A problem of plane integral geometry

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(1)Sel¸cuk J. Appl. Math. Vol. 4, No. 2, pp. 123–128, 2003. Selcuk Journal of Applied Mathematics. A problem of plane integral geometry Mustafa Yıldız Department of Mathematics, Devrek Arts and Sciences Faculty, Karaelmas University, Incivez 67100, Zonguldak, Turkey; e-mail: mustafayildiz@karaelmas.edu.tr Received: October 30, 2003. Summary. In this work we deal with existence, uniqueness and stability of solutions to the two dimensional integral geometry problem for a family of regular curves of given curvature. Here a family of regular curves means that from each point of a domain D ⊂ R2 , a curve of the family passes in any direction v. Solvability of the problem is proved by using the method of Galerkin. Key words:. Integral geometry, method of Galerkin. 2000 Mathematics Subject Classification: 53C65 Let us assume that in a domain D, a family of regular curves is given by curvature such that curvature of the curve passing from each point x of the domain D, in any direction ϑ = (cos ϕ, sin ϕ) is K = f2 cos ϕ − f1 sinϕ. We assume that there is a curve passing from every x ∈ D in the arbitrary direction ϑ. Suppose lengths of these curves in D are upper-bounded by the same constant. Let us denote the family of these curves by {Ak }. If K = 0, it is obvious that the lines intersecting the domain D can be taken as {Ak }. Let us now consider the problem of finding λ(x), when the integral of each curve of the set {Ak } is given. This problem of integral geometry is equivalent to an inverse problem given in the domain Ω = D × (0, 2π) for the equation (1). ux1 cos ϕ + ux2 sin ϕ + K(x, ϕ)uϕ = λ(x).. This inverse problem is expressed as follows:.

(2) 124. M. Yıldız. Problem 1. From the equation (1), find a pair of functions (u, λ) satisfying the condition (2). u|Γ = u0 ,. Γ = ∂D × (0, 2π).. Let u(x, ϕ) be (3). λds,. u(x, ϕ) = γ+ (x,ϕ). where γ+ (x, ϕ) is the curve which starts from the point x in the direction ϑ and has the curvature K(x, ϕ). If we differentiate the equation (3) at the point x in the direction ϑ, the equation (1) is obtained. The problem (1)-(2) can be reduced to the integral geometry problem. To achieve this, it is necessary to integrate the equation (1) on the curve γ+ (x, ϕ). Now, let us replace the overdetermined problem (1)-(2). Which is equivalent to the problem of integral geometry with the following determined problem. Problem 2. From an equation (4) Lu = ux1 cos ϕ + ux2 sin ϕ + (f2 cos ϕ − f1 sin ϕ)uϕ = λ(x, ϕ), find the pair of functions (u, λ) which satisfy following conditions (5). u|Γ = u0 ,. (6). ˆ = 0. Lλ. ˆ is Where L ∂ ˆ = (cos ϕ) ∂ − f1 ∂ − (sin ϕ) ∂ − f2 ∂ L . ∂x2 ∂ϕ ∂x1 ∂ϕ ∂ϕ The equation (6) is written for generalized functions, and solution of the problem 2 is sought in the appropriate class of generalized ˆ and Γ (A) is the set of all functions u ∈ L2 (Γ ) functions. A = LL such that for each function u ∈ Γ (A) there exists a y ∈ L2 (Ω) 3 (here the inner satisfying < u, A∗ η >=< y, η > for ever η ∈ C0π ∗ product is a product of the space L2 , and A is the adjoint of A) (see. [1], [2]). Assume that Γ (A) ⊂ Γ (A) is such that for any u ∈ Γ (A) 3 (Γ ) with the following properties: there is a sequence {uk } ⊂ C0π 1. uk u weakly in L2 (Ω), 2. < Auk , uk >→< Au, u >. 3 (Ω) with respect the norm u Γ (A) is the closure of C0π Γ (A) = uL2 + AuL2 (here and in what follows, . denotes the norm in.

