REFRACTION PROBLEMS IN GEOMETRIC OPTICS March 13, 2015

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1 REFRACTION PROBLEMS IN GEOMETRIC OPTICS March 13, 2015 CRISTIAN E. GUTIÉRREZ Contents 1. Introduction 2 2. Snell s law of refraction In vector form erivation of the Snell law Surfaces with the uniform refracting property: far field case Uniform refraction: near field case Case 0 < κ < Case κ > Optimal mass transportation Application to the refractor problem κ < Refractor measure and weak solutions Solution of the refractor problem Solution to the refractor problem for κ < 1 with the Minkowski method Existence of solutions in the discrete case Solution in the general case Uniqueness discrete case Maxwell s equations General case Maxwell equations in integral form Boundary conditions at a surface of discontinuity Maxwell s equations in the absence of charges The wave equation ispersion equation Plane waves Fresnel formulas Rewriting the Fresnel Equations The Poynting vector Polarization Estimation of the Fresnel coefficients Application to the far field refractor problem with loss of energy 48 References 51 March 13, 2015 The author was partially supported by NSF grant MS

2 2 C. E. GUTIÉRREZ 1. Introduction These are notes expanding the material of a mini course I taught on Monge- Ampère type equations and geometric optics in June 2012 at Cetraro. The purpose of these lectures was to explain basic facts about the Monge-Ampère equation and the application of these ideas to pose and solve problems in geometric optics concerning refraction with input and output energies. These problems are basically of two types: the far field and the near field. In the far field case the goal is to send radiation into a set of directions and in the second is to send radiation to a specific target set. The far field case can be treated with optimal transportation methods, Section 3, but the near field does not. Instead in this case, a method based on the Minkowski method in convex geometry can be used. In fact, with this method one can treat both far and near fields problems. The method is illustrated in Section 4 in the far field case. Refraction and reflection occur simultaneously, in other words, if an incident ray is refracted there is always a percentage of the incident ray that is internally reflected. The fractions of energy refracted and reflected are given by the Fresnel formulas. Beginning with Maxwell s equations this is developed and explained in Section 5. The application to the refraction problem with loss of energy is described in Subsection These notes include the some of my work with Qingbo Huang and Henok Mawi, and the relevant references are [GH09], [GH08], and [GH13]. The equations describing the solutions to these problems are Monge-Ampère type equations whose derivation is included in [GH08], and [GH13]. The notes are organized as follows. Section 2 contains the Snell law of refraction in vector form. This is simple but essential for the developments of the results. In particular, we introduce in this section the notion of surfaces having the uniform refraction property used later to give the definition of refractors. Section 3 contains basic facts about optimal mass transport and its application to solve the refractor problem in the far field case. Section 4 introduces a method to solve the far field refractor problem that has its roots in the Minkowski method. Section 5 contains a detailed development of Maxwell s equations, the so called boundary conditions explaining the propagation across two different materials, and its application to deduce the Fresnel formulas. These formulas are written in a convenient form that is used finally to propose, in Subsection 5.13, a model for the refractor problem with loss of energy in internal reflection. It is a pleasure to thank the CIME for their support in the organization of this series of lectures and in particular, to Elvira Mascolo and Pietro Zecca for their great support and help that made this series possible. 2. Snell s law of refraction 2.1. In vector form. Suppose Γ is a surface in R 3 that separates two media I and II that are homogeneous and isotropic. Let v 1 and v 2 be the velocities of propagation of light in the media I and II respectively. The index of refraction of the medium I is by definition n 1 = c/v 1, where c is the velocity of propagation of light in the

3 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 3 vacuum, and similarly n 2 = c/v 2. If a ray of light having direction x S 2 and traveling through the medium I hits Γ at the point P, then this ray is refracted in the direction m S 2 through the medium II according with the Snell law in vector form: 2.1) n 1 x ν) = n 2 m ν), where ν is the unit normal to the surface to Γ at P going towards the medium II. The derivation of this formula is in Subsection 2.2. This has several consequences: a) the vectors x, m, ν are all on the same plane called plane of incidence); b) the well known Snell law in scalar form n 1 sin θ 1 = n 2 sin θ 2, where θ 1 is the angle between x and ν the angle of incidence), θ 2 the angle between m and ν the angle of refraction), From 2.1), n 1 x n 2 m) ν = 0, which means that the vector n 1 x n 2 m is parallel to the normal vector ν. If we set κ = n 2 /n 1, then 2.2) x κ m = λν, for some λ R. Taking dot products we get λ = cos θ 1 κ cos θ 2, cos θ 1 = x ν > 0, and cos θ 2 = m ν = 1 κ 2 [1 x ν) 2 ], so 2.3) λ = x ν κ 1 κ 2 1 x ν) 2. Refraction behaves differently for κ < 1 and for κ > κ < 1. This means v 1 < v 2, and so waves propagate in medium II faster than in medium I, or equivalently, medium I is denser than medium II. In this case the refracted rays tend to bent away from the normal, that is the case for example, when medium I is glass and medium II is air. Indeed, from the scalar Snell law, sin θ 1 = κ sin θ 2 < sin θ 2 and so θ 1 < θ 2. For this reason, the maximum angle of refraction θ 2 is π/2 which, from the Snell law in scalar form, is achieved when sin θ 1 = n 2 /n 1 = κ. So there cannot be refraction when the incidence angle θ 1 is beyond this critical value, that is, we must have 0 θ 1 θ c = arcsin κ. Once again from the Snell law in scalar form, 2.4) θ 2 θ 1 = arcsinκ 1 sin θ 1 ) θ 1 and it is easy to verify that this quantity is strictly increasing for θ 1 [0, θ c ], and therefore 0 θ 2 θ 1 π 2 θ c. We then have x m = cosθ 2 θ 1 ) cosπ/2 θ c ) = κ, and therefore obtain the following physical constraint for refraction: 2.5) if κ = n 2 /n 1 < 1 and a ray of direction x through medium I is refracted into medium II in the direction m, then m x κ. Since the refraction angle depends on the frequency of the radiation, we assume radiation is monochromatic. If θ 1 > θ c, then the phenomenon of total internal reflection occurs, see Figure 1c).

4 4 C. E. GUTIÉRREZ Notice also that in this case λ > 0 in 2.3). Conversely, given x, m S 2 with x m κ and κ < 1, it follows from 2.4) that there exists a hyperplane refracting any ray through medium I with direction x into a ray of direction m in medium II. x Θ 1 x Θ c Θ 2 m Π 2 m Ν a) x κm parallel to ν Ν b) Critical angle Θ c x m Ν c) Internal reflection Figure 1. Snell s law κ < 1, e.g., glass to air κ > 1. In this case, waves propagate in medium I faster than in medium II, and the refracted rays tend to bent towards the normal. By the Snell law, the maximum angle of refraction denoted by θ c is achieved when θ 1 = π/2, and θ c = arcsin1/κ). Once again from the Snell law in scalar form 2.6) θ 1 θ 2 = arcsinκ sin θ 2 ) θ 2 which is strictly increasing for θ 2 [0, θ c], and 0 θ 1 θ 2 π 2 θ c. We therefore obtain the following physical constraint for the case κ > 1: 2.7) if a ray with direction x traveling through medium I is refracted into a ray in medium II with direction m, then m x 1/κ.

