Smallest area surface evolving with unit areal speed


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1 Smallest area surface evolving with unit areal speed Constantin Udrişte, Ionel Ţevy Abstract. The theory of smallest area surfaces evolving with unit areal speed is a particular case of the theory of surfaces of minimum area subject to various constraints. Based on our recent results, such problems can be solved using the twotime maximum principle in a controlled evolution. Section 1 studies a controlled dynamics problem (smallest area surface evolving with unit areal speed) via the twotime maximum principle. The evolution PDE is of flow type and the adjoint PDE is of divergence type. Section analyzes the smallest area surfaces evolving with unit areal speed, avoiding an obstacle. Section 3 reconsiders the same problem for touching an obstacle, detailing the results for the cylinder and the sphere. M.S.C. 010: 49J0, 49K0, 49N90, 90C9. Key words: multitime maximum principle, smallest area surface, minimal surfaces, controlled dynamics with obstacle. 1 Smallest area surface evolving with unit areal speed, passing through two points The minimal surfaces are characterized by zero mean curvature. They include, but are not limited to, surfaces of minimum area subject to various constraints. Minimal surfaces have become an area of intense mathematical and scientific study over the past 15 years, specifically in the areas of molecular engineering and materials sciences due to their anticipated nanotechnology applications (see [1][5]). Let Ω 0τ be a bidimensional interval fixed by the diagonal opposite points 0, τ R +. Looking for surfaces x i (t) = x i (t 1, t ), (t 1, t ) Ω 0τ, i = 1,, 3, that evolve with unit areal speed and relies transversally on two curves Γ 0 and Γ 1, let us show that a minimum area surface (sheet) is a solution of a special PDE system, via the optimal control theory (multitime maximum principle, see [11][]). An example is a planar quadrilateral (totally geodesic surface in R 3 ) fixed by the origin x i (0) = x i 0 on Γ 0 and passing through the diagonal terminal point x i (τ) = x i 1 on Γ 1. Balkan Journal of Geometry and Its Applications, Vol.16, No.1, 011, pp c Balkan Society of Geometers, Geometry Balkan Press 011.
2 156 Constantin Udrişte, Ionel Ţevy In R 3 we introduce the twotime controlled dynamics (P DE) t α (t) = ui α(t), t = (t 1, t ) Ω 0τ, i = 1,, 3; α = 1,, x i (0) = x i 0, x i (τ) = x i 1, where u i α(t) represents two openloop C 1 control vectors, noncollinear, eventually fixed on the boundary Ω 0τ. The complete integrability conditions of the (PDE) system, restrict the set of controls to { U = u = (u α ) = (u i α) } ui 1 t (t) = ui t 1 (t). A solution of (P DE) system is a surface (sheet) σ : x i = x i (t 1, t ). Suppose x(0) = x 0 belongs to the image Γ 0 of a curve in R 3 and τ = (τ 1, τ ) is the twotime when the sheet x(t 1, t ) reaches the curve Γ 1 in R 3, at x(τ) = x 1, with Γ 0 and Γ 1 transversal to σ. On the other hand, we remark that the area of the sheet σ is σ dσ = Ω 0τ ( u1 u < u 1, u > ) 1 dt 1 dt. This surface integral generates the cost functional (J) J(u( )) = dσ. Of course, the maximization of J(u( )) is equivalent to minimization of the area, under the constraint (PDE). Suppose the control set U is restricted to the hypersurface in R 3 R 3 corresponding to unit areal speed produced by two linearly independent vectors u α = (u 1 α, u α, u 3 α), α = 1,, in R 3, i.e., or in short U : δ ij u i 1u j 1 δ klu k u l (δ ij u i 1u j ) = 1 U : q(u) = u 1 u < u 1, u > 1 = 0 (see the Gram determinant). Then the surface integral dσ = dt 1 dt = τ 1 τ σ Ω 0τ represents the area of the bidimensional interval Ω 0τ whose image as (PDE) solution is the surface (sheet) σ : x i = x i (t 1, t ). We need to find the point τ(τ 1, τ ) such that x(τ) = x 1 and τ 1 τ = min. Twotime optimal control problem of smallest area surface evolving with unit areal speed: Find subject to max u( ) σ J(u( )) t α (t) = ui α(t), i = 1,, 3; α = 1, ; q(u) = 0,
