Classification of Quadratic Surfaces

Transcription

1 Classification of Quadratic Surfaces Pauline Rüegg-Reymond June 4, 202 Part I Classification of Quadratic Surfaces Context We are studying the surface formed by unshearable inextensible helices at equilibrium with a given reference state. A helix on this surface is given by its strains u R 3. The strains of the reference state helix are denoted by û. The strain-energy density of a helix given by u is the quadratic function where K R 3 3 is assumed to be of the form with K K 2. W (u û) = (u û) K (u û) () 2 K 0 K 3 K = 0 K 2 K 23 K 3 K 23 K 3 A helical rod also has stresses m R 3 related to strains u through balance laws, which are equivalent to m = µ u + µ 2 e 3 (2) for some scalars µ and µ 2 and e 3 = (0, 0, ), and constitutive relation m = K (u û). (3) Every helix u at equilibrium, with reference state û, is such that there is some scalars µ, µ 2 with µ u + µ 2 e 3 = K (u û) µ u = K (u û ) + K 3 (u 3 û 3 ) µ u 2 = K 2 (u 2 û 2 ) + K 23 (u 3 û 3 ) µ u 3 + µ 2 = K 3 (u û ) + K 23 (u û ) + K 3 (u 3 û 3 ) (4) Assuming u and u 2 are not zero at the same time, we can rewrite this surface (K 2 K ) u u 2 + K 23 u u 3 K 3 u 2 u 3 (K 2 û 2 + K 23 û 3 ) u + (K û + K 3 û 3 ) u 2 = 0. (5)

2 Since this is a quadratic surface, we will study further their properties. But before going to general cases, let us observe that the u 3 axis is included in (5) for any values of û and K components. 2 Quadratic Surfaces A quadratic surface is a set of points (x, y, z) R 3 satisfying Ax 2 + By 2 + Cz 2 + 2Dxy + 2Exz + 2F yz + 2Gx + 2Hy + 2Iz + J = 0 (6) for some A, B, C, D, E, F, G, H, I, J R with A, B, C, D, E, F not all zero. We will denote by Q the quadratic form Q (x, y, z) = Ax 2 + By 2 + Cz 2 + 2Dxy + 2Exz + 2F yz and by M Q its matrix M S will denote the matrix A D E M Q = D B F. E F C A D E G M S = D B F H E F C I. G H I J The signature of a matrix M R n n is the couple (α, β, γ) where α is the number of strictly positive eigenvalues of M, β the number of strictly negative eigenvalues of M and γ the number of zero eigenvalues of M Q. The rank of M is α + β. 2. Classification Tables [2] classifies quadratic surfaces in the following way : Rank of M Q Signature of M Q Surface 3 (3, 0, 0) or (0, 3, 0) Ellipsoid 3 (2,, 0) or (, 2, 0) or 2-sheeted hyperboloid or cone 2 (2, 0, ) or (0, 2, ) Elliptic paraboloid or elliptic cylinder 2 (,, ) Hyperbolic paraboloid or hyperbolic cylinder (, 0, 2) or (0,, 2) Parabolic cylinder A proof for this classification is provided. [] provides a more detailed classification without proof : 2