5 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 5 Notice also that in this case λ < 0 in 2.3). On the other hand, by 2.6), if x, m S 2 with x m 1/κ and κ > 1, then there exists a hyperplane refracting any ray of direction x through medium I into a ray with direction m in medium II. We summarize the above discussion on the physical constraints of refraction in the following lemma. Lemma 2.1. Let n 1 and n 2 be the indices of refraction of two media I and II, respectively, and κ = n 2 /n 1. Then a light ray in medium I with direction x S 2 is refracted by some surface into a light ray with direction m S 2 in medium II if and only if m x κ, when κ < 1; and if and only if m x 1/κ, when κ > κ = 1. This corresponds to reflection. It means 2.8) x m = λ ν. Taking dot products with x and then with m yields 1 m x = λ x ν and x m 1 = λ m ν, then x ν = m ν. Also taking dot product with x in 2.8) then yields λ = 2 x ν. Therefore m = x 2x ν)ν erivation of the Snell law. At time t, ψx, y, z, t) = 0 denotes a surface that separates the part of the space that is at rest with the part of the space that is disturbed by the electric and magnetic fields. For each t fixed the surface defined by ψx, y, z, t) = 0 is called a wave front. The light rays are the orthogonal trajectories to the wave fronts. We assume that ψ t 0 and so we can solve ψx, y, z, t) = 0 in t obtaining that φx, y, z) = c t, where c is the speed of light in vacuum. Therefore, it t runs, then we get that the wave fronts are the level sets of the function φx, y, z). Let us assume that the wave fronts travel in an homogenous and isotropic medium I with refractive index n 1 = c/v 1 ), v 1 is the speed of propagation in medium I. This wave front is transmitted to another homogeneous and isotropic medium II having refractive index n 2. Let Σ be the surface in 3-d separating the media I and II, and suppose it is given by the equations x = f ξ, η), y = gξ, η) and z = hξ, η). Let φ 1 x, y, z) = c t be the wave front in medium I and φ 2 x, y, z) = c t be the wave front, to be determined, in medium II. On the surface Σ the two wave fronts agree, that is, φ 1 f ξ, η), gξ, η), hξ, η) ) = φ2 f ξ, η), gξ, η), hξ, η) ). ifferentiating this equation with respect to ξ and η yields φ1 x φ ) 2 φ1 f ξ + x y φ ) 2 φ1 g ξ + y z φ ) 2 h ξ = 0 z φ1 x φ ) 2 φ1 f η + x y φ ) 2 φ1 g η + y z φ ) 2 h η = 0. z

6 6 C. E. GUTIÉRREZ This means that the vector φ 1 φ 2 is perpendicular to both vectors f ξ, g ξ, h ξ ) and f η, g η, h η ), and therefore it is normal to the surface Σ. Let ν be the outer normal at the surface Σ. Then we have ) 2.9) φ1 φ 2 = λ ν, for some scalar λ. A light ray γ 1 t) in medium I has constant speed v 1 and a light ray γ 2 t) in II constant speed v 2. Say γ 1 t) is the light ray in medium I and γ 2 t) the light ray in medium II. So we have φ 1 γ 1 t)) = c t and φ 2 γ 2 t)) = c t. ifferentiating with respect to t yields φ i γ i t)) γ t) = c, i = 1, 2. Let θ 1 i be the angle between the vectors φ i γ i t)), γ t). Since the rays are orthogonal trajectories we get that 1 θ i = 0 for i = 1, 2. If γ i is parametrized so that γ t) = v i i, we then obtain that φ i γ i t)) = c v i = n i. If we let x = φ 1γ 1 t)) φ 1 γ 1 t)) and m = φ 2γ 2 t)), we then obtain from 2.9) that φ 2 γ 2 t)) which is equivalent to 2.1). n 1 x n 2 m = λ ν 2.3. Surfaces with the uniform refracting property: far field case. Let m S 2 be fixed, and we ask the following: if rays of light emanate from the origin inside medium I, what is the surface Γ, interface of the media I and II, that refracts all these rays into rays parallel to m? Suppose Γ is parameterized by the polar representation ρx)x where ρ > 0 and x S 2. Consider a curve on Γ given by rt) = ρxt)) xt) for xt) S 2. According to 2.2), the tangent vector r t) to Γ satisfies r t) xt) κ m) = 0. That is, [ρxt))] xt) + ρxt))x t) ) xt) κ m) = 0, which yields ρxt))1 κm xt)) ) = 0. Therefore 2.10) ρx) = b 1 κ m x for x S 2 and for some b R. To understand the surface given by 2.10), we distinguish two cases κ < 1 and κ > Case κ = 1. When κ = 1 we see this is a paraboloid. Indeed, let m = e n, then a point X = ρx)x is on the surface 2.10) if X = b x n. The distance from X to the plane x n = b is b x n, and the distance from X to 0 is X. So this is a paraboloid with focus at 0, directrix plane x n = b and axis in the direction e n Case κ < 1. For b > 0, we will see that the surface Γ given by 2.10) is an ellipsoid of revolution about the axis of direction m. Suppose for simplicity that m = e n, the nth-coordinate vector. If y = y, y n ) R n is a point on Γ, then y = ρx)x

7 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 7 with x = y/ y. From 2.10), y κ y n = b, that is, y 2 + y 2 n = κ y n + b) 2 which yields y κ 2 )y 2 n 2κby n = b 2. This surface Γ can be written in the form 2.11) ) 2 y y 2 n κb 1 κ 2 ) 2 + ) 2 = 1 b b 1 κ 2 1 κ 2 which is an ellipsoid of revolution about the y n axis with foci 0, 0) and 0, 2κb/1 y κ 2 )). Since y = κy n + b and the physical constraint for refraction 2.5), y e n κ is equivalent to y n κb. That is, for refraction to occur y must be in the upper 1 κ2 part of the ellipsoid 2.11); we denote this semi-ellipsoid by Ee n, b), see Figure 2. To verify that Ee n, b) has the uniform refracting property, that is, it refracts any ray emanating from the origin in the direction ) e n, we check that 2.2) holds ) at each y y y point. Indeed, if y Ee n, b), then y κe n 1 κ > 0, and y y κe n e n 0, and so y y κe n is an outward normal to Ee n, b) at y. Rotating the coordinates, it is easy to see that the surface given by 2.10) with κ < 1 and b > 0 is an ellipsoid of revolution about the axis of direction m with foci 0 and 2κb m. Moreover, the semi-ellipsoid Em, b) given by 1 κ2 2.12) Em, b) = {ρx)x : ρx) = b 1 κ m x, x Sn 1, x m κ}, has the uniform refracting property, any ray emanating from the origin O is refracted in the direction m Case κ > 1. ue to the physical constraint of refraction 2.7), we must have b < 0 in 2.10). efine for b > ) Hm, b) = {ρx)x : ρx) = b κ m x 1, x Sn 1, x m 1/κ}. We claim that Hm, b) is the sheet with opening in direction m of a hyperboloid of revolution of two sheets about the axis of direction m. To prove the claim, set for simplicity m = e n. If y = y, y n ) He n, b), then y = ρx)x with x = y/ y. From 2.13), κ y n y = b, and therefore y 2 + y 2 n = κ y n b) 2 which yields y 2 κ 2 1) y n κb κ 2 1 ) 2 κb κ 2 1 ) 2 = b 2. Thus, any point y on He n, b)

8 8 C. E. GUTIÉRREZ Figure 2. Only half of the ellipse refracts in the direction m = e n satisfies the equation 2.14) y n κb ) 2 κ 2 1 y 2 ) 2 ) 2 = 1 b b κ 2 1 κ2 1 which represents a hyperboloid of revolution of two sheets about the y n axis with foci 0, 0) and 0, 2κb/κ 2 1)). Moreover, the upper sheet of this hyperboloid of revolution is given by y n = κb κ b 1 + κ 2 1 y 2 b/ κ2 1 ) 2