3 Smallest area surface evolving with unit areal speed 157 u(t) U, t Ω 0τ ; x(0) = x 0, x(τ) = x 1. To solve the previous problem we apply the multitime maximum principle [11][]. In general notations, we have x = (x i ), u = (u α ), u α = (u i α), p = (p α ), p α = (p α i ), α = 1, ; i = 1,, 3 u 1 = (u 1 1, u 1, u 3 1), u = (u 1, u, u 3 ), p 1 = (p 1 1, p 1, p 1 3), p = (p 1, p, p 3) and the control Hamiltonian is X α (x(t), u(t)) = u α (t), X 0 (x(t), u(t)) = 1, q(u) = 0 H(x, p, u) = p α i X i α(x, u) + p 0 X 0 (x, u) µq(u), where µ(t) is a Lagrange multiplier. Taking p 0 = 1, we have H(x, p, u) = p α i ui α 1 µq(u). The adjoint dynamics says (ADJ) p α i t α = H = 0. The general solution of the adjoint PDE system is p 1 i = F i t, p i = F i, i = 1,, 3, t1 where F i are arbitrary C functions. We have to maximize the Hamiltonian H(x, p, u) = p α i ui α 1 µq(u) with respect to the control u. The critical points u of H are solutions of the algebraic system p 1 i µ( u u i 1 < u 1, u > u i ) = 0 (1) p i µ( u 1 u i < u 1, u > u i 1) = 0 u 1 u < u 1, u > 1 = 0. The system (1) is equivalent to the system () u i 1p 1 i = µ, ui 1p i = 0, ui p 1 i = 0, ui p i = µ u 1 u < u 1, u > 1 = 0. In this way u 1 is orthogonal to p, and u is orthogonal to p 1. Also (3) δ ij p 1 i p 1 j = 4µ u, δ ij p 1 i p j = 4µ < u 1, u >, δ ij p i p j = 4µ u 1. The first two relations of (1) are equivalent to µu i 1 = u 1 p 1 i + < u 1, u > p i, µu i = u p i + < u 1, u > p 1 i or, via the relations (3), we obtain the unique solution u i 1 = p 8µ 3 p1 i < p1, p > 8µ 3 p i, u i = p1 8µ 3 p i < p1, p > 8µ 3 p 1 i, depending on the parameter µ(t). The functions F i are constrained by the complete integrability conditions ui 1 t = ui t 1.
4 158 Constantin Udrişte, Ionel Ţevy Lemma 1.1. The Lagrange multiplier µ(t) is a constant. Proof. The relations () can be written in the form p α i ui β = µδα β. By differentiation with respect to t α, we find pα i t α ui β + p α i u i βα = µ t β and, via (ADJ), we can write µ t β = 1 pα i u i βα. On the other hand, the relations (), the condition q(u) = 0 and techniques from differential geometry give p α i ui βα = 0, i.e., µ(t) = constant. Corollary 1.. The smallest area surface evolving with unit areal speed is a minimal surface. Introducing u i 1 and u i in the restriction U : q(u) = 0, we find the areal speed in the dual variables p 1 p < p 1, p > = 16µ 4 (t) = constant. Consequently, we have the following Theorem 1.3. The smallest area surface evolving with unit areal speed is a minimal surface, solution of the PDE system (P DE) t 1 = p 8µ 3 p1 i < p1, p > 8µ 3 p i, t = p1 8µ 3 p i < p1, p > 8µ 3 p 1 i, x(0) = x 0, x(τ) = x 1 ; µ = constant; (ADJ) where B means boundary condition. p α i t α = 0, B(p(t)) Ω 0τ = 0, In fact the strongest restriction on the surface is the existence of a global holonomic frame {u 1, u }. 1.1 Constant dual variables A very special solution of the adjoint (divergence) PDE is the constant solution p β i (t) = cβ i (constants) 0. Particularly, if p 1 (0), p (0) are linearly independent, then p 1, p rest linearly independent during the evolution. Also, by the initial condition p 1 (0) p (0) < p 1 (0), p (0) > = 1, the constants vectors p 1, p satisfy p 1 p < p 1, p > = 1 throughout, where p 1 = δ ij p 1 i p1 j, p = δ ij p i p j, < p1, p >= δ ij p 1 i p j. If we select p1, p with = p 1 p < p 1, p > = 1, then µ = ± 1. We select µ = 1. The vectors u α( ) = u α0 = (u i α0) are constant in time. Cauchy problem for (PDE) system has the solution x i (t) = u i 10t 1 + u i 0t + x i 0, i = 1,, 3.
