How to create materials with a desired refraction coefficient?

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1 How to create materials with a desired refraction coefficient? A. G. Ramm Mathematics Department, Kansas State University, Manhattan, KS66506, USA ramm@math.ksu.edu ramm

2 Abstract The novel points in this work are: 1) Asymptotic and numerical methods for solving wave scattering problem by many small bodies embedded in a non-homogeneous medium. 2) Derivation of the equation for the field in the limit a 0, where a is the characteristic size of the bodies (particles), and their number M = M(a) tends to infinity at a suitable rate. Multiple scattering is taken into account. 3) A recipe for creating materials with a desired refraction coefficient by embedding many small particles in a given material. 4) Discussion of possible applications: a) creating materials with negative refraction, b) creating materials with a desired radiation pattern.

3 Basic results 1. A rigorous method for solving many-body wave scattering problem for small impedance particles. 2. Formulation on this basis a recipe for creating materials with a desired refraction coefficient. This is done by distributing many small particles with a prescribed boundary impedance in a given material. 3. Formulation of the technological problem to be solved for this theory to be immediately applicable practically. 4. Materials with a desired radiation pattern can be created. (*) A.G.Ramm, Scattering of Acoustic and Electromagnetic Waves by Small Bodies of Arbitrary Shapes.Applications to Creating New Engineered Materials, Momentum Press, New York, 2013.

4 Recipe for creating materials with a desired refraction coefficient: Step 1. If 2 u + k 2 n 2 (x)u = 0, k = const > 0, then n(x) is called the refraction coefficient. Given the original coefficient n 2 0 (x) and the desired coefficient n 2 (x), calculate function p(x) by formula p(x) = k 2 [n 2 0(x) n 2 (x)]. This step is trivial.

5 Step 2. Given p(x) = 4πh(x)N(x), calculate the functions h(x) and N(x). These functions satisfy the following restrictions: Im h(x) 0, N(x) 0. This step is also trivial, and it has many solutions. For example, one can fix an arbitrary N(x) > 0, and then find h(x) = h 1 (x) + ih 2 (x), where h 1 =Re h, h 2 = Im h, by the formulas h 1 (x) = p 1(x) 4πN(x), h 2(x) = p 2(x) 4πN(x), where p 1 =Re p, p 2 =Im p. The condition Im h 0 holds if Im p 0, i.e., Im [n 2 0 (x) n2 (x)] 0.

6 Step 3. Prepare M = 1 a D N(x)dx[1 + o(1)] small balls 2 κ B m (x m, a) with the boundary impedances ζ m = h(xm) a, κ 0 κ < 1, where the points x m, 1 m M, are distributed in D according to formula N ( ) = 1 a N(x)dx[1 + o(1)], D is an arbitrary 2 κ open subset. Embed in D M balls B m (x m, a) with boundary impedance ζ m, d = O(a (2 κ)/3 ), d = min j m x m x j. The material, obtained after the embedding of these M small balls will have the desired refraction coefficient n 2 (x) with an error that tends to zero as a 0. Step 3 is the only non-trivial step in this recipe from the practical point of view.

7 Technological problems The first technological problem is: How can one embed many, namely M = M(a), small balls in a given material so that the centers of the balls (points x m ) are distributed as desired? Physicists know how to solve this problem. The second technological problem is: How does one prepare a ball B m of small radius a with a desired boundary impedance ζ m = h(xm) a, 0 κ < 1, κ where h(x), Imh 0, is a given function?

8 in the absence of the embedded particles L 0 u 0 := [ 2 + k 2 n 2 0(x)]u 0 := [ 2 + k 2 q 0 (x)]u 0 = 0 in R 3, u 0 = e ikα x + v 0, lim r r(u r iku) = 0. Im n 2 0(x) 0, α S 2, k = const > 0. L 0 G = δ(x y) in R 3. n 2 0 (x) = 1 k 2 q 0 (x), q 0 (x) = k 2 k 2 n 2 0 (x), Im q 0(x) 0, n 2 0 (x) = 1 in D := R 3 \ D, q 0 (x) = 0 in D := R 3 \ D.