9 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 9 b and satisfies κy n b > 0, and hence has polar equation ρx) = κ e n x 1. Similarly, b the lower sheet satisfies κy n b < 0 and has polar equation ρx) = κ e n x + 1. For a general m, by a rotation, we obtain that Hm, b) is the sheet with opening in direction m of a hyperboloid of revolution of two sheets about the axis of direction m with foci 0, 0) and 2κb κ 2 1 m. Notice that the focus 0, 0) is outside the region enclosed by Hm, b) and the 2κb y focus m is inside that region. The vector κm is an inward normal to κ 2 1 y Hm, b) at y, because by 2.13) κm y ) y 2κb κ 2 1 m y ) 2κ2 b κ 2 1 2κb κm y + y κ 2 1 = 2κb bκ 1) b = κ + 1 κ + 1 > 0. Clearly, κm y ) m κ 1 and κm y ) y > 0. Therefore, Hm, b) satisfies y y y the uniform refraction property. We remark that one has to use H e n, b) to uniformly refract in the direction e n, and due to the physical constraint 2.7), the lower sheet of the hyperboloid of equation 2.14) cannot refract in the direction e n. From the above discussion, we have proved the following. Lemma 2.2. Let n 1 and n 2 be the indexes of refraction of two media I and II, respectively, and κ = n 2 /n 1. Assume that the origin O is inside medium I, and Em, b), Hm, b) are defined by 2.12) and 2.13), respectively. We have: i) If κ < 1 and Em, b) is the interface of media I and II, then Em, b) refracts all rays emitted from O into rays in medium II with direction m. ii) If κ > 1 and Hm, b) separates media I and II, then Hm, b) refracts all rays emitted from O into rays in medium II with direction m Uniform refraction: near field case. The question we ask is: given a point O inside medium I and a point P inside medium II, find an interface surface S between media I and II that refracts all rays emanating from the point O into the point P. Suppose O is the origin, and let Xt) be a curve on S. By the Snell law of refraction the tangent vector X t) satisfies Xt) X t) Xt) κ P Xt) ) = 0. P Xt) That is, Xt) + κ P Xt) = 0.

10 10 C. E. GUTIÉRREZ Therefore S is the Cartesian oval 2.15) X + κ X P = b. Since f X) = X + κ X P is a convex function, the oval is a convex set. We need to find and analyze the polar equation of the oval. Write X = ρx)x with x S n 1. Then writing κ ρx)x P = b ρx), squaring this quantity and solving the quadratic equation yields 2.16) ρx) = b κ2 x P) ± b κ 2 x P) 2 1 κ 2 )b 2 κ 2 P 2 ) 1 κ 2. Set 2.17) t) = b κ 2 t) 2 1 κ 2 )b 2 κ 2 P 2 ) Case 0 < κ < 1. We have 2.18) x P) > κ 2 x P b) 2, if x P < P. If b P, then O and P are inside or on the oval, and so the oval cannot refract rays to P. If the oval is non empty, then κ P b. In case κ P = b, the oval reduces to the point O. The only interesting case is then κ P < b < P. From the equation of the oval we get that ρx) b. So we now should decide which values ± to take in the definition of ρx). Let ρ + and ρ be the corresponding ρ s. We claim that ρ + x) > b and ρ x) b. Indeed, ρ + x) = b κ2 x P) + x P) 1 κ 2 b κ2 x P) + κ b x P 1 κ 2 = b + κ2 b x P) + κ b x P 1 κ 2 b. The equality ρ + x) = b holds only if x P = P and b = x P. So ρ + x) > b if κ P < b < P. Similarly, ρ x) = b κ2 x P) x P) 1 κ 2 b κ2 x P) κ b x P 1 κ 2 = b + κ2 b x P) κ b x P 1 κ 2 b. So the claim is proved. Therefore the polar equation of the oval is then given by 2.19) hx, P, b) = ρ x) = b κ2 x P) x P) 1 κ 2.

11 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 11 ) P hx, P, b)x From the physical constraint for refraction we must have x κ, P hx, P, b)x and from the equation of the oval we then get that to have refraction we need 2.20) x P b. Concluding with the case κ < 1, given P R n and κ P < b < P, keeping in mind 2.19) and 2.20), a refracting oval is the set where OP, b) = {hx, P, b) x : x S n 1, x P b} hx, P, b) = b κ2 x P) b κ 2 x P) 2 1 κ 2 )b 2 κ 2 P 2 ) 1 κ 2. 0 P 0 P a) X + 2/3 X P = , P = 2, 0) b) X + 2/3 X P = 1.7, P = 2, 0) Figure 3. Cartesian ovals κ < 1, e.g., glass to air Remark 2.3. If P, then the oval converges to an ellipsoid which is the surface having the uniform refraction property in the far field case, see Subsection

12 12 C. E. GUTIÉRREZ In fact, if m = P/ P and b = κ P + C with C positive constant we have b 2 κ 2 P 2 hx, P, b) = b κ 2 x P + x P) C2κ P + C) = κ P κ 2 x m P + C) + κ P κ 2 x m P + C) 2 1 κ 2 )C2κ P + C) 2κC κ κ 2 x m) + κ κ 2 x m) = C 2 1 κx m as P Case κ > 1. In this case we must have P b, and in case b = P the oval reduces to the point P. Also b < κ P, since otherwise the points 0, P are inside the oval or 0 is on the oval, and therefore there cannot be refraction if b κ P. So to have refraction we must have P < b < κ P and so the point P is inside the oval and 0 is outside the oval. Rewriting ρ in 2.16) we get that ρ ± x) = κ2 x P b) ± κ 2 x P b) 2 κ 2 1)κ 2 P 2 b 2 ) κ 2 1 for x P) 0 which amounts x P b + κ 2 1)κ 2 P 2 b 2 ) κ 2. Notice that ρ ± x) < 0 for κ 2 x P b < 0. We have that ρ x) ρ + x) κ2 P b) + P ) = κ 2 1 κ P + b < b. To have refraction, by the physical constraint we need to have κ + 1 P xρ ± x) x P xρ ± x) 1/κ, which is equivalent to κ2 x P b κ 2 1)ρ ± x). Therefore, the physical constraint is satisfied only by ρ. Therefore if κ > 1, refraction only occurs when P < b < κ P, and the refracting piece of the oval is then given by 2.21) OP, b) = with b + hx, P, b)x : x P κ2 1)κ 2 P 2 b 2 ) 2.22) hx, P, b) = ρ x) = κ2 x P b) κ 2 x P b) 2 κ 2 1)κ 2 P 2 b 2 ). κ 2 1 Remark 2.4. If P, then the oval OP, b) converges to the semi hyperboloid appearing in the far field refraction problem when κ > 1, see Subsection Indeed, let m = P P Sn 1 and b = κ P a with a > 0 a constant. For x ΓP, b) we κ 2

13 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 13 have b + κ 2 1)κ 2 P 2 b 2 ) κ 2 P = κ P a + κ 2 1)κ 2 P 2 κ P a) 2 ) κ 2 P as P. On the other hand, if x m > 1/κ, we get hx, P, b) = as P. a2κ P a) 1 κ κ 2 P x m κ P + a) + κ 2 P x m κ P + a) 2 κ 2 1)a2κ P a) a2κ κ 2 x m κ + κ 2 x m κ) = a 2 κx m 1, 0 P 0 P a) X + 3/2 X P = , P = 2, 0) b) X + 3/2 X P = 2.7, P = 2, 0) Figure 4. Cartesian ovals κ > 1, e.g., air to glass 3. Optimal mass transportation Let, be two domains on S n 1 or domains in a manifold might be contained in one manifold and in another) with = 0. Let N be a multi-valued mapping from onto such that Nx) is singlevalued a.e. on. We also assume that {x : Nx) = } = 0. For F, we set T F) = N 1 F) = {x : Nx) F }. We say N is measurable if T F) is Lebesgue measurable for any Borel set F.