5 Smallest area surface evolving with unit areal speed 159 depending on six arbitrary constants. We fix these constants by the following conditions: x i (τ) = x i 1 which implies det[u 1, u, x 0 x 1 ] = 0, q(u) = 1, and finally, the transversality conditions (τ) p(0) T x0 Γ 0, p(τ) T x1 Γ 1, which show that the optimal plane is orthogonal to Γ 0 and Γ 1, and so the tangent lines T x0 Γ 0 and T x1 Γ 1 are parallel. That is why, the optimum sheet transversal to the curves Γ 0, Γ 1 is a planar quadrilateral fixed by the starting point x i (0) = x i 0 on Γ 0 and the terminal point x i (τ) = x i 1 on Γ 1. Remark 1.4. In the previous hypothesis, the surfaces evolving with unit areal speed and having a minimum area are planar sheets. Remark 1.5. In case of i = 1,, 3 and α = 1,, 3 we obtain a problem of controlled diffeomorphisms, having as result a set which include the affine unimodular group. Remark 1.6. The (PDE) system can be replaced by the Pfaff system dx i = u i αdt α. If this system is completely integrable, then we have the previous theory. If this system is not completely integrable, then we apply the theory in [13]. 1. Nonconstant dual variables We start from the minimal revolution surface of Cartesian equation z = 1 ch (cx + c 1 ) c c y, where Let us find a parametrization 1 + z x + z y = x = x(t 1, t ), y = y(t 1, t ), z = 1 c satisfying (unit areal speed) Taking x t = 0, we find which can be realized for x t 1 = ch 4 (cx + c 1 ) ch (cx + c 1 ) c y. ch (cx(t 1, t ) + c 1 ) c y (t 1, t ) (1 + z x + z y)(x t 1y t x t y t 1) = 1. 1 x t 1y t = 1 + zx + zy, 1 ch (cx + c 1 ), y t = ch (cx + c 1 ) c y. We find cx(t 1 ) + sh (cx(t 1 ) + c 1 ) = 4ct 1 + c
6 160 Constantin Udrişte, Ionel Ţevy y(t 1, t ) = ch (cx(t 1 ) + c 1 ) sin(ct + ϕ(t 1 )). In particular, for c = 1, c 1 = c = 0, ϕ(t 1 ) = 0, the implicit equation define the function x(t 1 ), and we obtain the parametrization x = x(t 1 ), y = ch x(t 1 ) sin t, z = ch x(t 1 ) cos t. To show that this nonplanar minimal surface is a solution to the optimal problem in Section 1, we evidentiate the optimal control ( ) 1 sh x sin t sh x cos t u 1 = ch, x ch, x ch x and the adjoint vectors u = ( 0, ch x cos t, ch x sin t ). p 1 = µ ch x u 1, p = µ ch x u. Moreover, the adjoint PDEs give µ = constant. Two time evolution with unit areal speed, passing through two points.1 Two time evolution avoiding an obstacle Let us apply the multitime maximum principle to search the smallest area surface satisfying the following conditions: it evolves above the rectangle Ω 0τ with unit areal speed, it contains two diagonal points x(0) and x(τ), and it avoids an obstacle A whose boundary is A. For that we start with the controlled dynamics problem and its solution in Section 1. Suppose x(t) / A for t Ω 0τ. In this hypothesis, the multitime maximum principle applies, and hence the initial dynamics (PDE) and the adjoint dynamics (ADJ) are those in Section 1. Particularly, if the dual variables are constants, then the evolution (P DE) shows that the surface of evolution is a planar quadrilateral (starting from origin x i (0) = x i 0 and ending at a terminal point x i (τ) = x 1. We accept that the boundary of the obstacle is a surface (manifold). Also, to simplify, we accept as obstacle a dimensional cylinder (that supports a global tangent frame {u 1, u }) or a dimensional sphere (that does not support a global tangent frame because any continuous vector field on such sphere vanishes somewhere). Remark.1. Of all the solids having a given volume, the sphere is the one with the smallest surface area; of all solids having a given surface area, the sphere is the one having the greatest volume. These properties define the sphere uniquely and can be seen by observing soap bubbles. A soap bubble will enclose a fixed volume and due to surface tension it will try to minimize its surface area. This is why a free floating soap bubble approximates a sphere (though external forces such as gravity will distort the bubble s shape slightly). The sphere has the smallest total mean curvature among all convex solids with a given surface area. The sphere has constant positive mean curvature. The sphere is the only imbedded surface without boundary or singularities with constant positive mean curvature.
