Filomat 30:1 (2016), 157–168 DOI 10.2298/FIL1601157C
Published by Faculty of Sciences and Mathematics, University of Niˇs, Serbia Available at: http://www.pmf.ni.ac.rs/filomat
Half Inverse Problem for the Impulsive Diffusion Operator with Discontinuous Coefficient Yas¸ar C ¸ akmaka , Seval Is¸ıkb b Cumhuriyet
a Cumhuriyet University, Faculty of Sciences, Department of Mathematics, 58140 Sivas, Turkey University, Faculty of Education, Department of Secondary School Science and Mathematics Education, 58140, Sivas, Turkey
Abstract. The half inverse problem is to construct coefficients of the operator in a whole interval by using one spectrum and potential known in a semi interval. In this paper, by using the Hocstadt-Lieberman and Yang-Zettl’s methods we show that if p (x) and q(x) are known on the interval (π/2, π), then only one spectrum suffices to determine p (x) , q(x) functions and β, h coefficients on the interval (0, π) for impulsive diffusion operator with discontinuous coefficient.
1. Introduction Inverse spectral problem is recovering the operator from its given spectral datas. These problems are of great importance in applied mathematics and physics, for example, vibration of a string, quantum mechanics etc. Inverse spectral problems for regular or singular Sturm-Liouville and diffusion operators are investigated in [1 − 32] . First results on half inverse problems for regular Sturm-Liouville operator were given by Hochstadt and Lieberman in [33]. In later years, half inverse problems for various Sturm-Liouville operators and diffusion operators, i.e., quadratic pencils of Sturm-Liouville operators, were studied by authors [34 − 44] . In this paper, we denote the problem L = L p, q, α, β, γ, h, H of the form `y (x) = −y00 (x) + 2λp (x) + q (x) y (x) = λ2 ρ (x) y (x) , x ∈ (0, π) (1) with the boundary conditions U y := y0 (0) − hy (0) = 0, V y := y0 (π) + Hy (π) = 0 and the discontinuity conditions π π + 0 = βy − 0 y 2 2 π π π y0 + 0 = β−1 y0 − 0 + γy −0 2 2 2 2010 Mathematics Subject Classification. Primary 34A55 ; Secondary 34B24, 34L05 Keywords. impulsive diffusion operator, inverse spectral problem, half inverse problem Received: 21 February 2014; Accepted: 10 February 2015 Communicated by Dragan S. Djordjevi´c Email addresses:
[email protected] (Yas¸ar C ¸ akmak),
[email protected] (Seval Is¸ık)
(2)
(3)
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where real-valued functions p (x) ∈ W21 (0, π) , q (x) ∈ L2 (0, π), λ is the spectral parameter, α, β, γ are real 2 numbers, β > 0, β − 1 + γ2 , 0 and 1, (x) ρ = α2 ,
0 0 into account, (17) and (18) implies that β = e β.
(17) ω(π) α
and integrate again with respect to
(18)
Y. C ¸ akmak, S. Is¸ık / Filomat 30:1 (2016), 157–168
161 π Theorem 3.2. Let {λn } be the spectrum of both L and e L. If p (x) = e p (x) and q (x) = e q (x) on , π , then h = e h, 2 β=e β, p (x) = e p (x) and q (x) = e q (x) almost everywhere on (0, π) .
e (x, λ) of equations (1) and (4) , respectively, with the Proof. It is clear from [24] that the solutions ϕ (x, λ) , ϕ 0 e (0, λ) = 1, ϕ (0, λ) = h, ϕ e0 (0, λ) = e initial h can be expressed in the integral forms on conditions ϕ (0, λ) = ϕ π , 0, 2 Zx ϕ (x, λ) = cos (λx − ω (x)) +
Zx A (x, t) cos λtdt +
0
(19)
e B (x, t) sin λtdt.