9 M L 0 u M = 0 in Ω := R 3 \ D m ; D m = B m (x m, a) m=1 u M N = ζ mu M on S m := D m, ζ m = h(x m) a κ, 0 κ 1, u M = u 0 + v M, where N is the outer unit normal to S m, and h(x) C(D) is an arbitrary function, h = h 1 + ih 2, h 2 0, ζ m is boundary impedance, d := min m j dist (x m, x j ).

10 Basic assumptions Let n 2 0 := max x R 3 n2 0 (x). We assume that: a d λ a d, kn 0 a 1. N ( ) := 1 = 1 a 2 κ N(x)dx[1 + o(1)], a 0. ( ) x m Here N(x)a (2 κ) 0 is the density of the distribution of particles, d is the minimal distance between neighboring particles, d = O(a (2 κ)/3 ). ( ) M = M(a) O(a (2 κ) ), 0 κ < 1. Since d 3 = O(M), relation (**) follows from (*).

11 Representation of the solution M u M (x) = u 0 (x) + G(x, t)σ m (t)dt = S m m=1 M M = u 0 (x) + G(x, x m )Q m + J m. m=1 m=1 Q m := σ m (t)dt, S m J m := [G(x, t) G(x, x m )]σ m (t)dt, S m I m := G(x, x m )Q m. Basic result: J m I m, a 0; x x m a.

12 Impedance bc (boundary condition) For the impedance boundary condition the limiting field u solves the equation: u(x) = u 0 (x) G(x, y)p(y)u(y)dy, D where p(y) = 4πN(y)h(y), ζ m = h(x m) a κ, 0 κ < 1. If the small bodies D m are of arbitrary shape, such that S m = ca 2, then the factor 4π is replaced by the factor c. This factor may depend on m if the small bodies are not identical.

13 Effective field 1 If J m I m, then, as a 0, one has u M (x) = u 0 (x) + M G(x, x m )Q m, x x m a. m=1 Define effective field acting on the m-th particle: u e := u (m) e := u M (x) G(x, t)σ m (t)dt, S m If x x m a, then u e u M as a 0. We prove below that Q m 4πu e (x m ) h(x m )a 2 κ.

14 Effective field 2 The equation for the effective field u e (x), as a 0, is u e (x) = u 0 (x) 4π M G(x, x m )u e (x m ) h m a 2 κ, m=1 where the term with x m is dropped if x x m a, h m := h(x m ). Here h m are known, but u m := u e (x m ) are unknown. To calculate u m one can use a linear algebraic system (LAS): u j = u 0j 4πa 2 κ M m=1,m j G(x j, x m )h m u m, 1 j M. The order of this system is substantially reduced: see next slide.

15 Reduction of the order of LAS. To reduce the order M of this system, consider a partition of D into a union of small cubes p, 1 p P, P M, y p p, diam p d. Then a linear algebraic system (LAS) for u p is P u q = u 0q 4π G(y q, y p )h(y p ) u p N(y p ) p, p q ( ) where 1 q P, P M, u q = u(y q ), u 0q = u 0 (y q ), a 2 κ x m p 1 = N(y p ) p. The LAS (*) is used for efficient numerical solution of many-body scattering problems when the scatterers are small.

16 How efficient can this reduction be? Let the small particles be distributed in a cube with side L = 10 1 m, a = 10 8 m, d = 10 6 m. Then M ( L d )3 = Let the side b of the partition cubes p be b = a 1/6 = m. Then P = ( L b )3 = 10. The reduction of the order M of the LAS in this example is from to 10. Of course, there is a question of the accuracy of approximation of the solution of the original LAS by the solution of the reduced order LAS. If b = a 1/4 = 10 2 m, then P = ( L b )3 = In this case the reduction of the order M of the LAS is from to 10 3.

17 Asymptotic formula for Q m u en ζ m u e + A mσ m σ m ζ m T m σ m = 0 on S m. 2 G(s, t) A m σ m := 2 σ m (t)dt, T m σ m := G(s, t)σ m (t)dt. S m N s S m 1 G(x, y) = [1 + O( x y )], x y 0. 4π x y 4 3 πa3 u e (x m ) ζ m 4πa 2 σ m (t)dt u e (x m ) = Q m +ζ m ds S m S m 4π s t, Aσ m dt = σ m dt, S m S m

18 If ζ m = h(xm) a κ S m ds 4π s t = a 4 3 πa3 u e (x m ) 4πζ m u e (x m )a 2 = Q m (1 + ζ m a). Q m = a3 [ 4π 3 u e(x m ) 4πu e (x m )ζ m a 1 ] 1 + ζ m a, κ < 1, then Q m 4πu e (x m ) h(x m )a 2 κ..