7 Smallest area surface evolving with unit areal speed 161. Two time evolution touching an obstacle The points 0 s 0 s 1 τ generates a decomposition Ω 0s0, Ω 0s1 \ Ω 0s0, Ω 0τ \ Ω 0s1 of the bidimensional interval Ω 0τ. To simplify the problem, suppose the sheet x(t) / A for t Ω 0s0 (Ω 0τ \ Ω 0s1 ) is a union of two planar quadrilaterals (one starting from x(0) and ending in x(s 0 ) and the other starting from x(s 1 ) and ending at x(τ)). If x(t) A for t Ω 0s1 \Ω 0s0, then we need the study in Section 3 and Section 4, which evidentiates the controls and the dual variables capable to keep the evolution on the obstacle. Furthemore, suitable smoothness conditions on boundaries are necessary. 3 Touching, approaching and leaving a cylinder 3.1 Touching a cylinder Let us take the cylinder C : (x 1 ) + (x ) r as obstacle. Suppose x(t) C for t Ω 0s1 \ Ω 0s0. In this case we use the modified version of twotime maximum principle. We introduce the set N = R 3 \ C : f(x) = r ((x 1 ) + (x ) ) 0, where x = (x 1, x, x 3 ) and we build the functions c α (x, u) = f (x)xi α(x, u), α = 1,, i.e., c α (x, u) = (x 1 u 1 α + x u α). Let us use the twotime maximum principle using the constraints c 1 (x, u) = (x 1 u x u 1) = 0, c (x, u) = (x 1 u 1 + x u ) = 0 Then the adjoint equations q(u) = u 1 u < u 1, u > 1 = 0. are reduced to p α i (t) = H tα + λγ (t) c γ (ADJ p α 1 ) t α (t) = λγ (t)( u 1 p α γ), t α (t) = λγ (t)( u γ), The critical point condition with respect to the control u is i.e., or (4) H u i 1 H u = λγ c γ u, = λ 1 c 1 u i, 1 H u i = λ c u i, p 1 i = λ1 ( x i ) + µ( u u i 1 < u 1, u > u i ), i = 1, p 1 3 = µ( u u 3 1 < u 1, u > u 3 ), p i = λ ( x i ) + µ( u 1 u i < u 1, u > u i 1), i = 1,, p 3 = µ( u 1 u 3 < u 1, u > u 3 1). p α 3 (t) = 0. tα
8 16 Constantin Udrişte, Ionel Ţevy We recall that x(t) C means (x 1 ) + (x ) = r. Consequently, x i p 1 i = λ 1 ( r ), x i p i = λ ( r ), i = 1,. To develop further our ideas, we accept that the cylinder C is represented by the parametrization x 1 = r cos θ 1, x = r sin θ 1, x 3 = θ. We use the partial velocities (orthogonal vectors) Then the area formula x 1 θ 1 = r sin θ1, x θ 1 = r cos θ1, x3 θ 1 = 0 σ C x 1 x x3 = 0, = 0, θ θ θ = 1. dσ = r θ 1 θ 0 0 dθ 1 dθ = rθ 1 θ = t 1 t suggests to take t 1 = r θ 1, t = θ. On the other hand, the evolution PDEs are transformed in θ 1 = xi t 1 t 1 θ 1 + xi t t θ 1 = r xi t 1 = rui 1 θ = xi t 1 t 1 θ + xi t t θ = xi t = ui, evidentiating the controls u 1 : u 1 1 = sin θ 1, u 1 = cos θ 1, u 3 1 = 0 u : u 1 = 0, u = 0, u 3 = 1, with < u 1, u >= 0, u 1 = 1, u = 1. Using the equalities (4), we obtain u i 1p 1 i = µ, u i 1p i = 0, ui p 1 i = 0, ui p i = µ, which confirm that u 1 is orthogonal to p and u is orthogonal to p 1. The relations p 1 i = λ1 ( x i ) + µu i 1, i = 1,, p 1 3 = 0 produce p 1 1 = λ 1 r cos θ 1 µ sin θ 1, p 1 = λ 1 r sin θ 1 + µ cos θ 1, p 1 3 = 0. Similarly, the relations p i = λ ( x i ) + µu i, i = 1,, p 3 = µ give p 1 = λ r cos θ 1, p = λ r sin θ 1, p 3 = µ. Using the partial derivative operators θ 1 = r t 1, the adjoint equations (ADJ ) become p 1 1 θ 1 + r p 1 θ = λγ ( ru 1 γ), θ = t, p 1 θ 1 + r p θ = λγ ( ru γ), p 1 3 θ 1 + r p 3 θ = 0.