3 Rank of M Q Rank of M S Signature of M Q Sign of M S Surface 3 4 (3, 0, 0) or (0, 3, 0) Real ellipsoid 3 4 (3, 0, 0) or (0, 3, 0) + Imaginary ellipsoid 3 3 (3, 0, 0) or (0, 3, 0) Imaginary elliptic cone 3 4 (2,, 0) or (, 2, 0) 2-sheeted hyperboloid 3 4 (2,, 0) or (, 2, 0) + -sheeted hyperboloid 3 3 (2,, 0) or (, 2, 0) Real elliptic cone 2 4 (2, 0, ) or (0, 2, ) Elliptic paraboloid 2 3 (2, 0, ) or (0, 2, ) Elliptic cylinder 2 2 (2, 0, ) or (0, 2, ) Imaginary intersecting planes 2 4 (,, ) + Hyperbolic paraboloid 2 3 (,, ) Hyperbolic cylinder 2 2 (,, ) Real intersecting planes 3 Parabolic cylinder 2 Parallel planes Coincident planes 3 Our surface Our surface has A = B = C = 0, D = 2 (K 2 K ), E = 2 K 23, F = 2 K 3, G = 2 (K 2û 2 + K 23 û 3 ), H = 2 (K û + K 3 û 3 ) and I = J = 0. So M Q = and its characteristic polynomial is 0 2 (K 2 K ) 2 K 23 2 (K 2 K ) 0 2 K 3 2 K 23 2 K 3 0 (7) P MQ (λ) = λ 3 + (K 2 K ) 2 + K K λ (K 2 K ) K 3 K 23. (8) 4 For further simplification we will denote α = (K2 K)2 +K 2 3 +K and β = (K2 K)K3K23 4. So in order to study the eigenvalues of M Q, we will study the roots of the polynomial P (λ) = λ 3 αλ + β. (9) We can observe that α 0. The case α = 0 implies K 2 K = K 3 = K 23 = 0 and therefore M Q = 0, which means that equation (5) does not define a quadratic surface anymore. We will study this degenerate case later, but in a first time, we will assume that α > 0. Furthermore, if λ, λ 2 and λ 3 are the eigenvalues of M Q, we have P (λ) = (λ λ ) (λ λ 2 ) (λ λ 2 ) = λ 3 (λ + λ 2 + λ 3 ) λ 2 + (λ λ 2 + λ λ 3 + λ 2 λ 3 ) λ λ λ 2 λ 3. (0) So λ λ 2 + λ λ 3 + λ 2 λ 3 < 0 which implies there is one positive and one negative eigenvalue. The third one can be either positive, negative or zero. M S = 0 2 (K 2 K ) 2 K 23 2 (K 2û 2 + K 23 û 3 ) 2 (K 2 K ) 0 2 K 3 2 (K û + K 3 û 3 ) 2 K 23 2 K () 2 (K 2û 2 + K 23 û 3 ) 2 (K û + K 3 û 3 ) 0 0 3

4 and its determinant is ( ) 2 K2 K 3 û 2 K K 23 û M S = (2) 4 We have M S 0 for any values of the components of K and û. 3. Case : K K 2, K 3 0 K 23 and K 2 K 3 û 2 K K 23 û Under the above assumptions, M Q has rank 3, M S has rank 4 and its determinant is strictly positive. We remarked before that M Q has one positive and one negative eigenvalue so its signature is either (2,, 0) or (, 2, 0). So in this case, the surface is a -sheeted hyperboloid. 3.2 Case 2: K K 2, K 3 0 K 23 and K 2 K 3 û 2 = K K 23 û This case is similar to case, but M S has rank 3 and its determinant is 0. The surface is then a real elliptic cone. 3.3 Case 3: Either K = K 2 or K 3 = 0 or K 23 = 0 and K 2 K 3 û 2 K K 23 û In this case, M Q has rank 2 and M S has rank 4 and its determinant is strictly positive. So the surface is a hyperbolic paraboloid. 3.4 Case 4: K = K 2, K 3 0 K 23 and K 2 K 3 û 2 = K K 23 û M Q has rank 2 and K 3 û 2 = K 23 û so û = K3 K 23 û 2. So K 23 2 (K û 2 + K 23 û 3 ) M S = K K K 23 (K û 2 + K 23 û 3 ) 2 K 23 2 K (K K û 2 + K 23 û 3 ) 3 2 K 23 (K û 2 + K 23 û 3 ) 0 0 so it has rank 2. In (9) we have α = K2 3 +K and β = 0. So P (λ) = λ 3 αλ = λ ( λ + α ) ( λ α ). (3) So the two nonzero eigenvalue of M Q are α and α. (,, ). So the surface has shape of real intersecting planes. The signature of M Q is therefore 3.5 Case 5: K K 2 and either K 3 = 0 or K 23 = 0 and K 2 K 3 û 2 = K K 23 û In this case, M Q has rank 2. We have either K 3 = 0 = K û and M S = 0 2 (K 2 K ) 2 K 23 2 (K 2û 2 + K 23 û 3 ) 2 (K 2 K ) K (K 2û 2 + K 23 û 3 )