19 Asymptotic formula for σ m u M = u e + σ m S m dt 4π x t = u e + σ ma 2, x x m = O(a). x u en h(x m) a κ u e σ m h(x m) a κ σ m a = 0 σ m = u e N h(x m )u e (x m )a κ 1 + h(x m )a 1 κ, ha 1 κ = o(1), u en a κ. If κ < 1, a 0, then σ m h(x m )u e (x m )a κ.

20 Why is I m J m? G(x, x m )Q m = I m a2 κ d, J m a a 2 κ d d, a d 1. Thus, J m /I m = O(a/d). Consequently, I m J m if a d λ. Formula for calculating the field u M (x) is: M u M (x) = u 0 (x) 4π G(x, x m )h(x m )u e (x m )a 2 κ. m=1 This formula is valid everywhere outside small particles. Since the input from one small particle into u M is not more than O(a 1 κ ), it tends to zero as a 0. Remember that κ [0, 1).

21 Limiting procedure as a 0 4π a 2 κ M G(x, x m )h(x m )u e (x m )a 2 κ = 4π m=1 P G(x, y (p) )h(y (p) )u e (y (p) ) p=1 P 1 = 4π G(x, y (p) )h(y (p) )u e (y (p) )N(y (p) ) p (1+ε p ) x m p p=1 G(x, y)p(y)u(y)dy, p(y) := 4πh(y)N(y). D N ( p ) = a (2 κ) p N(x)dx[1 + o(1)], a 0. u(x) = u 0 (x) G(x, y)p(y)u(y)dy. D

22 An auxiliary lemma Lemma. If f C(D) and x m are distributed in D so that N ( ) = 1 N(x)dx [1 + o(1)], a 0, ϕ(a) for any subdomain D, where ϕ(a) 0 is a continuous, monotone, strictly growing function, ϕ(0) = 0, then lim f(x m )ϕ(a) = f(x)n(x)dx. a 0 x m D Remark:This lemma holds for bounded f with the set of discontinuities of Lebesgue s measure zero. It can be generalized to a class of unbounded f. D

23 Proof of the Lemma Proof. Let D = p p be a partition of D into a union of small cubes p, having no common interior points. Let p denote the volume of p, δ := max p diam p, and y (p) be the center of the cube p. One has lim a 0 x m D f(x m )ϕ(a) = lim a 0 y (p) p f(y (p) )ϕ(a) x m p 1 = lim f(y (p) )N(y (p) ) p [1 + o(1)] = f(x)n(x)dx. a 0 D The last equality holds since the preceding sum is a Riemannian sum for the continuous function f(x)n(x) in the bounded domain D. Thus, the Lemma is proved.

24 New equation for the limiting effective field u(x) = u 0 (x) G(x, y)p(y)u(y)dy, D p(x) = 4πh(x)N(x). Lu := [ 2 + k 2 n 2 (x)]u = 0, k 2 n 2 (x) = k 2 q(x). L = L 0 p(x) := 2 + k 2 q 0 (x) 4πh(x)N(x). The new refraction coefficient n 2 (x) and the new potential q(x) are in 1-to-1 correspondence: n 2 (x) = 1 k 2 q(x); q(x) = q 0 (x) + p(x). k 2 [n 2 0(x) n 2 (x)] = p(x).

25 Creating materials with a desired n 2 (x) Step 1. {n 2 (x), n 2 0(x)} p(x) = k 2[ n 2 0(x) n 2 (x) ]. Step 2. Given p(x) = p 1 + ip 2, find {h(x), N(x)}. Here h(x) = h 1 (x) + ih 2 (x), N(x) 0, h 2 (x) 0. We have p(x) = 4πN(x)h(x). Thus, h 1 (x) = p 1(x) 4πN(x), h 2(x) = p 2(x) 4πN(x). There are many solutions, because N(x) 0 can be arbitrary. For example, one can take N(x) = const > 0.