(20)
0
Zx
Zx e (x, t) cos λtdt + A
e (x, λ) = cos λx − ω e (x) + ϕ
B (x, t) sin λtdt
0
0
e (x, t) , e where the kernels A B (x, t) have properties similar to those of A (x, t) , B (x, t) . Using (19) and (20) , we find that 1 e (x) cos (2λx − θ (x)) + cos ω (x) − ω 2 Zx Zx e (x, t) cos (λx − ω (x)) cos λtdt e (x) cos λtdt + A + A (x, t) cos λx − ω
e (x, λ) = ϕ (x, λ) ϕ
0
0
Zx
Zx
+
e (x) dt + B (x, t) sin λt cos λx − ω 0
e B (x, t) sin λt cos (λx − ω (x)) dt 0
x x x x Z Z Z Z e e + A (x, t) cos λtdt A (x, t) cos λtdt + B (x, t) sin λtdt B (x, t) sin λtdt 0
0
0
0
x x x x Z Z Z Z e e + A (x, t) cos λtdt B (x, t) sin λtdt + A (x, t) cos λtdt B (x, t) sin λtdt 0
0
0
0
e (x) . where θ (x) = ω (x) + ω e (x, t) evenly and B (x, t) , e By extending the range of A (x, t) , A B (x, t) oddly with respect to the argument t, we can write 1 e (x, λ) = e (x) cos (2λx − θ (x)) + cos ω (x) − ω ϕ (x, λ) ϕ 2 x Z Zx 1 + Hc (x, t) cos (2λt − θ (t)) dt − Hs (x, t) sin (2λt − θ (t)) dt 2 0
0
where e (x, x − 2t) cos [θ (t) − ω (x)] e (x) + 2A Hc (x, t) = 2A (x, x − 2t) cos θ (t) − ω e (x) − 2e −2B (x, x − 2t) sin θ (t) − ω B (x, x − 2t) sin [θ (t) − ω (x)] +K1 (x, t) cos θ (t) − K2 (x, t) cos θ (t) − M1 (x, t) sin θ (t) + M2 (x, t) sin θ (t) ,
(21)
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e (x, x − 2t) sin [θ (t) − ω (x)] e (x) + 2A Hs (x, t) = 2A (x, x − 2t) sin θ (t) − ω e (x) + 2e +2B (x, x − 2t) cos θ (t) − ω B (x, x − 2t) cos [θ (t) − ω (x)] +K1 (x, t) sin θ (t) − K2 (x, t) sin θ (t) + M1 (x, t) cos θ (t) − M2 (x, t) cos θ (t) , x−2t Z
K1 (x, t) =
Zx
e (x, s + 2t) ds + A (x, s) A −x
e (x, s − 2t) ds, A (x, s) A 2t−x
x−2t Z
K2 (x, t) = −
Zx B (x, s) e B (x, s + 2t) ds −
−x
B (x, s) e B (x, s − 2t) ds, 2t−x
Zx
x−2t Z
A (x, s) e B (x, s + 2t) ds −
M1 (x, t) = −x
A (x, s) e B (x, s − 2t) ds, 2t−x
x−2t Z
M2 (x, t) = −
Zx
e (x, s + 2t) ds + B (x, s) A −x
e (x, s − 2t) ds. B (x, s) A 2t−x
Now, let us write the equations −ϕ00 (x, λ) + 2λp (x) + q (x) ϕ (x, λ) = λ2 ρ (x) ϕ (x, λ)
(22)
and e (x, λ) = λ2 ρ (x) ϕ e (x, λ) . −e ϕ00 (x, λ) + 2λe p (x) + e q (x) ϕ
(23)
e (x, λ) and (23) with ϕ (x, λ) , second subtracting them side by side and First, by multiplying (22) with ϕ then integrating on (0, π) , we get π/2 Z
π/2 e (x, λ) dx = ϕ e0 (x, λ) ϕ (x, λ) − ϕ0 (x, λ) ϕ e (x, λ) 0 + |ππ/2 2λ e p (x) − p (x) + e q (x) − q (x) ϕ (x, λ) ϕ
0
from the hypothesis p (x) = e p (x), q (x) = e q (x) on we obtain
π , π and the initial conditions ϕ (0, λ) = 1, ϕ0 (0, λ) = 0, 2
π/2 Z