14 14 C. E. GUTIÉRREZ For example, if N = u with u convex, then N is measurable. Because for u convex we have m ux 0 ) iff x 0 u m), where u is the Legendre transformation of u see exercises). Therefore u) 1 F) = u F), and since u F) is Lebesgue measurable for each Borel set F, we get that u is measurable. Suppose g L 1 ) is nonnegative and Γ on is a finite Radon measure satisfying the conservation condition 3.1) gx) dx = Γ ) > 0. Notice that the set function µf) = gx) dx is Borel measure because N is T F) single valued a.e., and measurable. Indeed, if S is the set of measure zero such that Nx) is single valued for all x \ S, and F j is a sequence of disjoint Borel sets, then T F i ) T F j ) = 0 for i j, then µ is σ-additive. Therefore µ is a finite Borel measure and so is regular. We say N is measure preserving from gx)dx to Γ if for any Borel F 3.2) gx) dx = ΓF). T F) Lemma 3.1. N is a measure preserving mapping from gx)dx to Γ if and only if for any v C ) 3.3) vnx))gx) dx = vm) dγm). We remark that vnx)) is well defined for x \S where Nx) is single-valued on \ S and S = 0, and vnx))gx) dx is understood as vnx))gx) dx. \S Proof. Let N be a measure preserving mapping. To show 3.3), it suffices to prove it for v = χ F, the characteristic function of a Borel set F, because for each v continuous there exists a sequence of simple functions converging uniformly to v. It is easy to verify that χ T F) x) = χ F Nx)) for x \ S. Therefore by 3.2) χ F m) dγ = g dx = χ F Nx))gx) dx. \S T F) \S) To prove the converse, assume that 3.3) holds. We will show that for any Borel set E 3.4) g dx ΓE). T E) Indeed, let us first assume that E = G is open, then given a compact set K G, choose v C ) such that 0 v 1, v = 1 on K, and v = 0 outside G. By 3.3), one gets gx) dx vnx))gx) dx ΓG), T K)

15 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 15 for each compact K G. Since µ is regular, 3.4) follows for E open. For a general Borel set E, since Γ is also regular, given ɛ > 0 there exists G open E G with ΓG \ E) < ɛ. Then gx) dx gx) dx ΓG) = ΓE) + ΓG \ E) < ΓE) + ɛ T E) T G) and so 3.4) follows. We next prove that equality holds in 3.4). First notice that {x : Nx) } T F)) c T F c ), for any set F. Then applying 3.4) to \ F with F Borel set yields gx) dx gx) dx Γ \ F) = Γ ) ΓF). {x :Nx) } T F)) c T F c ) Since {x : Nx) = } = 0, we have gx) dx = gx) dx = gx) dx gx) dx. {x :Nx) } T F)) c T F)) c T F) So from the conservation condition 3.1), we obtain the reverse inequality in 3.4). Consider the general cost function cx, m) Lip ), the space of Lipschitz functions on, and the set of admissible functions K = {u, v) : u C), v C ), ux) + vm) cx, m), x, m }. efine the dual functional I for u, v) C) C ) Iu, v) = ux)gx) dx + vm) dγ, and define the c- and c - transforms u c m) = inf [cx, m) ux)], m ; v c x) = inf [cx, m) vm)], x. x m efinition 3.2. A function φ C) is c-concave if for x 0, there exist m 0 and b R such that φx) cx, m 0 ) b on with equality at x = x 0. Obviously v c is c-concave for any v C ). We collect the following properties: 1) For any u C) and v C ), v c Lip) and u c Lip ) with Lipschitz constants bounded uniformly by the Lipschitz constant of c. Indeed, x 0 the point where the minimum is attained) u c m 1 ) u c m 2 ) u c m 1 ) cx 0, m 2 ) ux 0 )) cx 0, m 1 ) ux 0 ) cx 0, m 2 ) + ux 0 ) K m 1 m 2. 2) If u, v) K, then vm) u c m) and ux) v c x). Also v c, v), u, u c ) K.

16 16 C. E. GUTIÉRREZ 3) φ is c-concave iff φ = φ c ) c. Indeed, if φx) cx, m 0 ) b on and the equality holds at x = x 0, then b = φ c m 0 ). So φx 0 ) = cx 0, m 0 ) φ c m 0 ) which yields φx 0 ) φ c ) c x 0 ). On the other hand, from the definitions of c and c transforms we always have that φ c ) c φ for any φ. efinition 3.3. Given a function φx), the c-normal mapping of φ is defined by N c,φ x) = {m : φx) + φ c m) = cx, m)}, for x, and T c,φ m) = N 1 c,φ m) = {x : m N c,φx)}. 3.5) We assume that the cost function cx, m) satisfies the following: For any c-concave function φ, N c,φ x) is single-valued a.e. on and N c,φ is Lebesgue measurable. Notice that if cx, m) = x m, then N c,φ x) = φx), where φ is the superdifferential of φ φx) = {m R n : φy) φx) + m y x) y Ω}, and we have φx) = φ)x). Lemma 3.4. Suppose that cx, m) satisfies the assumption 3.5). Then i) If φ is c-concave and N c,φ is measure preserving from gx)dx to Γ, then φ, φ c ) is a maximizer of Iu, v) in K. ii) If φx) is c-concave and φ, φ c ) maximizes Iu, v) in K, then N c,φ is measure preserving from gx)dx to Γ. Proof. First prove i). Given u, v) K, obviously ux) + vn c,φ x)) cx, N c,φ x)) = φx) + φ c N c,φ x)), a.e. x on. Integrating the above inequality with respect to g dx yields ug dx + vn c,φ x))gx) dx φg dx + φ c N c,φ x))gx) dx. By Lemma 3.1, it yields Iu, v) Iφ, φ c ) and from 2) above the conclusion follows. To prove ii), let ψ = φ c, and for v C ), let ψ θ m) = ψm) + θ vm) where ɛ 0 < θ ɛ 0 with ɛ 0 small, and let φ θ = ψ θ ) c. We shall prove that 3.6) lim θ 0 Iφ θ, ψ θ ) Iφ, ψ) θ = vn c,φ x)) g dx + vm) dγ. Since φ θ, ψ θ ) K, we have Iφ θ, ψ θ ) Iφ, ψ) for ɛ 0 < θ ɛ 0, and hence the existence of the limit 3.6) implies it must be zero. Therefore the measure preserving property of N c,φ follows from Lemma 3.1. To prove 3.6) we write Iφ θ, ψ θ ) Iφ, ψ) θ = φ θ φ g dx + vm) dγ. θ