(3) A problem of plane integral geometry. 125. L2 (Ω)). It is clear that Γ (A) ⊂ Γ (A) ⊂ Γ (A). Furthermore the 0 ⊂ Γ (A) ⊂ L (Ω). Instead of following inclusion holds: Γ (A) ∩ H1,π 2 the problem (4)-(6), let us consider a problem (7). Lu = λ + f,. (8). ˆ = 0. u|Γ = 0, Lλ. In the problem (7)-(8), we have to find the pair of functions (u, λ) in the domain Ω. The problem (4)-(6) can be reduced to the problem (7)-(8). ¯ ∀x ∈ D ¯ f1x + f2x > 0 and Theorem 1. If f1 (x), f2 (x) ∈ C 2 (D), 1 2 f ∈ H2,π (Ω) then the problem (7)-(8) has a unique solution, (u, λ) such that satisfying the conditions u ∈ Γ (A) ∩ H1,π (Ω), λ ∈ L2 (Ω), and the inequality uH1,π (Ω) + λ ≤ C(f + fϕ ) holds where C > 0 depends on D. Proof. First of all we will prove the uniqueness of the solution. For this, we show that the corresponding homogeneous linear problem has only trivial solution satisfying the conditions of the theorem. If ˆ to the equation (7) and take into account of the equations we apply L (8, we find Au = 0. Since the solution of the equation u ∈ Γ (A), there 3 and exists such a sequence {uk } ⊂ C0π 1. In the space L2 , uk u, since we have Au = 0 in (8), 2. < Auk , uk >→ 0 (k → ∞) is obtained. It is easy to show that ∂ ∂ ∂ ∂uk (Luk ) − (Luk )uk , (9) −Auk uk = ∂ϕ ∂l ∂l ∂ϕ (10) and. ∂ ∂ ∂ ∂ ∂ ≡ (cos ϕ) − (sin ϕ) , − f1 + f2 ∂l ∂x2 ∂ϕ ∂x1 ∂ϕ −2 < Auk , uk > = [(ukx1 + ukϕ f2 )2 + (ukx2 − ukϕ f1 )2 Ω. (11). + u2kϕ (f1x1 + f2x1 )]dΩ,. since the divergent terms are all zero. It is clear that, if f1x1 + f2x2 > 0, then the quadratic form is positive in J(∇uk ), (ukx1 + ukϕ f2 ), (ukx2 + ukϕ f1 ), ukϕ , where J(∇uk ) = (ukx1 + ukϕ f2 )2 + (ukx2 + ukϕ f1 )2 + u2kϕ (f1x1 + f2x2 )..

(4) 126. M. Yıldız. Since the domain D is bounded and the function uk is zero on ∂D × (0, 2π), we have u2 ≤ C J(∇uk )dΩ, where C > 0 depends on D. Ω. By the equality (11) and the definition of Γ (A) we get 2. . 2. u = lim uk ≤ C lim. (12). k→∞. k→∞. J(∇uk )dΩ Ω. = −2C lim < Auk , uk >= 0. k→∞. Inequality (12) gives lim < Auk , uk >= 0 and so u = 0 i.e. k→∞. u = 0. Equation (7) implies F = 0 and u = 0 then we obtain λ = 0. Thus, the system of homogeneous linear problem has only trivial solution u = 0, λ = 0. This completes the proof of the theorem. Secondly we will prove existence of the solution of our problem in the same set. Let us assume {w1 , w2 , · · · , wn , · · ·} complete and or3 . thonormal system in the space L2 (Ω) where for each i ∈ N wi ∈ C0π Let Pn be the orthonormal projection of L2 (Ω) on Mn which is the linear span of the system {w1 , w2 , · · · , wn , · · ·}. From the expressions (7)-(8), we obtain the problem ˆ = F, u|Γ = 0. Au = Lf. (13). An approximate solution of order N of the problem (13) uN =. N . αN i wi (x, ϕ);. αN = (αN1 , αN2 , · · · , αNN ),. i=1. is defined as a solution to the following problem: Find the vector αN from the system of linear algebraic equations ˆ l = 1, 2, · · · , N, dΩ = dxdϕ. (14) L(Lu N − f )wl dΩ = 0, Ω. We shall prove ( see [5]) that there exists a unique solution αN of the system (14) for any f ∈ H2,π (Ω) under the hypotheses of the Theorem. To this end, it suffices to prove that the homogeneous version of system (14) has only trivial solution. Assume the contrary. Let the homogeneous version of system (14) have a nonzero solution αN = (αN1 , αN2 , · · · , αNN ). In the system (14) with f = 0, substituting αN for αN , multiplying the first equation by −2αNl and summing with respect to l from 1 to N , we obtain ˆ N uN dΩ = 0. (15) −2 LLu Ω.