e (x, λ) dx +e e (π, λ) − ϕ e0 (π, λ) ϕ (π, λ) = 0. (24) 2λ e p (x) − p (x) + e q (x) − q (x) ϕ (x, λ) ϕ h − h + ϕ0 (π, λ) ϕ
0
Let P (x) := e p (x) − p (x) , Q (x) := e q (x) − q (x) and π/2 Z
H (λ) := e h−h+
e (x, λ) dx. (2λP (x) + Q (x)) ϕ (x, λ) ϕ 0
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It is clear from the properties of ϕ (x, λ), ϕ0 (x, λ) and the boundary conditions (2) that the first term in (24) vanishes and thus H (λn ) = 0
(25)
for each eigenvalue λn . Let us define π/2 Z
π/2 Z
e (x, λ) dx, Q (x) ϕ (x, λ) ϕ
e (x, λ) dx, H2 (λ) = P (x) ϕ (x, λ) ϕ
H1 (λ) =
0
0
then equation (25) can be rewritten as e h − h + 2λn H1 (λn ) + H2 (λn ) = 0.
(26)
From (21) and (24), we obtain |H (λ)| ≤ (C1 + C2 |λ|) exp (τπ)
(27)
for all complex λ, where C1 , C2 > 0 is constant. If we denote Φ (λ) :=
H (λ) , ∆ (λ)
(28)
then Φ (λ) is an entire function with respect to λ. It follows from (11) and (27) that Φ (λ) = O (1)
(29)
for sufficiently large |λ| . Using Liouville’s Theorem, we obtain Φ (λ) = C, for all λ where C is a constant. Now, we can rewrite the equation H (λ) = C∆ (λ) as π/2
Z e e (x, λ) dx = (2λP (x) + Q (x)) ϕ (x, λ) ϕ h−h + 0
p(π) p(π) ω(π) − − (π) C −β+ λα − α sin λµ+ (π) − ω(π) + β λα − sin λµ + α α α o ω(π) ω(π) + + (π) − − (π) + (π) +Hβ cos λµ − α + Hβ cos λµ + α + O exp τµ . n
By the Riemann-Lebesgue Lemma, the limit of the left side of the above equality exists for λ → ∞, λ ∈ R. Therefore, we get that C = 0. Then e h − h + 2λH1 (λ) + H2 (λ) = 0 for all λ. (30) By virtue of (21) , π/2 Z
π/2 Z
2H1 (λ) =
e (x) dx + P (x) cos ω (x) − ω 0
π/2 Z
P1 (t) cos (2λt − θ (t)) dt −
0
P2 (t) sin (2λt − θ (t)) dt, 0
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where π/2 Z
π/2 Z
P (x) Hc (x, t) dx, P2 (t) =
P1 (t) = P (t) +
P (x) Hs (x, t) dx.
(31)
t
t
If we change the order of integration, apply partial integration and take P1 (π/2) = P (π/2) and P2 (π/2) = 0 into account, we get e (x) dx + P (x) cos ω (x) − ω
2H1 (λ) =
π/2 Z
π/2 Z
π/2 Z
2iλt
T1 (t) e
0
0
0 π/2 Z
=
T2 (t) e−2iλt dt
dt −
P (π/2) e (x) dx + P (x) cos ω (x) − ω sin [λπ − θ (π/2)] 2λ
(32)
0 π/2 Z
i P2 (0) + − 2λ 2λ
i T10 (t) e2iλt dt − 2λ
π/2 Z
T20 (t) e−2iλt dt,
0
0
where T1 (t) =
P1 (t) + iP2 (t) −iθ(t) P1 (t) − iP2 (t) iθ(t) e , T2 (t) = e . 2 2
Similarly, we get π/2 Z
2H2 (λ) =
π/2 Z
e (x) dx + Q (x) cos ω (x) − ω 0
π/2 Z
Q1 (t) cos (2λt − θ (t)) dt − 0
Q2 (t) sin (2λt − θ (t)) dt, 0
where π/2 Z
Q1 (t) = Q (t) +
π/2 Z
Q (x) Hc (x, t) dx, Q2 (t) = t
Q (x) Hs (x, t) dx.