5 or K 23 = 0 = K 2 û 2 and 0 2 (K 2 K ) 0 0 M S = 2 (K 2 K ) 0 2 K 3 2 (K û + K 3 û 3 ) 0 2 K (K û + K 3 û 3 ) 0 0 So the rank of M S is 2. As in previous case, The signature of M Q is (,, ) so we have again intersecting planes. 3.6 Case 6: K = K 2 and either K 3 = 0 or K 23 = 0 and K 2 K 3 û 2 K K 23 û Same as case Case 7: K = K 2 and either K 3 = 0 or K 23 = 0 and K 2 K 3 û 2 = K K 23 û M Q has rank 2. Since K 2 K 3 û 2 = K K 23 û, we have either K 3 = 0 = K û and M S = or K 23 = 0 = K 2 û 2 and So M S has also rank K 23 2 (K 2û 2 + K 23 û 3 ) K (K 2û 2 + K 23 û 3 ) M S = K 3 2 (K û + K 3 û 3 ) 0 2 K (K û + K 3 û 3 ) 0 0 As in case 4 nonzero eigenvalue of M Q are α and α. So the surface has shape of real intersecting planes. 3.8 Case 8: K K 2, K 3 = 0 = K 23 = 0 and K û 0 K 2 û 2 M Q has rank 2 and K 2 K 3 û 2 = K K 23 û so M S = 0. It is easy to verify that M S has rank (K 2 K ) 0 2 K 2û 2 M S = 2 (K 2 K ) K û K 2û 2 2 K û 0 0 As in case 4, the two nonzero eigenvalue of M Q are α = K2 K 2 and α. So we have a hyperbolic cylinder. 3.9 Case 9: K K 2, K 3 = 0 = K 23 = 0 and either K û = 0 or K 2 û 2 = 0 Then M Q and M S have rank 2. The nonzero eigenvalues of M Q still have opposite signs so we have intersecting planes. 5

6 3.0 Case 0: K K 2, K 3 = 0 = K 23 = 0 and K û = 0 = K 2 û 2 Same as case Non quadratic case : K = K 2, K 3 = 0 = K 23 = 0 With these assumptions, the equation (5) becomes K û u 2 K 2 û 2 u = 0. (4) Provided at least one of K û, K 2 û 2 is nonzero, this is a plane. Otherwise, this is not a surface. 4 Summarizing table We set κ = K 2 K 3 û 2 K K 23 û. H stands for one-sheeted hyperboloid, HP for hyperbolic paraboloid, EC for elliptic cone, IP for intersecting planes and HC for hyperbolic cylinder. κ K 2 K K 3 K 23 K û K 2 û 2 Surface u 3 -axis H HP HP HP HP HP ( ) ( ) EC 0, 0, Kû K 3 + û 3 = 0, 0, K2û2 K 23 + û 3 ( ) IP 0, 0, Kû K 3 + û 3 ( ) IP 0, 0, K2û2 K 23 + û HC IP IP IP ( Whole ) axis IP 0, 0, Kû K 3 + û 3 = ( ) IP 0, 0, Kû K 3 + û 3 ( ) IP 0, 0, K2û2 K 23 + û Plane Plane 5 X-space ( 0, 0, K2û2 K 23 + û 3 ) We set x = Pu + w with and d b P = d b 0 0 c a w = c a bc+ad 2bd 6

7 where a = Kû+K3û3 K 2 K, b = K3 K 2 K, c = K2û2+K23û3 K 2 K and d = K23 K 2 K. This transforms (5) into equation in the canonical form 4 x2 4 x2 2 bdx 2 (bc ad)2 3 + = 0. (5) 4bd However, this equation is of not useful in the study limiting cases : K 2 = K is not allowed because of factor bd multiplying x 3, and K 23 = 0 and K 23 = 0 are forbidden by the division of the constant coefficient by b and d. We can still see what M Q and M S look like in this configuration : and M Q = bd M S = bd We see immediately that det M S = 64 (bc ad)2 0. (bc ad) 2 4bd But bc ad = K3K2û2 K23Kû (K 2 K ) 2 so if K K 2, det M S in x-space is zero if and only if det M S in u-space is zero.. 6 M-space Remember that m = K (u û) u = K m + û. So equation (5) is 0 = ξ K (u û) = Au m = A ( K m + û ) m = AK m m + Aû m (6) where 0 0 A = K = K 2 K 3 K23 2 K 3 K 23 K 2 K 3 K 3 K 23 K K 3 K3 2 K K 23 det K K 3 K 2 K K 23 K K 2 so (6) becomes ( K3 K 23 m 2 + K 3 K 23 m ( K3 2 K 2 ) ) 23 + (K 2 K ) K 3 m m 2 + K K 23 m m 3 K 2 K 3 m 2 m 3 det K û 2 m + û m 2 = 0. Note that positive definiteness of K ensures det K = K K 2 K 3 K 2 K3 2 K K23 2 > 0. In particular, we cannot have K = 0 and K 2 = 0 at the same time. We have in m-space 7

