(1) on the interval (0,T)
1 11
with the conditions Ua(y)=V (y)=0 andUa(y)=V (y)=0, respectively. Then,
1 1 1 2
according to Eq.(10), the functions ∆a(λ) := −Ua(Z ) and ∆a (λ) := Ua(Z ) are
1 1 2 11 1 1
the characteristic functions of La and La , respectively. And
1 11
a
Ua/2(Z )=Ua(Z )− Z (t,λ)dσ (t), k =1,2,
1 k 1 k k 1
Za/2
hence
∆a/2(λ)=∆a(λ)+ a Z (t,λ)dσ (t),
1 1 a/2 2 1 (33)
∆a/2(λ)=∆a (λ)− a Z (t,λ)dσ (t).
11 11 R a/2 1 1
Let us use (33) for a=T. Since ∆T1(λ)=∆R1(λ), ∆T11(λ)=∆11(λ), it follows from
(32)-(33) that
∆T/2(λ)=∆˜T/2(λ), ∆T/2(λ)=∆˜T/2(λ).
1 1 11 11
Repeating preceding arguments subsequently for a = T/2,T/4,T/8,..., we con-
clude that p(x)=p˜(x) and q(x)a=.e.q˜(x) on (0,T). Theorem 2 is proved. ✷
Proof of Theorem 1
Define the functions
R (x,λ):=Φ(x,λ)ϕ˜′(x,λ)−Φ˜′(x,λ)ϕ(x,λ),
1 (34)
R (x,λ):=ϕ(x,λ)Φ˜(x,λ)−ϕ˜(x,λ)Φ(x,λ).
2
SinceΛ ∩Ξ=∅wecaninferthatΛ ∩Λ =∅. Otherwise,ifacertainλ∈Λ ∩Λ
1 1 2 1 2
then λ∈Ξ. Thus λ∈Λ ∩Ξ; this leads to a contradiction to the assumption that
1
Λ ∩Ξ = ∅. Moreover, Eqs. M(λ) = M˜(λ), M(λ) = ∆2(λ), and M˜(λ) = ∆˜2(λ)
1 ∆1(λ) ∆˜1(λ)
imply that
∆ (λ)=∆˜ (λ), ∆ (λ)=∆˜ (λ).
1 1 2 2
It follows from (11) and (34) that
1
R (x,λ)= ψ(x,λ)ϕ˜′(x,λ)−ψ˜′(x,λ)ϕ(x,λ) ,
1
∆ (λ)
1
(cid:16) (cid:17)
1
R (x,λ)= ϕ(x,λ)ψ˜(x,λ)−ϕ˜(x,λ)ψ(x,λ) .
2
∆ (λ)
1
(cid:16) (cid:17)
The above equations imply that for each fixed x, the functions R (x,λ) are mero-
k
morphic in λ with possible poles only at λ=λ . On the other hand, taking (12)
n1
into account, we also get
1
R (x,λ)= θ(x,λ)ϕ˜′(x,λ)−θ˜′(x,λ)ϕ(x,λ) , (35)
1
ω(λ)
(cid:16) (cid:17)
1
R (x,λ)= ϕ(x,λ)θ˜(x,λ)−ϕ˜(x,λ)θ(x,λ) . (36)
2
ω(λ)
(cid:16) (cid:17)
The assumption that Λ ∩Ξ=∅ tells us that the functions R (x,λ) are regular at
1 k
λ=λ . Thus, for each fixed x, the functions R (x,λ) are entire in λ. Using (17)
n1 k
and (20), we can obtain for x≥T/2:
R (x,λ)−δ Ω (x)=o(1), |ρ|→∞, ρ∈Π ,
k 1k 2 δ
RECOVERING NONLOCAL DIFFERENTIAL PENCILS 9
where
x x
exp[i (p(t)−p˜(t))dt]+exp[−i (p(t)−p˜(t))dt]
Ω (x)= 0 0 .
2
2
R R
Also, using (27)-(28), we obtain for x≥T/2:
R (x,λ)=O(1), |ρ|→∞, ρ∈G .
k δ
Therefore, R (x,λ) ≡ Ω (x), R (x,λ) ≡ 0 for x ≥ T/2. Using the asymptotic
1 2 2
expression (20) for ϕ(x,λ) and ϕ˜(x,λ), we have
x
exp −i(λx− p(t)dt) [1+o(1)]
(cid:18) Z0 (cid:19)
x
=Ω (x)exp −i(λx− p˜(t)dt) [1+o(1)].
2
(cid:18) Z0 (cid:19)
Thus
x
exp i (p(t)−p˜(t))dt =Ω (x),
2
(cid:18) Z0 (cid:19)
which deduces for x≥T/2,
x
(p(t)−p˜(t))dt=0.
Z0
this yields
T
p(x)=p˜(x) for x∈ ,T .
2
(cid:20) (cid:21)
x
At this case we have (p(t) − p˜(t))dt = 0 for x ≥ T/2. Thus Ω (x) = 1 for
0 2
x≥T/2. Together with (14) and (34), it yields
R
ϕ(x,λ)=ϕ˜(x,λ), ψ(x,λ)=ψ˜(x,λ), p(x)=p˜(x), q(x)a=.e.q˜(x), x≥T/2.