(5) A problem of plane integral geometry. Where uN = . N. i=1. 127. αN i wi . By (11) and (15),. [(uN x1 + uN ϕ f2 )2 + (uN x2 + uN ϕ f1 )2 + u2N ϕ (f1x1 + f2x2 )] = 0.. Ω. Therefore since the quadratic form J(∇uk ) is positive definite and uN = 0 on Γ , we have uN = 0 in D, and αNl = 0 because {wi } is linearly independent. This contradicts the condition αNl

(6) = 0. Thus, system (14) has a unique solution αN for any f ∈ H2,π (Ω). Now we estimate the solution uN of system (14) in terms of f . For this purpose, we multiply the first equation of the system (14) by −2αNl and sum the obtain relations with respect to l from to N . Then we have ˆ ˆ dΩ. (16) −2 uN L(LuN )dΩ = −2 uN Lf Ω. Ω. Since uN |Γ = 0, the right hand side of (16) as follows: . . δ0 2. 2 ˆ −2 uN Lf dΩ ≤ |fϕ | dΩ + |∇x,1 uN |2 dΩ. δ0 2 Ω. Ω. Ω. Then, using (9), (16) and Schwartz inequality for sufficiently large δ0 > 0, we get δ0 2 2 2 |∇x,ϕ uN | dΩ ≤ fϕ dΩ + |∇x uN |2 dΩ δ0 2 Ω. or. . Ω. 2. . |LuN | dΩ ≤ C, Ω. Ω. . 2. |∇x uN | dΩ ≤ C ⇒ Ω. u2N dΩ ≤ C.. Ω. This implies that the sets of functions {uN } and {LuN } are bounded in L2 . Since L2 (Ω) is a Hilbert space, the sets {uN } and {LuN } are weakly compact in L2 (Ω). Therefore, they have subsequences such (k → ∞) weakly converge. If we take into that uNk u, LuNk λ consideration the fact that the operator L is weakly closed, we see = Lu. Since uN |Γ = 0, we have u|Γ = 0. Multiplying the that λ equations ˆ uN |Γ = 0, uN u ⇒ u|Γ = 0, L(Lu N − f )w1 dΩ = 0, Ω.

(7) 128. M. Yıldız. by β ∈ R1 , and taking into account the condition uN |∂D×(0,2π) = 0, ˆ l on β, for N ≥ 1, we obtain and calculating the action of Lw ˆ (LuN − f )L(βw l )dΩ = 0. Ω. Since the linear span of {wl } is dense on the space H20 (Ω), we get ˆ (Lu − f )LηdΩ = 0, (17) Ω. for every η ∈ H20 (Ω). If we put λ = Lu−f , we obtain λ ≤ cuH 1 + ˆ = 0 in the sense of generalized functions. f L2 . From (17), we get Lλ To complete the proof of the theorem, for N → ∞ we have to show < AuN , uN >→< Au, u >. From the equation (16), we obtain PN AuN = PN F . Since the system {w1 , w2 , · · · wN } is orthogonal in L2 , if we take the system sin nx as the system {w1 , w2 , · · · wN } PN F converges strongly to F in the sense of L2 , when N → ∞ in other words we get N → ∞ ⇒ PN AuN → F = Au. Then, since we have uN u and PN AuN → Au in the L2 when N → ∞, for N → ∞ we obtain < PN AuN , uN >→< Au, u >. By the definitions of the functions PN and uN (since the operator PN is self adjoint in L2 (Ω), see [3], [4]) we obtain < PN AuN , uN >=< Au, PN∗ uN >=< AuN , PN uN >=< AuN , uN > . This implies that for N → ∞ we have < AuN , uN >→< Au, u > which is the required expression to complete the proof of the theorem. References 1. Amirov, A. Kh. (1986): Existence And Uniqueness Theorems For The Solution Of An Inverse Problem for The Transport Equation, Sib. Mat. J., 27, 783–800. 2. Amirov, A. Kh. (1995):Inverse Problem For a Kinetic Equation, J. Inv IllPosed Problems, 3, 351–357. 3. Kolmogorov, A.N. (1970): Introductory Real Analysis . Prentice-Hall International, Inc., London. 4. Kreysizg, E. (1978): Introductory Functional Analysis With Applications. John Wiley-Sons. Inc, Canada, 257–258. 5. Mikhailov, V. P. (1978): Partial Differential Equations. Mir Publishers, Moskow..

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