On the Stability of a Differential-Difference Analogue of a Two-Dimensional Problem of Integral Geometry

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1 Filomat 3:3 18, Published b Facult of Sciences and Mathematics, Universit of Niš, Serbia Available at: On the Stabilit of a Differential-Difference Analogue of a Two-Dimensional Problem of Integral Geometr Galitdin Bakanov a a B.Sattarkhanov Street, 9, 161, Department of Mathematics, Akhmet Yasawi International Kazakh-Turkish Universit, Turkestan, Kazakhstan Abstract. A differential - difference analogue of a two-dimensional problem of integral geometr with a weight function is studied. A stabilit estimate for the problem considered is obtained. 1. Introduction A great deal of work has been devoted to integral geometr problem, that is, the problems of determining a function from its integrals along a given famil of curves. One of the stimuli for studing such problems is their connection with multidimensional inverse problems for differential equations, 4, 5]. Some inverse problems for hperbolic equations were shown to reduce to integral geometr problems and, in particular, a problem of integral geometr was considered in the case of shift-invariant curves. Mukhometov 3] showed the uniqueness and estimated the stabilit of the solution of a two-dimensional integral geometr on the whole. His results were mainl based on the reduction of the two-dimensional integral geometr problem Vγ, z = U, ρ,, zds, γ, l],z, l] 1 Kγ,z where U C D, ρ,, z is a known function, to the boundar value problem W cos θ + W ρ sin θ =,,, z Ω 1, ρ W ξ γ, η γ, z = V γ, z, V z, z =, γ, z, l]. 3 Here D is a bounded simpl connected domain in the plane with a smooth boundar Γ : = ξ z, = η z, z, l], ξ = ξl, η = ηl, 4 1 Mathematics Subject Classification. Primar 65N6; Secondar 65N, 65N1 Kewords. Integral geometr problem, differential-difference problem, stabilit, uniqueness Received: 8 December; Revised: 8 March 17; Accepted: April 17 Communicated b Allaberen Ashralev Research supported b the grant of Akhmet Yassawi Universit, Turkestan, Kazakhstan address: galitdin.bakanov@au.edu.kz Galitdin Bakanov

2 where the parameter is the length of the curve Γ : G. Bakanov / Filomat 3:3 18, Ω 1 = Ω \ { ξγ, ηγ, z : z, l] }, Ω = D, l]. K,, z is the part of the curve K γ, z included between the points, D and ξ z, η z, z, l] ; W,, z = U, ρ,, z ds, K,,z θ,, z is an angle between the tangent to K,, z at the point, and the -ais and the variable parameter s is the curve length.. Differential-Difference Problem Suppose that the requirements on the famil of curves K γ, z and the domain D necessar for the problem 1 to reduce to problem, 3 are met 3]. Assume also that ever line parallel to either the or the ais intersects the boundar of D at no more than two points. Let a 1 = a = inf {}, b 1 = sup {},, D, D inf {}, b = sup {},, D, D h j = b j a j /N j, j = 1, ; h 3 = l/n 3, and the N j, j = 1,, 3, are natural numbers. Suppose that < ε < min { b 1 a 1 /3, b a /3 }, D ε =, D : min ρ,, α, β > ε α,β Γ, R h = { i, j : i = a 1 + ih 1, j = a + jh, i = 1,..., N 1, j = 1,..., N }. A neighborhood B ih 1, jh of the point a1 + ih 1, a + jh is defined as the five-point set { a1 + ih 1, a + jh, a1 + i ± 1h 1, a + j ± 1h }. A set D ε h consists of all points a 1 + ih 1, a + jh such that their neighborhoods B ih 1, jh are contained in D ε R h. A set Γ ε h is made of all points a 1 + ih 1, a + jh D ε h such that B ih 1, jh D ε R h \D ε. h Finall, ε h = B ih 1, jh, Dh = R h D. Γ ε h From here on we suppose that the coefficients and the solution of problem, 3 have the following properties: W,, z C 3 Ω ε, θ,, z C Ω ε, Ω ε = D ε, l], ρ,, z C Ω, ρ,, z > C >, > ρ 1 ρ. We consider the differential-difference problem of finding the functions Φ i,j z, U i,j which satisf the equation