9 Smallest area surface evolving with unit areal speed 163 Replacing p 1, p, it follows the PDE system (r λ1 λ + r θ1 θ + µ)cos θ1 + µ θ 1 sin θ1 = 0 (r λ1 λ + r θ1 θ + µ)sin θ1 + µ θ 1 cos θ1 = 0 and hence the parameters λ γ and µ are determined as solution of PDE system in the next theorem Theorem 3.1. We consider the problem of smallest area surface, evolving with unit areal speed, touching a cylinder. Then the evolution on the cylinder is characterized by the parameters λ γ and µ related by the PDEs gives The particular solution r λ1 λ + r θ1 θ + µ = 0, µ θ 1 = 0. λ 1 = µθ1 + k 1, λ = k = constant, µ = constant r p 1 1(θ) = (µθ 1 + k 1 ) cos θ 1 µ sin θ 1 p 1 (θ) = (µθ 1 + k 1 ) sin θ 1 + µ cos θ 1, p 1 3 = 0; p 1(θ) = k r cos θ 1, p (θ) = k r sin θ 1, p 3(θ) = µ, for θ = (θ 1, θ ). The evolution (PDE) on the interval ω 0 θ 1 ω 1 shows that the surface of evolution is a cylindric quadrilateral fixed by the initial point x i (ω 0, θ ) = x i 0 generator and with the terminal point x i (ω 1, θ ) = x i 1 generator. 3. Approaching and leaving the cylinder Now we must put together the previous results. So suppose x(t) N = R 3 \ C for t Ω 0s0 (Ω 0τ \ Ω 0s1 ) and x(t) C for t Ω 0s1 \ Ω 0s0. For t Ω 0s0 (Ω 0τ \ Ω 0s1 ), the sheet of evolution x( ) consists in two pieces. Particularly, it can be a union of two planar sheets. Suppose the first planar sheet touchs the cylinder at the point x(s 0 ). In this case, we can take p 1 1 = cos φ 0, p 1 = sin φ 0, p 1 3 = 0, for the tangency angle φ 0 from initial point x 0, and p 1 = 0, p = 0, p 3 = 1. By the jump conditions, the vectors p 1 ( ), p ( ) are continuous when the evolution sheet x( ) hits the boundary C at the twotime s 0. In other words, we must have the identities k 1 cos θ 1 0 sin θ θ 1 0 cos θ 1 0 = cos φ 0 k 1 sin θ cos θ θ 1 0 sin θ 1 0 = sin φ 0 k rcos θ 1 0 = 0, k rsin θ 1 0 = 0, µ = 1,
10 164 Constantin Udrişte, Ionel Ţevy i.e., k 1 = θ0, 1 θ0 1 + φ 0 = π, k = 0, µ = 1. The last two equalities show that the optimal (particularly, planar) sheet is tangent to the cylinder along the generator x 1 = x 1 (s 0 ), x = x (s 0 ), x 3 R. Let us analyse what happen with the evolution sheet as it leaves the boundary C at the point x(s 1 ). We then have p 1 1(θ 1 1, θ ) = θ 1 0 cos θ 1 1 sin θ θ 1 1 cos θ 1 1, p 1 (θ 1 1, θ ) = θ 1 0sin θ 1 1 cos θ θ 1 1 sin θ 1 1, p 1 3(θ 1 1, θ ) = 0; p 1(θ 1 1, θ ) = 0, p (θ 1 1, θ ) = 0, p 3(θ 1 1, θ ) = 1. The formulas for k 1, λ 1, λ imply λ 1 (θ1, 1 θ ) = θ1 0 θ1 1, λ (θ 1 r 1, θ ) = k. The jump theory gives p α (θ1 1+, θ ) = p α (θ1 1, θ ) λ α (θ1, 1 θ ) f(x(θ1, 1 θ )) for f(x) = r (x 1 ) (x ). Then cos θ1 1 λ 1 (θ1, 1 θ ) f(x(θ1, 1 θ )) = (θ1 1 θ0) 1 sin θ In this way, p 1 1(θ 1+ 1, θ ) = sin θ 1 1, p 1 (θ 1+ 1, θ ) = cos θ 1 1, and so the planar sheet of evolution is tangent to the boundary C along the generator by the point x(s 1 ). If we apply the usual twotime maximum principle after x( ) leaves the cylinder C, we find p 1 1 = constant = cos φ 1, p 1 = constant = sin φ 1 ; p 1 = 0, p = 0. Therefore cos φ 1 = sin θ 1 1, sin φ 1 = cos θ 1 1 and so φ 1 + θ 1 1 = π, k = 0. Open Problem. What happen when the surface in the exterior of the cylinder is a nonplanar minimal sheet? 4 Touching, approaching and leaving a sphere 4.1 Touching a sphere Let us take as obstacle the sphere B : f(x) = r δ ij x i x j 0, x = (x 1, x, x 3 ), i, j = 1,, 3. Suppose x(t) B for t Ω 0s1 \Ω 0s0. In this case we use the modified version of twotime maximum principle. We introduce the set N = R 3 \ B : f(x) = r δ ij x i x j 0 and we build the functions c α (x, u) = f (x)xi α(x, u), α = 1,, i.e., c α (x, u) = δ ij x i u j α. Let us use the twotime maximum principle using the constraints c 1 (x, u) = δ ij x i u j 1 = 0, c (x, u) = δ ij x i u j = 0 q(u) = u 1 u < u 1, u > 1 = 0.