17 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 17 By Lebesgue dominated convergence theorem, to show 3.6), it is enough to show that φ θx) φx) is uniformly bounded, and φ θx) φx) vn c,φ x)) θ θ for all x \ S, where N c,φ x) is single-valued on \ S and S = 0. Let us first prove the uniform boundedness. Fix x, we have by continuity that φ θ x) = cx, m θ ) ψ θ m θ ) for some m θ. Since φ is c-concave there exists m 1 and b R such that φy) cy, m 1 ) b for all y with equality when y = x. This implies that b = φ c m 1 ) and so φx) = cx, m 1 ) ψm 1 ). Hence φ θ x) φx) = cx, m θ ) ψm θ ) θ vm θ ) φx) by 3) above. We also have ψ c x) θ vm θ ) φx) = φ c ) c x) θ vm θ ) φx) θ vm θ ), φ θ x) φx) = φ θ x) cx, m 1 ) + ψm 1 ) = φ θ x) cx, m 1 ) + ψ θ m 1 ) θ vm 1 ) Then we get φ θ x) ψ θ ) c x) θ vm 1 ) = θ vm 1 ). θ vm θ ) φ θ x) φx) θ vm 1 ). Moreover, if x \S, then m 1 = N c,φ x) since ψ = φ c. To finish the proof, we show that m θ converges to m 1 as θ 0. Otherwise, there exists a sequence m θk such that m θk m m 1. So φx) = lim θ 0 φ θ x) = cx, m ) ψm ), which yields m N c,φ x). We then get m 1 = m, a contradiction. The proof is complete. Lemma 3.5. There exists a c-concave φ such that Proof. Let Iφ, φ c ) = sup{iu, v) : u, v) K}. I 0 = sup{iu, v) : u, v) K}, and let u k, v k ) K be a sequence such that Iu k, v k ) I 0. Set ū k = v k ) c and v k = ū k ) c. From property 2) above, ū k, v k ) K, u k ū k, v k v k, and so Iū k, v k ) I 0. Let c k = min ū k and define u k = ū k c k, v k = v k + c k. Obviously u, k v ) K and by the mass conservation condition on gdx and Γ, k equation 3.2), Iū k, v k ) = Iu, k v ). Since ū k k are uniformly Lipschitz, u are uniformly bounded. In addition, v = ū k k) c + c k = u k )c and consequently v are also k k uniformly bounded. By Arzelá-Ascoli s theorem, u, k v ) contains a subsequence k converging uniformly to φ, ψ) on. We then obtain that φ, ψ) K and I 0 = sup{iu, v) : u, v) K} = Iφ, ψ). Notice that this shows in particular that the supremum of I over K is finite. From property 2) above, ψ c, ψ c ) c ) is the sought maximizer of Iu, v), and ψ c is c-concave.

18 18 C. E. GUTIÉRREZ Lemma 3.6. Suppose that cx, m) satisfies the assumption 3.5). Let φ, φ c ) with φ c-concave be a maximizer of Iu, v) in K. Then inf cx, sx))gx) dx is attained at s S s = N c,φ, where S is the class of measure preserving mappings from gx)dx to Γ. Moreover 3.7) inf s S cx, sx))gx) dx = sup{iu, v) : u, v) K}. Proof. Let ψ = φ c. For s S, we have ) cx, sx))gx) dx φx) + ψsx)) gx) dx = φx)gx) dx + ψsx))gx) dx = φx)gx) dx + ψm) dγ = Iφ, ψ) = φx) + ψnc,φ x)) ) gx) dx, from Lemma 3.4ii) = cx, N c,φ x))gx) dx. Obviously, for any c-concave function φ, N c,φ has the following converging property C): if m k N c,φ x k ), x k x 0 and m k m 0, then m 0 N c,φ x 0 ). Lemma 3.7. Assume that cx, m) satisfies the assumption 3.5) and that g dx > 0 for G any open G. Then the minimizing mapping of inf s S cx, sx))gx) dx is unique in the class of measure preserving mappings from gx)dx to Γ with the converging property C). Proof. From Lemmas 3.5 and 3.6, let N c,φ be a minimizing mapping associated with a maximizer φ, φ c ) of Iu, v) with φ c-concave. Suppose that N 0 is another minimizing mapping with the converging property C). Clearly cx, N0 x)) φx) φ c N 0 x)) ) gx) dx ) = inf cx, sx))gx) dx φx)gx) dx + φ c m) dγ = 0, s S and since φx)+φ c N 0 x)) cx, N 0 x)), it follows that φx)+φ c N 0 x)) = cx, N 0 x)) on the set {x : gx) > 0} which is dense in. Hence from 3.5) and the converging property C), we get N 0 x) = N c,φ x) a.e. on. We remark from the above proof that if gx) > 0 on, then the minimizing mapping of inf s S cx, sx))gx) dx is unique in the class of measure preserving mappings from gx)dx to Γ.

19 REFRACTION PROBLEMS IN GEOMETRIC OPTICS Application to the refractor problem κ < 1. Let n 1 and n 2 be the indexes of refraction of two homogeneous and isotropic media I and II, respectively. Suppose that from a point O inside medium I light emanates with intensity f x) for x Ω. We want to construct a refracting surface R parameterized as R = {ρx)x : x Ω}, separating media I and II, and such that all rays refracted by R into medium II have directions in Ω and the prescribed illumination intensity received in the direction m Ω is f m). We first introduce the notions of refractor mapping and measure, and weak solution. In the next section we then convert the refractor problem into an optimal mass transport problem from Ω to Ω with the cost function log 1 1 κ x m and establish existence and uniqueness of weak solutions. Let Ω, Ω be two domains on S n 1, the illumination intensity of the emitting beam is given by nonnegative f x) L 1 Ω), and the prescribed illumination intensity of the refracted beam is given by a nonnegative Radon measure µ on Ω. Throughout this section, we assume that Ω = 0 and the physical constraint 3.1) inf x Ω,m Ω x m κ. We further suppose that the total energy conservation 3.2) f x) dx = µω ) > 0, and for any open set G Ω 3.3) Ω G f x) dx > 0, where dx denotes the surface measure on S n Refractor measure and weak solutions. We begin with the notions of refractor and supporting semi-ellipsoid. efinition 3.8. A surface R parameterized by ρx)x with ρ CΩ) is a refractor from Ω to Ω for the case κ < 1 often simply called as refractor in this section) if for any b x 0 Ω there exists a semi-ellipsoid Em, b) with m Ω such that ρx 0 ) = 1 κ m x 0 b and ρx) for all x Ω. Such Em, b) is called a supporting semi-ellipsoid 1 κ m x of R at the point ρx 0 )x 0. From the definition, any refractor is globally Lipschitz on Ω. efinition 3.9. Given a refractor R = {ρx)x : x Ω}, the refractor mapping of R is the multi-valued map defined by for x 0 Ω N R x 0 ) = {m Ω : Em, b) supports R at ρx 0 )x 0 for some b > 0}.