(33)
t
By changing the order of integration, we obtain π/2 Z
2H2 (λ) =
π/2 Z
e (x) dx + Q (x) cos ω (x) − ω 0
π/2 Z
R1 (t) e
2iλt
dt +
0
R2 (t) e−2iλt dt
(34)
0
where R1 (t) =
Q1 (t) − iQ2 (t) iθ(t) Q1 (t) + iQ2 (t) −iθ(t) e , R2 (t) = e . 2 2
If (32) and (34) are substituted into (30) , we get π/2
π/2
Z Z e e (x) dx + Q (x) cos ω (x) − ω e (x) dx + P (π/2) sin (λπ − θ (π/2)) h − h + 2λ P (x) cos ω (x) − ω 0 π/2 Z
−P2 (0) + i
0 π/2 Z
T20 (t) e−2iλt dt +
T10 (t) e2iλt dt − i 0
π/2 Z
0
R1 (t) e2iλt dt + 0
(35)
π/2 Z
R2 (t) e−2iλt dt = 0. 0
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Using the Riemann-Lebesgue Lemma for λ → ∞, λ ∈ R in (35) , then it follows that π/2 Z e (x) dx = 0 P (x) cos ω (x) − ω 0 P (π/2) = 0 π/2 Z e e (x) dx = 0 2 h − h + Q (x) cos ω (x) − ω
(36)
0
and π/2 Z
R1 (t) +
iT10
(t) e
2iλt
dt +
0
π/2 Z
R2 (t) − iT20 (t) e−2iλt dt = 0
0
for all complex number λ. n o π π , it follows Since the system e±2iλt : λ ∈ R is entire in L2 − , 2 2 0 R1 (t) + iT1 (t) = 0 R2 (t) − iT0 (t) = 0 2 which yields the following system (t) (t) 0 (t) − P02 (t) + i Q2 (t) + P2 (t) θ0 (t) + P01 (t) = 0 Q1 + P1 θ . Q1 (t) + P1 (t) θ0 (t) − P0 (t) − i Q2 (t) + P2 (t) θ0 (t) + P0 (t) = 0 2
1
And hence, 0 0 Q1 (t) + P1 (t) θ (t) − P2 (t) = 0 . Q2 (t) + P2 (t) θ0 (t) + P0 (t) = 0 1
(37)
Substituting (31) and (33) into system (37) and taking P (π/2) = 0 into account, it yields Q (t) P (t) P0 (t)
π/2 Z
=−
π/2 Z
θ0 (t) Hc (x, t) −
Hc (x, t) Q (x) dx − t
t
! ∂Hs (x, t) P (x) dx − (θ0 (t) + Hs (t, t)) P (t) ∂t
π/2 Z
=−
P0 (x) dx
.
t π/2 Z
=−
π/2 Z
θ0 (t) Hs (x, t) +
Hs (x, t) Q (x) dx − t
If we denote that F (t) = (Q (t) , P (t) , P0 (t))T
t
! ∂Hc (x, t) P (x) dx + Hc (t, t) P (t) ∂t
(38)
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and H (x, t) θ0 (t) H (x, t) − ∂Hs (x, t) c c ∂t 0 0 K (x, t) = ∂Hc (x, t) Hs (x, t) θ0 (t) Hs (x, t) + ∂t
− (θ0 (t) + Hs (t, t)) , 1 Hc (t, t)
equation (38) can be reduced to a vector form π/2 Z
K (x, t) F (x) dx = 0 for 0 < t