11 Smallest area surface evolving with unit areal speed 165 Then the condition is reduced to p α i (t) = H tα + λγ (t) c γ (ADJ ) p α i t α (t) = λγ (t)( u i γ). The condition of critical point for the control u becomes or (5) H u i 1 = λ 1 c 1 u i, 1 H u i = λ c u i, p 1 i = λ1 ( x i ) + µ( u u i 1 < u 1, u > u i ) p i = λ ( x i ) + µ( u 1 u i < u 1, u > u i 1). We recall that x(t) B means δ ij x i x j = r. Consequently, x i p 1 i = λ 1 ( r ), x i p i = λ ( r ). H u = λγ c γ u, i.e., To develop further our ideas, we accept that the sphere B is represented by the parametrization x 1 = r cos θ 1 cos θ, x = r sin θ 1 cos θ, x 3 = r sin θ. We use the partial velocities (orthogonal vectors) x 1 θ 1 = r sin θ1 cos θ, x θ 1 = r cos θ1 cos θ, x3 θ 1 = 0 x 1 θ = r cos θ1 sin θ, x θ = r sin θ1 sin θ, x3 θ = r cos θ. Then the area formula θ 1 θ dσ = r cos θ dθ 1 dθ = r θ 1 sin θ = t 1 t σ B 0 0 suggests to take t 1 = rθ 1, t = r sin θ. On the other hand, the evolution P DEs are transformed in θ 1 = xi t 1 t 1 θ 1 + xi t t θ 1 = r xi t 1 = rui 1 θ = xi t 1 t 1 θ + xi t xi t = r cos θ θ t = r cos θ u i, evidentiating the controls u 1 : u 1 1 = sin θ 1 cos θ, u 1 = cos θ 1 cos θ, u 3 1 = 0 u : u 1 = cos θ 1 tan θ, u = sin θ 1 tan θ, u 3 = 1,
12 166 Constantin Udrişte, Ionel Ţevy with < u 1, u >= 0, u 1 = cos θ, u = 1. Using the equalities (5), we obtain cos θ u i 1p 1 i = µ, ui 1p i = 0, ui p 1 i = 0, ui p i = µ, which confirm that u 1 is orthogonal to p and u is orthogonal to p 1. We have or explicitely p 1 i = λ 1 ( x i ) + µ u u i 1, p i = λ ( x i ) + µ u 1 u i, i = 1,, 3, p 1 1 = λ 1 ( r cos θ 1 cos θ sin θ1 ) µ cos θ p 1 = λ 1 ( r sin θ 1 cos θ cos θ1 ) + µ cos θ p 1 3 = λ 1 ( r sin θ ) p 1 = λ ( r cos θ 1 cos θ ) µ cos θ 1 sin θ cos θ p = λ ( r sin θ 1 cos θ ) µ sin θ 1 sin θ cos θ Using the partial derivative operators p 3 = λ ( r sin θ ) + µ (cos θ ). θ 1 = r t 1, the adjoint equations (ADJ ) become θ = r cos θ t, p 1 i θ 1 cos θ + p i θ = λγ ( ru i γ cos θ ). Replacing p 1, p, it follows the PDE system λ 1 θ 1 ( r cos θ1 ( cos θ ) ) µ θ 1 sin θ1 + λ θ ( r cos θ1 cos θ ) µ θ cos θ1 sin θ cos θ 4µ cos θ 1 (cos θ ) = 0, λ 1 θ 1 ( r sin θ1 ( cos θ ) ) + µ cos θ1 θ1 + λ θ ( r sin θ1 cos θ ) µ θ sin θ1 sin θ cos θ 4µ sin θ 1 (cos θ ) = 0, λ 1 θ 1 ( r sin θ cos θ ) + λ θ ( r sin θ ) + µ θ (cos θ ) 4µ cos θ sin θ = 0, equivalent to λ 1 θ 1 ( r cos θ ) + λ µ ( r) θ θ sin θ 4µ cos θ = 0, µ θ 1 = 0, λ 1 ( r sin θ cos θ ) + λ θ1 θ ( r sin θ ) µ θ cos θ + 4µ cos θ sin θ = 0 or to the system in the next
13 Smallest area surface evolving with unit areal speed 167 Theorem 4.1. We consider the problem of smallest area surface, evolving with unit areal speed, touching a sphere. Then the evolution on the sphere is characterized by the parameters λ γ and µ related by the PDEs λ 1 θ 1 ( r cos θ ) + λ θ ( r) µ cos θ = 0, It follows µ = constant. The particular solution produces λ 1 = µθ1 + k 1, λ = k r r µ θ 1 = 0, µ θ = 0. p 1 1(θ) = (µθ 1 + k 1 )( cos θ 1 cos θ sin θ1 ) µ cos θ p 1 (θ) = (µθ 1 + k 1 )( sin θ 1 cos θ ) + µ p 1 3(θ) = (µθ 1 + k 1 )( sin θ ); cos θ1 cos θ p 1(θ) = k (cos θ 1 cos θ ) µ cos θ 1 sin θ cos θ p (θ) = k (sin θ 1 cos θ ) µ sin θ 1 sin θ cos θ p 3(θ) = k (sin θ ) + µ (cos θ ), θ = (θ 1, θ ). 