3 G. Bakanov / Filomat 3:3 18, A C + B C = U i,j, a1 + ih 1, a + jh Dh, z, l], 5 and the boundar condition Φ i,j z = F i,j z, a1 + ih 1, a + jh ε, z, l] h, 6 where Φ i,j z = Φ i, j, z = Φa 1 + ih 1, a + jh, z, u i,j = u i, j = ua 1 + ih 1, a + jh, i =, N 1, j =, N ; = Φ i+1,j Φ i 1,j /h 1, Φ = Φ i,j+1 Φ i,j 1 /h, A = cos θ i,j z, B = sin θ i,j z, θ i,j z = θ a 1 + ih 1, a + jh, z, C = ρ a 1 + ih 1, a + jh, z. We note that in the differential-difference formulation information on the solution is given not onl on the boundar Γ but also in its ε neighborhood, because the partial derivatives θ z, W z, W z, W have singularities of the tpe ξ z + η z ] 1 in a neighborhood of an arbitrar point ξ z, η z, z see 1, 3]. 3. The Main Result Theorem 3.1. Suppose that problem 5-6, with the functions θ i,j z, ρ,, z such that θ i,j = θ i,j l, ρ,, z C 1 Ω, ρ,, z > C > and > ρ 1 ρ, has a solution Φ i,j z C 1, l], Φ i,j = Φ i,j l. Then, for all N j > 9, j = 1,, the following inequalit holds: U i,j h 1h c 1 D ε h 1 F h 1 + F h + ε h F h 1 + h dz, where c 1 is a constant dependent on the function ρ,, z and the curves famil K γ, z. Proof. Multipling both sides of 5 b C + AΦ B, we get A C B + AΦ C + B C Φ =. 7 Using the product differentiation formula, we rewrite 7 as follows: A = C B + AΦ C + B C Φ = A C B + AΦ C + B ] C Φ ] A C B + AΦ C + B C Φ = ] + BΦ + AΦ A B C + AΦ B A C + B C Φ ] + AΦ + BΦ = B A ] + BΦ + AΦ A B C 1 C + AB 1 C A 1 C Φ + B 1 Φ ABΦ + C C C C A + AB + Φ AB A Φ + Φ AB + B Φ + B Φ ABΦ Φ. 8

4 At the same time, rearranging 7 ields G. Bakanov / Filomat 3:3 18, A = C B + AΦ C + B C Φ = C B + AΦ A C + B 1 C AC + B C C = AB + A Φ Sum 9 with 8 and denote ABΦ + C B Φ B C 1 C ABΦ A C + Φ + Φ A C BC B Φ + A Φ + ABΦ K = sin θ = sin θ cos θ = AB, E = cos θ = cos θ sin θ = A B. As a result, we obtain + 1 C C K C Φ E + Using the formula + Φ C K Φ 1 C B C AC ] 1 C C B + Φ + A Φ 1 AB C C 1 C 1 C ABΦ. 9 Φ + ] + BΦ + AΦ =. 1 A B where uυ = u υ + uυ + h 1 u υ ], f = f i+1,j f i,j /h1, f = f i,j f i 1,j /h1, we have ] Φ Φ Φ Φ = Φ + Φ Similarl, we can write Φ ] Φ Φ = Φ + It follows from 11, 1 that + h 1 + h Φ ] Φ Φ Φ Φ Φ = Φ ] Φ. 11 Φ. 1 ] ȳ h 1 Φ ] Φ + h Φ. 13 ȳ Taking into account that Φ =, Φ = Φ, 14

5 combining 1 with 13 and 14, we arrive at I 1 + Here I 1 = ] ] Φ + BΦ + AΦ + Φ A B G. Bakanov / Filomat 3:3 18, Φ ] + 1 C C K C Φ E + 1 C C K Φ. h 1 Φ ] Φ + h The epression for I 1 is a quadratic form with respect to and, whose determinant is Then, from the condition C 1. C > C 1 C the positive definiteness of the quadratic form I 1 follows. Using the inequalit ac b a + b + c a + c + + a c + 4b Φ =. 15 for a positivel definite quadratic form a + b + c, we have I 1 C 1 C + Φ. 16 Due to 5, we obtain + Φ = U i,j C + B AΦ. 17 Since C = ρ,, z, ρ,, z > C >, there eists a positive constant c 1 such that C 1 C >, 1 C 1 C C dz 1 c 1 >. 18 Sum 15 over i, j with the use of 6 and integrate it with respect to z. Then using 16, 17, 18, the inequalit ab a + b / and the periodicit of the functions Φ i,j z, θ i,j z in z, we can transform the resulting equation so as to obtain the desired estimate U i,j h 1h c 1 D ε h 1 ε h F h 1 + F h + F h 1 + h dz. Here c 1 is a constant dependent on the function ρ,, z and on the famil of curves K γ, z. Thus, the proof of Theorem 3.1 is completed.

6 G. Bakanov / Filomat 3:3 18, References 1] S.I. Kabanikhin, G.B. Bakanov, On the stabilit of a finite-difference analogue of a two dimensional problem of integral geometr, Soviet Math. Doklad ] M.M. Lavrent ev, V.G. Romanov, Three linearized inverse problems for hperbolic equations, Soviet Math. Doklad ] R.G. Mukhometov, On a problem of integral geometr, Math. Problems Geophs in Russian. 4] V.G. Romanov, Integral Geometr and Inverse Problems for Hperbolic Equations, Springer-Verlag, Berlin, ] V.G. Romanov, Integral geometr on the geodesics of an isotropic Riemannian metric, Soviet Math. Doklad