20 20 C. E. GUTIÉRREZ Given m 0 Ω, the tracing mapping of R is defined by T R m 0 ) = N 1 R m 0) = {x Ω : m 0 N R x)}. efinition Given a refractor R = {ρx)x : x Ω}, the Legendre transform of R is defined by R = {ρ m)m : ρ 1 m) = inf x Ω ρx)1 κ x m), m Ω }. We now give some basic properties of Legendre transforms. Lemma Let R be a refractor from Ω to Ω. Then i) R is a refractor from Ω to Ω. ii) R = R ) = R. iii) If x 0 Ω and m 0 Ω, then x 0 N R m 0 ) iff m 0 N R x 0 ). Proof. Given m 0 Ω, ρx)1 κx m 0 ) must attain the maximum over Ω at some x 0 Ω. Then ρ m 0 ) = 1/[ρx 0 )1 κx 0 m 0 )]. We always have 3.4) ρ 1 m) = inf x Ω ρx)1 κm x) 1 ρx 0 )1 κx 0 m), m Ω. Hence Ex 0, 1/ρx 0 )) is a supporting semi-ellipsoid to R at ρ m 0 )m 0. Thus, i) is proved. To prove ii), from the definitions of Legendre transform and refractor mapping we have 3.5) ρx 0 ) ρ m 0 ) = 1 1 κm 0 x 0 for m 0 N R x 0 ). For x 0 Ω, there exists m 0 N R x 0 ) and so from 3.5) ρ m 0 ) = 3.4), ρ m)1 kx 0 m) attains the maximum 1/ρx 0 ) at m 0. Thus, ρ x 0 ) = inf m Ω 1 ρ m)1 kx 0 m) = 1 ρx 0 ) 1. 1/ρx 0 ) 1 κx 0 m 0. By To prove iii), we get from the proof of ii) that if m 0 N R x 0 ), then the semiellipsoid Ex 0, 1/ρx 0 )) supports R at ρ m 0 )m 0 and so x 0 N R m 0 ). On the other hand, if x 0 N R m 0 ), we get that m 0 N R x 0 ), and since R = R, m 0 N R x 0 ). The next two lemmas discuss the refractor measure. Lemma C = {F Ω : T R F) is Lebesgue measurable} is a σ-algebra containing all Borel sets in Ω.

21 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 21 Proof. Obviously, T R ) = and T R Ω ) = Ω. Since T R i=1 F i) = i=1 T RF i ), C is closed under countable unions. Clearly for F Ω 3.6) T R F c ) = {x Ω : N R x) F c } = {x Ω : N R x) F = } {x Ω : N R x) F c, N R x) F } = [T R F)] c [T R F c ) T R F)]. If x T R F c ) T R F) Ω, then R parameterized by ρ has two distinct supporting semi-ellipsoids Em 1, b 1 ) and Em 2, b 2 ) at ρx)x. We show that ρx)x is a singular point of R. Otherwise, if R has the tangent hyperplane Π at ρx)x, then Π must coincide both with the tangent hyperplane of Em 1, b 1 ) and that of Em 2, b 2 ) at ρx)x. It follows from the Snell law that m 1 = m 2. Therefore, the area measure of T R F c ) T R F) is 0. So C is closed under complements, and we have proved that C is a σ-algebra. To prove that C contains all Borel subsets, it suffices to show that T R K) is compact if K Ω is compact. Let x i T R K) for i 1. There exists m i N R x i ) K. Let Em i, b i ) be the supporting semi-ellipsoid to R at ρx i )x i. We have 3.7) ρx)1 κm i x) b i for x Ω, where the equality in 3.7) occurs at x = x i. Assume that a 1 ρx) a 2 on Ω for some constants a 2 a 1 > 0. By 3.7) and 3.1), a 1 1 κ) b i a 2 1 κ 2 ). Assume through subsequence that x i x 0, m i m 0 K, b i b 0, as i. By taking limit in 3.7), one obtains that the semi-ellipsoid Em 0, b 0 ) supports R at ρx 0 )x 0 and x 0 T R m 0 ). This proves T R K) is compact. To show that C is closed by complements, it is enough to notice the formula that T R Ω \ F) = [T R Ω ) \ T R F)] [T R Ω \ F) T R F)]. Lemma Given a nonnegative f L 1 Ω), the set function M R, f F) = f dx is a finite Borel measure defined on C and is called the refractor measure associated with R and f. Proof. Let {F i } be a sequence of pairwise disjoint sets in C. Let H i=1 1 = T R F 1 ), and H k = T R F k ) \ k 1T i=1 RF i ), for k 2. Since H i H j = for i j and H k=1 k = T k=1 RF k ), it is easy to get M R, f k=1 F k) = f dx = f dx. k=1 H k T R F) k=1 H k

22 22 C. E. GUTIÉRREZ Observe that T R F k ) \ H k = T R F k ) k 1T i=1 RF i )) is a subset of the singular set of R and has area measure 0 for k 2. Therefore, f dx = M H R, f F k ) and the k σ-additivity of M R, f follows. The notion of weak solutions is introduced through the conservation of energy. efinition A refractor R is a weak solution of the refractor problem for the case κ < 1 with emitting illumination intensity f x) on Ω and prescribed refracted illumination intensity µ on Ω if for any Borel set F Ω 3.8) M R, f F) = f dx = µf). T R F) 3.3. Solution of the refractor problem. We introduce the cost cx, m) = log 1 1 κ x m for x Ω and m Ω where we assume Ω Ω κ. From efinitions 3.2 and 3.8, R = {ρx)x : x Ω} is a refractor iff log ρ is c-concave. Using efinitions 3.3 and 3.9 we get that N c,φ x) = N R x), R = {ρx)x : x Ω}, ρz) = e φz). Furthermore, log ρ = log ρ) c, log ρ = log ρ ) c by Remark 3) after efinition 3.2, and N R x 0 ) = N c,log ρ x 0 ) by 3.5). By the Snell law and Lemma 3.12, cx, m) satisfies 3.5). From the definitions, R is a weak solution of the refractor problem iff log ρ is c-concave and N c,log ρ is a measure preserving mapping from f x)dx to µ. By Lemma 3.5, there exists a c-concave φx) such that φ, φ c ) maximizes Iu, v) = u f dx + v dµm) Ω Ω in K = {u, v) CΩ) CΩ ) : ux) + vm) cx, m), for x Ω, m Ω }. Then by Lemma 3.4, N c,φ x) is a measure preserving mapping from f dx to µ. Therefore, R = {e φx) x : x Ω} is a weak solution of the refractor problem. It remains to prove the uniqueness of solutions up to dilations. Let R i = {ρ i x)x : x Ω}, i = 1, 2, be two weak solutions of the refractor problem. Obviously, N c,log ρi have the converging property C) stated before Lemma 3.7. It follows from Lemmas 3.4, 3.6 and 3.7 that N c,log ρ1 x) = N c,log ρ2 x) a.e. on Ω. That is, N R1 x) = N R2 x) a.e. on Ω. From the Snell law ν i x) = x κn R i x) is the unit x κn Ri x) normal to R i towards medium II at ρ i x)x where R i is differentiable. So ν 1 x) = ν 2 x) a.e. and consequently ρ 1 x) = C ρ 2 x) for some C > 0.