4. Approaching and leaving the sphere Now we must put together the previous results. So suppose x(t) N = R 3 \ B for t Ω 0s0 (Ω 0τ \ Ω 0s1 ) and x(t) C for t Ω 0s1 \ Ω 0s0. For t Ω 0s0 (Ω 0τ \ Ω 0s1 ), the sheet of evolution x( ) consists in two pieces. Particularly, it can be a union of two planar sheets. Suppose the first planar sheet touchs the sphere at the point x(s 0 ). In this case, we can take p 1 1 = cos φ 0, p 1 = sin φ 0, p 1 3 = 0, for the tangency angle φ 0 from the initial point x 0, and p 1 = 0, p = 0, p 3 = 1. By the jump conditions, the vectors p 1 ( ), p ( ) are continuous when the evolution sheet x( ) hits the boundary B at the twotime s 0. In other words, we must have the identities (µθ k 1 )( cos θ 1 0 cos θ 0) µ sin θ1 0 cos θ 0 (µθ k 1 )( sin θ 1 0 cos θ 0) + µ cos θ1 0 cos θ 0 (µθ k 1 )( sin θ 0) = 0; = cos φ 0 = sin φ 0 k (cos θ 1 0 cos θ 0) µ cos θ 1 0 sin θ 0 cos θ 0 = 0 k (sin θ 1 0 cos θ 0) µ sin θ 1 0 sin θ 0 cos θ 0 = 0 k (sin θ 0) + µ (cos θ 0) = 1,
14 168 Constantin Udrişte, Ionel Ţevy i.e., k 1 = θ0, 1 θ0 = 0, θ0 1 + φ 0 = π, k = 0, µ = 1. The last two equalities show that the optimal (particularly, planar) sheet is tangent to the sphere at the point (x 1 (s 0 ), x (s 0 ), x 3 (s 0 )). Let us analyse what happen with the evolution sheet as it leaves the boundary B at the point x(s 1 ). We then have p 1 1(θ 1 1, θ 1 ) = (θ1 1 θ 0 1)( cos θ 1 1 cos θ 1) µ sin θ1 1 cos θ 1 p 1 (θ 1 1, θ 1 ) = (θ1 1 θ 0 1)( sin θ 1 1 cos θ 1) + µ cos θ1 1 cos θ 1 p 1 3(θ 1 1, θ 1 ) = 0, p 1(θ 1 1, θ 1 ) = 0, p (θ 1 1, θ 1 ) = 0, p 3(θ 1 1, θ 1 ) = 1. The formulas for k 1, λ 1, λ imply λ 1 (θ1, 1 θ ) = θ1 0 θ1 1, λ (θ 1 r 1, θ ) = k. The jump theory gives p α (θ 1+ 1, θ+ 1 ) = pα (θ 1 1, θ 1 ) λα (θ 1 1, θ 1) f(x(θ 1 1, θ 1)) for f(x) = r δ ij x i x j. Then λ 1 (θ1, 1 θ1) f(x(θ 1, 1 θ1)) = (θ1 1 θ0) 1 cos θ 1 1 cos θ 1 sin θ 1 1 cos θ 1. sin θ 1 In this way, p 1 1(θ 1+ 1, θ+ 1 ) = sin θ1 1, p 1 (θ 1+ 1, θ+ 1 ) = cos θ1 1, and so the planar sheet of evolution is tangent to the boundary B at the point x(s 1 ). If we apply the usual twotime maximum principle after x( ) leaves the sphere B, we find p 1 1 = constant = cos φ 1, p 1 = constant = sin φ 1 ; p 1 = 0, p = 0. Therefore cos φ 1 = sin θ 1 1, sin φ 1 = cos θ 1 1 and so φ 1 + θ 1 1 = π, k = 0. Open Problem. What happens when the surface in the exterior of the sphere is a nonplanar minimal sheet? References [1] Y. Jung, K.T. Chu, S. Torquato, A variational level set approach for surface area minimization of triplyperiodic surfaces, Jour. Comp. Phys. 3 (007), [] W. Meeks III, H. Rosenberg, The maximum principle at infinity for minimal surfaces in flat threemanifolds, Comment. Math. Helvetici, 65 (1990), [3] A. Nabutovsky, R. Rotman, Curvaturefree upper bounds for the smallest area of a minimal surface, Geometric and Functional Analysis, 16, (006), [4] M. Pîrvan, C. Udrişte, Optimal control of electromagnetic energy, Balkan J. Geom. Appl. 15, 1(010), [5] W. Schief, On a maximum principle for minimal surfaces and their integrable discrete counterparts, Jour. Geom. Phys. 56, 9 (006),