26 Step 3. Embed N ( p ) = 1 a 2 κ p N(x)dx small particles in p, where p p = D. Physical properties of these particles are given by their boundary impedances ζ m = h(y(p) ) a for all x κ m p. The distance between neighboring particles is d = O(a 2 κ 3 ). Theorem. The resulting new material has the desired function n 2 (x) with the error which tends to 0 as a 0. Remark. The total volume V p of the embedded particles in the limit a 0 is zero. Proof. Since κ 0, one has: V p = lim a 0 O(a 3 /a 2 κ ) = lim a 0 O(a 1+κ ) = 0.

27 Technological problems The first technological problem is: How can one embed many, namely M = M(a), small balls in a given material so that the centers of the balls are points x m distributed as desired? The stereolitography process and chemical methods for growing small particles can be used. The second technological problem is: How does one prepare a ball B m of small radius a with boundary impedance ζ m = h(xm) a, 0 κ < 1, which has a desired frequency κ dependence? Remark: It is not necessary to have large boundary impedance: if 1 κ = 0, or κ = O( ln a ), then ζ m is bounded. However, if κ = 0, then M = O(a 2 ), so more particles have to be embedded.

28 Playing with numbers N 10 6 ; N 1 a 2 κ ; d a (2 κ)/3. N = 10 6 ; κ = 2/3, a = , d = N = 10 6 ; κ = 1/2, a = 10 4, d = The difference between the solution of the limiting integral equation for the effective field and the solution to the linear algebraic system for u e (x m ) is O(1/n), where 1/n is the side of a partition cube.

29 Spatial dispersion. Negative refraction u = a(k)e i[k r ω(k)t], k k + ω(k) ω(k) < δ k v g = k ω(k), v p = ω k k0. k k = k 0 := k k ; ω 2 n 2 c 2 = k 2, ωn c = k. [ n c + ω n c ω ] k ω = k 0. {v g = const v p, const > 0} negative refraction. n + ω n ω < 0

30 Isotropic medium If ω > 0, ω = ω(k), k := k, then v p v g = ω (k) ω k that ω ( k ) < 0. Indeed, < 0, provided v g := k ω(k) = ω (k)k 0, k ω(k) v p = ω (k) ω k, v p := ω k k0 k0 := k/k. Terminology : Negative refraction means v g is directed opposite to v p ; Negative index means that ɛ < 0 and µ < 0.

31 Inverse scattering with data at fixed energy and fixed incident direction This theory is a basis for preparing materials with a desired radiation pattern. [ 2 + k 2 q(x)]u = 0 in R 3, u = e ikα x + v := u 0 + v, ( ) v = A(β) eikr 1 + o, r = x, r r A(β) = 1 e ikβ x h(x)dx, 4π D x r := β, h(x) := q(x)u(x, α). Here α is a unit vector in the direction of propagation of the incident wave, A(β) is the scattering amplitude. We assume that α and k > 0 are fixed.

32 Inverse Scattering Problem with fixed energy and fixed incident direction IP(inverse problem): Given f(β) L 2 (S 2 ), α S 2, k > 0, and ɛ > 0, D R 3 ( a bounded domain), find q L 2 (D) such that f(β) A(β) L 2 (S 2 ) < ɛ. (1) A priori it is not clear that this problem has a solution. We prove that it has a solution. If this IP has a solution, then it has infinitely many solutions because small variations of q lead to small variations of A(β).

33 Claim 1. The set { D e ikβ x h(x)dx} h L 2 (D) is dense in L 2 (S 2 ) Corollary 1. Given f L 2 (S 2 ) and ɛ > 0, one can find h L 2 (D) such that f(β) + 1 4π D e ikβ x h(x)dx < ɛ. Claim 2. The set {q(x)u(x, α)} q L 2 (D) is dense in L 2 (D). Corollary 2. Given h L 2 (D) and ɛ > 0, one can find q L 2 (D) such that h(x) q(x)u(x, α) L 2 (D) < ɛ. Since the scattering amplitude A(β) = 1 4π D e ikβ x h(x)dx depends continuously on h, the inverse problem IP is solved by Claims 1,2.

34 Proof of Claim 1 Assume the contrary. Then ψ L 2 (S 2 ) such that 0 = dβψ(β) e ikβ x h(x)dx h L 2 (D). S 2 D Thus, dβψ(β)e ikβ x = 0 x R 3. S 2 Therefore, dλλ 2 dβe iλβ x δ(λ k) ψ(β) 0 S 2 k 2 = 0 x R 3. By the injectivity of the Fourier transform, one gets δ(λ k) ψ(β) k 2 = 0. Therefore, ψ(β) = 0. Claim 1 is proved.