23 REFRACTION PROBLEMS IN GEOMETRIC OPTICS Solution to the refractor problem for κ < 1 with the Minkowski method In this section we solve the refractor problem using a method having its roots in the Minkowski method from convex analysis, [Sch93, Section 7.1]. In addition, to the definitions and lemmas from Subsection 3.2, we have the following. Lemma 4.1. We have for a refractor R that i) [T R F)] c T R F c ) for all F Ω, with equality except for a set of measure zero. ii) The set C = {F Ω : T R F) is Lebesgue measurable} is a σ-algebra containing all Borel sets in Ω. Lemma 4.2. Let R j = {ρ j x)x : x Ω}, j 1 be refractors from Ω to Ω. Suppose that 0 < a 1 ρ j a 2 and ρ j ρ uniformly on Ω. Then: i) R := {ρx)x : x Ω} is a refractor from Ω to Ω. ii) For any compact set K Ω iii) For any open set G Ω, where S is the singular set of R. lim sup T Rj K) T R K). j T R G) lim inf j T R j G) S, Proof. i) Obviously ρ CΩ) and ρ > 0. Fix x o Ω. Then there exist m j Ω and b j > 0 such that Em j, b j ) supports R j at ρx o )x o and thus ρ j x o ) = for all x Ω. Consequently b j 1 κm j x o and ρ j x) b j 1 κm j x o a 2 and a 1 b j 1 κm j x b j 1 κm j x for all j and therefore a 1 1 κ) b j a 2 for all j. If need be by passing to a subsequence we obtain m o and b o such that m j m o Ω and b j b o. We claim Em o, b o ) supports R at ρx o )x o. Indeed and ρx o ) = lim j ρ j x o ) = lim j b j 1 κm j x o = for all x Ω. Thus R is a refractor. b o 1 κm o x o ρx) = lim j ρ j x) lim j b j 1 κm j x = b o 1 κm o x

24 24 C. E. GUTIÉRREZ ii) Let x o lim sup T Rj K). Without loss of generality assume that x o T Rj K) for all j 1. Then there exist m j N Rj x o ) K and b j such that ρ j x o ) = b j 1 κm j x o and ρ j x) b j 1 κm j x for all x Ω. As in the proof of i) we may assume that m j m o K and b j b o and conclude that Em o, b o ) supports R at ρx o )x o, proving that x o T R m o ). Hence x o T R K). iii) Let G be an open subset of Ω. By ii) lim sup T Rj G c ) T R G c ) as G c is compact. Also 4.1) lim sup[t Rj G)] c lim sup{[t Rj G)] c [T Rj G) T Rj G c )]} j j and by Lemma 4.1 the right hand side of 4.1) is equal to lim sup j T Rj G c ). By ii) we will then have lim sup[t Rj G)] c T R G c ) = {[T R G)] c [T R G) T R G c )]}. j Taking complements we obtain {lim sup[t Rj G)] c } c [T R G)] [T R G) T R G c )] c. j Consequently lim inf T R j G) [T R G)] [T R G) T R G c )] c j and thus [[T R G)] [T R G) T R G c )] c ] S lim inf j T R j G) S. But T R G) T R G c ) S. Thus T R G) T R G) S lim inf T R j G) S j as required. Remark 4.3 Invariance by dilations). Suppose that R is a refractor weak solution in the sense of efinition 3.14 with intensities f, µ and defined by ρx)x for x Ω. Then for each α > 0, the refractor αr defined by α ρx)x for x Ω is a weak solution in the sense of efinition 3.14 with the same intensities. In fact, Em, b) is a supporting ellipsoid to R at the point y if and only if Em, α b) is a supporting ellipsoid to αr at the point y. This means that T R m) = T αr m) for each m Ω.

25 REFRACTION PROBLEMS IN GEOMETRIC OPTICS Existence of solutions in the discrete case. Theorem 4.4. Let f L 1 Ω) with inf Ω f > 0, g 1,, g N positive numbers, m 1,, m N Ω distinct points, N 2, with x m j κ for all x Ω and 1 j N. Let µ = N j=1 g j δ mj, and assume the conservation of energy condition 4.2) f x) dx = µω ). Then there exists a refractor R such that a) Ω = N j=1 T Rm j ), b) T R m j ) f x) dx = g j for 1 j N. To prove the theorem, we prove first a sequence of lemmas. Ω Lemma 4.5. Let 4.3) W = b = 1, b 2,, b N ) : b j > 0, M Rb), f m j ) = f x) dx g j, j = 2,, N T Rb) m j ), where b i 4.4) ρx) = Rb)x) = min. 1 j N 1 κ x m j Then, with the assumptions of Theorem 4.4, we have a) W b) if b = 1, b 2,, b N ) W, then b j > 1 for j = 2,, N. 1 + κ Proof. a) If for some j 1, the semi-ellipsoid Em j, b) supports Rb) at some x Ω, b b then ρz) for all z Ω, and ρx) =. Since x m j κ, we 1 κ z m j 1 κ x m j have b 1 κ b 1 = ρx) κ x m j 1 κ x m 1 1 κ, and so b 1 + κ. Therefore, if b i > 1 + κ for 2 i N, then Em i, b i ) cannot be a supporting ellipsoid to Rb) at any x Ω. On the other hand, if x T R m j ), then m j N R x) and if x is not a singular point of Rb) there is a unique ellipsoid Em j, b) supporting R at x. But from the definition of R there is an ellipsoid Em k, b k ) that supports R at x, and so Em k, b k ) = Em j, b), i.e., k = j and b = b j. Consequently the set T R m j ) is contained in the set of singular points and therefore has measure zero. So M Rb), f m j ) = 0 < g j for j = 2,, N and so any point b = 1, b 2,, b N ) W as long as b i > 1 + κ for i = 2,, N. Claim 2. For each b W, T Rb) m 1 ) N i=2 T Rb)m i ) ) c. Otherwise, T Rb) m 1 ) N i=2 T Rb)m i ) which means that each point in T Rb) m 1 ) is singular, and therefore T Rb) m 1 ) = 0. This contradicts Claim 1, since g 1 > 0.

26 26 C. E. GUTIÉRREZ so Therefore, if b W, then we can pick x 0 T Rb) m 1 ) N i=2 T Rb)m i ) ) c and so ρx 0 ) = 1 1 κ x 0 m 1 < b i 1 κ x 0 m i, i = 2,, N b i > 1 κ x 0 m i 1 κ x 0 m 1 1 κ x 0 m i 1 κ 2 1 κ 1 κ 2 = κ. Lemma 4.6. If b j = b j 1,, bj N ) b 0 = b 0 1,, b0 N ) as j, then ρ j = Rb j ) ρ 0 = Rb 0 ) uniformly in Ω as j. Proof. Given y Ω, there exists 1 l N such that ρ 0 y) = as j. ρ j y) ρ 0 y) b j l 1 κ y m l b 0 l b j 1 κ y m l b 0 l 1 κ y m l. Hence l b0 l b l b0 l 1 κ y m l 1 κ 0, Lemma 4.7. Let δ > 0 and the region R δ = {1, b 2,, b N ) : b j δ, 2 j N}. The functions G Rb) m i ) := M Rb) m i ) are continuous for b R δ for i = 1, 2,, N. Proof. Let b j = 1, b j,, 2 bj ) R N δ with b j b 0 as j. By Lemma 4.6, ρ j ρ 0 b j l uniformly in Ω. Given x Ω, we have ρ j x) = for some 1 l N and 1 κ x m l so ρ j x) min{1, δ} 1 + κ. On the other hand, ρ jx) = min 1 l N 1 1 κ. Therefore min{1, δ} 1 + κ ρ j x) 1 1 κ b j l 1 κ x m l j 1 1 κ x m 1 for all x Ω and for all j. Let us fix 1 i N. Let G Ω be a neighborhood of m i such that m l G for l i. If x 0 T Rbj )G) and x 0 is not a singular point, then there exists a unique b m G and b > 0 such that ρ j x 0 ) = 1 κ x 0 m and ρ b jz) 1 κ z m for all x Ω. From the definition of Rb j ) and since x 0 is not singular, m = m l. Since m G, we get m = m i. Therefore T Rbj )G) T Rbj )m i ) S, where S is the set of singular points. By Lemma 4.2 and we therefore obtain T Rb0 )G) lim inf j T Rb j )G) S, T Rb0 )G) lim inf j T Rb j )m i ) S.