15 Smallest area surface evolving with unit areal speed 169 [6] C. Udrişte, Controllability and observability of multitime linear PDE systems, Proceedings of The Sixth Congress of Romanian Mathematicians, Bucharest, Romania, June 8  July 4, vol. 1, 007, [7] C. Udrişte, Equivalence of multitime optimal control problems, Balkan J. Geom. Appl. 15, 1(010), [8] C. Udrişte, Finsler optimal control and geometric dynamics, Mathematics and Computers In Science and Engineering, Proc. Amer. Conf. on Appl. Math., Cambridge, 008, [9] C. Udrişte, Lagrangians constructed from Hamiltonian systems, Mathematics a Computers in Business and Economics, Proc. of the 9th WSEAS Int. Conf. on Mathematics a Computers in Business and Economics(MCBE08), Bucharest, Romania, June 46, 008, [10] C. Udrişte, Multitime controllability, observability and bangbang principle, Journal of Optimization Theory and Applications, 139, 1(008), [11] C. Udrişte, Multitime maximum principle, Short Communication, International Congress of Mathematicians, Madrid, August 30, ICM Abstracts, 006, 47. [1] C. Udrişte, Multitime stochastic control theory, Selected Topics on Circuits, Systems, Electronics, Control&Signal Processing, Proc. of the 6th WSEAS Int. Conf. on Circuits, Systems, Electronics, Control&Signal Processing (CSECS 07), Cairo, Egypt, December 931, 007, [13] C. Udrişte, Nonholonomic approach of multitime maximum principle, Balkan J. Geom. Appl. 14, (009), [14] C. Udrişte, Simplified multitime maximum principle, Balkan J. Geom. Appl. 14, 1 (009), [15] C. Udrişte, V. Arsinte, C. Cipu, Von Neumann analysis of linearized discrete Tzitzeica PDE, Balkan J. Geom. Appl. 15, (010), [16] C. Udrişte, O. Dogaru, I. Ţevy, D. Bala, Elementary work, Newton law and Euler Lagrange equations, Balkan J. Geom. Appl. 15, (010), [17] C. Udrişte, L. Matei, LagrangeHamilton Theories (in Romanian), Monographs and Textbooks 8, Geometry Balkan Press, Bucharest, 008. [18] C. Udrişte, I. Ţevy, Multitime EulerLagrangeHamilton theory, WSEAS Trans. Math. 6, 6 (007), [19] C. Udrişte, I. Ţevy, Multitime dynamic programming for curvilinear integral actions, Journal of Optimization Theory and Applications, 146 (010), [0] C. Udrişte, I. Ţevy, Multitime EulerLagrange dynamics, Proc. of the 7th WSEAS Int. Conf. on Systems Theory and Scientific Computation (ISTASC 07), Vouliagmeni Beach, Athens, Greece, August 46 (007), [1] C. Udrişte, I. Ţevy, Multitime linearquadratic regulator problem based on curvilinear integral, Balkan J. Geom. Appl. 14, (009), [] C. Udrişte, I. Ţevy, V. Arsinte, Minimal surfaces between two points, J. Adv. Math. Studies, 3, (010), Authors address: Constantin Udrişte, Ionel Ţevy University POLITEHNICA of Bucharest, Faculty of Applied Sciences, Department of MathematicsInformatics I, Splaiul Independentei 313, RO Bucharest, Romania. ;
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