35 Proof of Claim 2 Given h L 2 (D), define u := u 0 g(x, y)h(y)dy, g := eik x y 4π x y, (2) D q(x) := h(x) u(x). (3) If q L 2 (D), then this q solves the problem, and u, defined in (2), is the scattering solution: u = u 0 g(x, y)q(y)u(y)dy, (4) and D A(β) = 1 e ikβ y h(y)dy. 4π D

36 If q is not in L 2 (D), then the null set N := {x : x D, u(x) = 0} is non-void. Let N δ := {x : u(x) < δ, x D}, D δ := D \ N δ. { h, in Dδ, Claim 3. h δ = such that h 0, in N δ, δ h L 2 (D) < cɛ, { hδ q δ := u δ, in D δ, q δ L (D), u δ := u 0 0, in N δ, D gh δdy. Proof of Claim 3. The set N is, generically, a line l = {x : u 1 (x) = 0, u 2 (x) = 0}, where u 1 = Ru and u 2 = Iu. Consider a tubular neighborhood of this line, ρ(x, l) δ. Let the origin O be chosen on l, s 3 be the Cartesian coordinate along the tangent to l, and s 1 = u 1, s 2 = u 2 are coordinates in the plane orthogonal to l, s j -axis is directed along u j l, j = 1, 2.

37 The Jacobian J of the transformation (x 1, x 2, x 3 ) (s 1, s 2, s 3 ) is nonsingular, J + J 1 c, because u 1 and u 2 are linearly independent. { Define h, in Dδ, h δ := u 0, in N δ, δ := u 0 D g(x, y)h δ(y)dy, { hδ q δ := u δ, in D δ, 0, in N δ. One has u δ = u 0 D ghdy + D g(x, y)(h h δ)dy, u δ (x) u(x) c N δ dy 4π x y δ I(δ), x D δ, c = max x N δ h(x). If one proves, that I(δ) = o(δ), δ 0, x D δ then q δ L (D), and Claim 3 is proved.

38 Claim 4: Proof of Claim 4. dy x y dy N δ y = c 1 N δ I(δ) = O(δ 2 ln(δ) ), δ 0. c2 δ 0 1 ds 3 ρ dρ ρ 2 + s c2 δ c2 = c 1 dρρ ln(s 3 + ρ 2 + s 23 δ ) 10 c 3 ρ ln 0 O(δ 2 ln(δ) ). 0 ( ) 1 dρ ρ

39 The condition u j l c > 0, j = 1, 2, implies that a tubular neighborhood of the line l, N δ = {x : u u 2 2 δ}, is included in a region {x : x c 2 δ} and includes a region {x : x c 2δ}. This follows from the estimates c 2ρ u(x) = u(ξ) (x ξ) c 2 ρ. Here ξ l, x is a point on a plane passing through ξ and orthogonal to l, ρ = x ξ, and δ > 0 is sufficiently small, so that the terms of order ρ 2 are negligible, c 2 = max ξ l u(ξ), c 2 = min ξ l u(ξ). Claim 4, and, therefore, Claim 2 are proved.

40 Calculation of h given f(β) and ɛ > 0 1. Let {φ j } be a basis in L 2 (D), Consider the problem: h n = ψ j (β) := 1 4π f(β) n j=1 n j=1 D c (n) j φ j, e ikβ x φ j (x)dx. c (n) j ψ j (β) = min. (5) A necessary condition for (5) is a linear system for c (n) j.

41 An analytical solution. 2. Let D = {x : x 1} := B, or B D, h = 0 in D \ B. One has: ( 1) l+1 f l,m h lm =, l L, π 2k g 1,l+ 1 (k) 2 0, l > L, (6) where g µ,ν (k) = 1 0 xµ+ 1 2 J ν (kx)dx (Bateman-Erdelyi book, formula (8.5.8)) and L is chosen so that f l,m 2 < ɛ 2. l>l