27 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 27 Thus T Rb0 )m i ) We next prove that f x) dx T Rb0 )G) lim inf j f x) dx T Rbj )m i ) f x) dx lim inf j T Rbj )m i ) by Fatou. lim sup f x) dx f x) dx. j T Rbj )m i ) T Rb0 )m i ) By Lemma 4.2 lim sup T Rbj )K) T Rb0 )K) j for each K compact. Hence f x) dx f x) dx. By reverse Fatou we have lim sup j lim sup j T Rbj )m i ) T Rbj )m i ) and therefore the lemma is proved. f x) dx T Rb0 )m i ) lim sup j T Rbj )m i ) Proof of Theorem 4.4. Fix b = 1, b 2,, b N ) W and let f x) dx W = {b = 1, b 2,, b N ) W : b j b j, j = 2,, N}. f x) dx W is compact. Let d : W R be given by db) = 1 + b b N ; d attains its minimum value in W at a point b = 1, b 2,, b N ) notice that the minimum is strictly positive by Lemma 4.5b)). We prove that Rb ) is the refractor that solves the problem. By conservation of energy it is enough to show that f x) dx = T Rb ) m j ) g j for j = 2,, N. Since b W, we have f x) dx g T Rb ) m j ) j for j = 2,, N. Suppose by contradiction that this inequality is strict for some j, suppose for example that 4.5) f x) dx < g 2. T Rb ) m 2 ) Let 0 < λ < 1 and b λ = 1, λb 2, b 3,, b N ). We claim that 4.6) T Rb λ ) m i ) \ set of measure zero T Rb )m i ) for i = 3, 4,, N and all 0 < λ < 1. Indeed, if x 0 T Rb point of Rb a ), then there is a unique ellipsoid λ λ ) m i ) is not a singular that supports Rb 1 κ x m ) at x λ 0 i

28 28 C. E. GUTIÉRREZ for some a > 0. Since Rb )x) = min b ) λ i λ 1 i N, there exists 1 j N such 1 κ x m i that Rb )x b λ λ 0) = ) j b. That is, the ellipsoid ) λ j supports Rb 1 κ x m j 1 κ x m ) at x λ 0. j a b Therefore, = ) λ j implying j = i and so a = b 1 κ x m i 1 κ x m ) λ i = b. Since i j b i Rb )x) Rb 1 κ x m )x) for all x, it follows that b i is a supporting λ i 1 κ x m i ellipsoid to Rb ) at x 0. This proves the claim. This implies that f x) dx f x) dx g i T Rb λ ) m i) T Rb ) m i ) for i = 3, 4,, N and all 0 < λ < 1. Finally from Lemma 4.7, inequality 4.5) holds for all λ sufficiently close to one, and therefore the point b W for all λ close to one. This is a contradiction λ because db ) < λ db ). Remark 4.8. Notice that the solution in Theorem 4.4 has the form given by formula 4.4), where b 1 = 1 and 1, b 2,, b N ) W. So from Lemma 4.5b), we have b i > 1/1 + κ) for i = 2,, N. This implies that inf Ω ρx) = α > Solution in the general case. Lemma 4.9. Let R = {ρx)x : x Ω} be a refractor from Ω to Ω such that inf x Ω ρx) = 1. Then there is a constant C, depending only on κ, such that sup ρx) C. x Ω Proof. Suppose inf x Ω ρx) is attained at x o Ω, and let Em, b) be a supporting semi-ellipsoid to R at ρx 0 )x 0. Then b b 1 = ρx 0 ) = and ρx) x Ω. 1 κ m x 0 1 κ m x From the first equation we get that b 1 + κ, and using this in the inequality we obtain ρx) 1 + κ for all x Ω 1 κ 2 which proves the lemma. Theorem Let f L 1 Ω) with inf Ω f > 0, and let µ be a Radon measure on Ω, such that f x) dx = µω ) Ω Then there exists a weak solution R of the refractor problem for the case κ < 1, with emitting illumination intensity f and prescribed refracted illumination intensity µ.

29 REFRACTION PROBLEMS IN GEOMETRIC OPTICS 29 Proof. Fix l N, l 2. Partition Ω into a finite number of disjoint Borel subsets ω l,..., 1 ωl k l such that diamω l) 1 i l. Choose points ml ω l and define a measure i i on Ω k l µ l = µω l i )δ m i. l Then k l µ l Ω ) = µω l i ) = µω ) = f x) dx. i=1 i=1 Ω If h CΩ ), then k l h dµ l h dµ = Ω Ω i=1 k l = i=1 k l = ω l i ω l i i=1 ω l i hx) dµ l hm l i )dµ ω l i ω l i hm l i ) hx))dµ. hx) dµ hx)dµ Since h CΩ ) and diamω l) < 1, we obtain that i l h dµ l h dµ as l Ω Ω and hence µ l converges weakly to µ. By Theorem 4.4, let R l = {ρ l x)x : x Ω} be the solution corresponding to µ l, that is, M Rl, f ω) = µ l ω) for every Borel subset ω of Ω. Notice that from Remark 4.8, inf Ω ρ l x) = α l > 0. In view of Remark 4.3, the refractor defined by the function ρ lx) solves the ρ l x) same problem and inf Ω α l = 1. So by normalizing ρ l, we may assume that inf Ω ρ l x) = 1. Then by Lemma 4.9) there exists a uniform bound C = Cκ) such that sup ρ l x) C for all l 1. x Ω α l

30 30 C. E. GUTIÉRREZ Also if x o, x 1 Ω and Em o, b o ) is a supporting semi ellipsoid to R l at ρ l x o )x o then for x 1 Ω we have b o 1 κ m o x 1 ρ l x 1 ) ρ l x o ) = 1 κ m o x o κ b 0 m o x 1 x 0 1 κm o x 1 )1 κm o x o ) C κ 1 κ x 1 x o. By changing the roles of x o and x 1 we conclude that b o ρ l x 1 ) ρ l x o ) C κ 1 κ x 1 x o for all l 1. κ b 0 m o x 1 x 0 ) 1 κm o x 1 )1 κm o x o ) Thus {ρ l : l 1} is an equicontinuous family which is bounded uniformly. Then by Arzelà - Ascoli Theorem, if need be by taking subsequence, we have that ρ l ρ uniformly on Ω. By Lemma 4.2i), R = {ρx)x : x Ω} is a refractor. We claim that M Rl, f converges weakly to M R, f. Indeed, if F is any closed subset of Ω then by Lemma 4.2ii) and reverse Fatou we have lim sup M Rl, f F) f x) dx f x) dx = M R, f F). l lim sup l T Rl F) T R F) Moreover for any open set G Ω we claim that 4.7) M Rl, f G) = f x) dx lim inf M R l, f G). l T R G) Indeed, from Lemma 4.2iii) we have M R, f G) = f x) dx T R G) lim inf l T Rl G) f x) dx = lim inf χ T Rl G)x) f x)dx Ω l lim inf l χ TRl G)x) f x) dx = lim inf M R l, f G), Ω l by Fatou s lemma. Consequently M Rl, f M R, f weakly. Since M Rl, f ω) = µ l ω) for each Borel set ω, it follows by uniqueness of the weak limit that M R, f = µ Uniqueness discrete case. With the method used in this section we show uniqueness when the target measure is discrete. Notice that uniqueness up to dilations) was already proved with the mass transport approach in Subsection 3.3. But the method we describe here is applicable to the near field problem which is not a mass transport problem.