Also, we obtain
Z (x,λ)=Z˜ (x,λ), k =1,2, x≥T/2.
k k
Since
ϕ(x,λ)=U (Z )Z (x,λ)−U (Z )Z (x,λ)
1 1 2 1 2 1
and
ϕ˜(x,λ)=U (Z˜ )Z˜ (x,λ)−U (Z˜ )Z˜ (x,λ)
1 1 2 1 2 1
we have
ϕ(x,λ)=∆ (λ)Z (x,λ)+∆ (λ)Z (x,λ)
11 2 1 1
and
ϕ˜(x,λ)=∆˜ (λ)Z˜ (x,λ)+∆˜ (λ)Z˜ (x,λ).
11 2 1 1
Taking x=T we get
ϕ(T,λ)=∆ (λ), ϕ˜(T,λ)=∆˜ (λ), ϕ′(T,λ)=∆ (λ), ϕ˜′(T,λ)=∆˜ (λ).
1 1 11 11
It follows from ϕ(x,λ)=ϕ˜(x,λ) for x≥T/2 that
∆ (λ)=∆˜ (λ), ∆ (λ)=∆˜ (λ).
1 1 11 11
a.e.
Using Theorem 2, we conclude that p(x)=p˜(x) and q(x) = q˜(x) on (0,T). Theo-
rem 1 is proved. ✷
10 C.F.YANGANDV.YURKO
5. Counterexamples
Example 1 (To illustrate that if condition S does not hold then Theorem 1 is
false)
Suppose that T = π, U (y) = y(0), U (y) = y(π/2), p(x) = p(x+π/2) and
1 2
q(x) = q(x + π/2) for x ∈ (0,π/2), p(x) 6≡ p(π − x) and q(x) 6≡ q(π − x) for
x∈(0,π).
Take p˜(x):=p(π−x) and q˜(x):=q(π−x) for x∈(0,π). ThenBVP L˜ : Eq.(1)
1
with p˜(x) = p(π −x) and q˜(x) = q(π −x), y˜(x) = y(π−x), and the conditions
U (y˜)=V (y˜)=0;
1 1
BVP L˜ : Eq.(1) with p˜(x) = p(π−x) and q˜(x) = q(π−x), y˜(x) = y(3π −x),
2 2
and the conditions U (y˜)=V (y˜)=0;
2 1
BVP L˜ : Eq.(1) with p˜(x) = p(π/2−x) (also equal to p(π −x)) and q˜(x) =
0
q(π/2−x) (also equal to q(π−x)), y˜(x) = y(π −x), and the conditions U (y˜) =
2 1
U (y˜)=0.
2
Here ∆ (λ) is the characteristic function for Eq.(1) with y(0) = 0 = y(π);
1
∆ (λ) is the characteristic function for Eq.(1) with y(π/2)=0=y(π); ω(λ) is the
2
characteristicfunction for Eq.(1) with y(0)=0=y(π/2). From the above fact the
following relations are true:
∆ (λ)=∆˜ (λ), ∆ (λ)=∆˜ (λ), ω(λ)=ω˜(λ),
1 1 2 2
and, in view of (13), M(λ)=M˜(λ).
Note that Λ , Λ , and Ξ are sets of zeros for characteristic functions ∆ (λ),
1 2 1
∆ (λ) and ω(λ), respectively. Since p(x) = p(x+π/2) and q(x) = q(x+π/2) for
2
x∈(0,π/2),thereholds ω(λ)=∆ (λ), i.e. Ξ=Λ . Thus forall λ∈Ξ(=Λ )then
2 2 2
it yields λ∈Λ , which implies that Ξ∩Λ 6=∅.
1 1
Now M(λ) = M˜(λ) and ω(λ) = ω˜(λ), but Ξ∩Λ 6= ∅. In Theorem 1 condition
1
S does not hold. In fact, at this case p(x) 6= p˜(x) := p(π−x) and q(x) 6= q˜(x) :=
q(π−x). This means, that the specification of M(λ) and ω(λ) does not uniquely
determine the functions p(x) and q(x).
Example 2 (To illustrate that even if condition S and M(λ) = M˜(λ) hold
without the assumption that ω(λ)=ω˜(λ) then Theorem 1 is false)
Suppose that T =π, U (y)=y(0), U (y)=y(π−α), where α∈(0,π/2).
1 2
Let p(x) 6≡ p(π − x) and q(x) 6≡ q(π − x), and (p(x),q(x)) ≡ (0,0) for x ∈
[0,α ]∪[π−α ,π], where α ∈(0,π/2). If α<α , then λ =πn/α, n∈Z\{0}.
0 0 0 0 n2
Choose a sufficiently small α < α such that Λ ∩Λ = ∅. Clearly, such choice
0 1 2
is possible. Then Λ ∩ Ξ = ∅, i.e. condition S holds. Otherwise, if a certain
1
λ∗ ∈Λ ∩Ξ, then λ∗ ∈Λ ∩Λ ; this contradicts to the fact that Λ ∩Λ =∅.
1 1 2 1 2
Takep˜(x):=p(π−x)andq˜(x):=q(π−x).Notethat∆ (λ)isthecharacteristic
2
function for the problem
−y′′(x)=λ2y(x), y(π−α)=0=y(π);
and ∆˜ (λ) is the characteristic function for the problem
2
−y′′(x)=λ2y(x), y(0)=0=y(α).
A simple calculation shows Λ =Λ˜ = πn,n∈Z\{0} .
2 2 α
At this case ∆ (λ) = ∆˜ (λ), ∆ (λ) = ∆˜ (λ), and consequently, M(λ) = M˜(λ).
1 1 2 (cid:8) 2 (cid:9)
Now in Theorem 1 condition S holds, and M(λ)=M˜(λ). In fact, in this example,