42 References I [0 ] A.G.Ramm, Scattering of Acoustic and Electromagnetic Waves by Small Bodies of Arbitrary Shapes.Applications to Creating New Engineered Materials, Momentum Press, New York, [1 ] A.G.Ramm, Wave scattering by small bodies of arbitrary shapes, World Sci. Publishers, Singapore, [2 ] A.G.Ramm, Inverse problems, Springer, New York, [3 ] A.G.Ramm, Distribution of particles which produces a smart material, J. Stat. Phys., 127, N5, (2007), [4 ] A.G.Ramm, Many-body wave scattering by small bodies and applications, J. Math. Phys., 48, N10, , (2007), pp [5 ] A.G.Ramm, Scattering by many small bodies and applications to condensed matter physics, Europ. Phys. Lett., 80, (2007), [6 ] A.G.Ramm, Inverse scattering problem with data at fixed energy and fixed incident direction, Nonlinear Analysis: Theory, Methods and Applications, 69, N4, (2008),

43 References II [7 ] A.G.Ramm, A recipe for making materials with negative refraction in acoustics, Phys. Lett A, 372/13, (2008), [8 ] A.G.Ramm, Distribution of particles which produces a desired radiation pattern, Physica B, 394, N2, (2007), [9 ] A.G.Ramm, Wave scattering by many small particles embedded in a medium, Phys. Lett. A, 372/17, (2008), [10 ] A.G.Ramm, Materials with the desired refraction coefficients can be made by embedding small particles, Phys. Lett. A, 370, 5-6, (2007), [11 ] A.G.Ramm, Wave scattering by small impedance particles in a medium, Phys. Lett. A 368, N1-2,(2007), [12 ] A.G.Ramm, Electromagnetic wave scattering by many conducting particles, Journ.Physics A, 41, (2008), [13 ] A.G.Ramm, Electromagnetic wave scattering by many small bodies, Phys. Lett. A, 372/23, (2008),

44 References III [14 ] A.G.Ramm, Creating wave-focusing materials, LAJSS (Latin-Amer. Jour. Solids and Structures), 5, (2008), [15 ] A.G.Ramm, Creating materials with desired properties, Oberwolfach Workshop Material properties, report 58/2007, pp [16] A.G.Ramm, A method for creating materials with a desired refraction coefficient, Internat. Journ. Mod. Phys B, 24, 27, (2010), [17] A.G.Ramm, Materials with a desired refraction coefficient can be created by embedding small particles into the given material, International Journal of Structural Changes in Solids (IJSCS), 2, N2, (2010), [18] A.G.Ramm, Electromagnetic wave scattering by many small bodies and creating materials with a desired refraction coefficient, Progress in Electromagnetic Research M (PIER M), 13, (2010), [19] A.G.Ramm, Electromagnetic wave scattering by many small perfectly conducting particles of an arbitrary shape, Optics Communications, 285, N18, (2012),

45 References IV [20] A.G.Ramm, Electromagnetic wave scattering by small impedance particles of an arbitrary shape, J. Appl. Math and Comput., (JAMC), 43, N1, (2013), [21] A.G.Ramm, Many-body wave scattering problems in the case of small scatterers, J. of Appl. Math and Comput., (JAMC), 41, N1, (2013), [22] A.G.Ramm, Scattering of electromagnetic waves by many nano-wires, Mathematics, 1, (2013), Open access Journal: [23] A.G.Ramm, N.Tran, A fast algorithm for solving scalar wave scattering problem by billions of particles, Jour. of Algorithms and Optimization, 3, N1, (2015), Open access Journal

46 References V [24] A.G.Ramm, M.I.Andriychuk, Application of the asymptotic solution to EM wave scattering problem to creating medium with a prescribed permeability, Journ. Appl. Math. and Computing, Application of the asymptotic solution to EM wave scattering problem to creating medium with a prescribed permeability, Journ Appl. Math. and Computing, (JAMC), 45, (2014), doi: /s [25] A.G.Ramm, M.I.Andriychuk, Calculation of electromagnetic wave scattering by a small impedance particle of an arbitrary shape, Math. Meth. in Natur. Phenomena, Calculation of electromagnetic wave scattering by a small impedance particle of an arbitrary shape, Math. Meth. in Natur. Phenomena (MMNP), 9, N5, (2014),

Inverse scattering problem with underdetermined data.

Math. Methods in Natur. Phenom. (MMNP), 9, N5, (2014), 244-253. Inverse scattering problem with underdetermined data. A. G. Ramm Mathematics epartment, Kansas State University, Manhattan, KS 66506-2602,