8 M Q = det K and M S = 2 det K 2 ( ) K 3 K 23 2 K 2 3 K (K 2 K ) K 3 2 K ( ) K 23 K 2 3 K (K 2 K ) K 3 K 3 K 23 2 K 2K 3 2 K K 23 2 K 2K 3 0 det K K ( ) 3K 23 2 det K K 2 3 K (K 2 K ) K 3 2 det K K ( ) K 23 2û2 K 2 3 K (K 2 K ) K 3 det K K 3K 23 2 det K K 2K 3 2û 2 det K K K 23 2 det K K 2K û2 2û 0 0 So det M S = 6(det K) 2 (K K 23 û K 3 K 2 û 2 ) 2. Determinant of M Q is det M Q = 4 (det K) 2 (K 2 K ) K 3 K 23. Therefore, the presence of K 3 in equation (5) expressed in m-space has no influence on classification of the surface. 7 Hyperbolic paraboloids We saw that when K 2 K 3 û 2 K K 23 û 0 and one or two of K 2 K, K 3 and K 23 are zero (but not K 3 and K 23 at the same time), the surface is a hyperbolic paraboloid. A hyperbolic paraboloid is a mapping of a plane so we are interested in finding this plane, that we will identify by its normal p n. It also has a saddle point s at which the normal direction to the surface is p n. There are two canonical equations of a hyperbolic paraboloid: and z = y2 b 2 x2 a 2 (7) In both these canonical forms, p n = (0, 0, ) T and s = (0, 0, 0). In cases K = K 2 and either K 3 or K 23 = 0, (5) becomes either z = xy. (8) or K 23 u u 3 (K 2 û 2 + K 23 û 3 ) u + K û u 2 = 0 ((K 2 û 2 + K 23 û 3 ) K 23 u 3 ) u = K û u 2 K 3 u 2 u 3 K 2 û 2 u + (K û + K 3 û 3 ) u 2 = 0 ((K û + K 3 û 3 ) K 3 u 3 ) u 2 = K 2 û 2 u. 8

9 ( So, in the first case, p n is parallel to the u 2 -axis and s = 0, 0, K2û2+K23û3 ( case, p n is parallel to the u -axis and s = 0, 0, Kû+K3û3 K 3 ). In cases with either K = K 2 or K 3 = 0 or K 23 = 0, we have K 23 ) and in the second or K 23 u u 3 K 3 u 2 u 3 (K 2 û 2 + K 23 û 3 ) u + (K û + K 3 û 3 ) u 2 = 0 (K 23 u K 3 u 2 ) u 3 = (K 2 û 2 + K 23 û 3 ) u (K û + K 3 û 3 ) u 2 or (K 2 K ) u u 2 + K 23 u u 3 (K 2 û 2 + K 23 û 3 ) u + K û u 2 = 0 ((K 2 K ) u 2 + K 23 u 3 ) u = (K 2 û 2 + K 23 û 3 ) u K û u 2 (K 2 K ) u u 2 K 3 u 2 u 3 K 2 û 2 u + (K û + K 3 û 3 ) u 2 = 0 ((K 2 K ) u K 3 u 3 ) u 2 = K 2 û 2 u (K û + K 3 û 3 ) u 2. So in all these cases, s = (0, 0, 0) and p n = ((K 2 û 2 + K 23 û 3 ), (K û + K 3 û 3 ), 0) T (with K 3 or K 23 = 0 in the second and third cases). 8 Intersecting planes When we get intersecting planes, we would like to know which line is in both planes. This case appears every time K 2 K 3 û 2 K K 23 û = 0 and one or two of K 2 K, K 3 and K 23 = 0, except when K 3 = K 23 = 0 and K û 0 K 2 û 2. As for hyperbolic paraboloid, there are two canonical equations for intersecting planes: and y 2 b 2 x2 a 2 = 0 (9) xy = 0. (20) In these forms, the intersection line is the z-axis, i.e. points (x, y, z) R 3 with x = 0 and y = 0. Suppose now K 2 K 3 û 2 K K 23 û = 0. Then (5) becomes - if K 2 K = 0 : ( ( )) K û (K 23 u K 3 u 2 ) u 3 + û 3 = 0 K 3 {( ) } so the intersection line is u, K23 K 3 u, Kû K 3 + û 3 : u R. - If K 3 = 0: u ((K 2 K ) u 2 + K 23 u 3 (K 2 û 2 + K 23 û 3 )) = 0 {( ) } so the intersection line is 0, K2û2+K23û3 K 2 K K23 K 2 K u 3, u 3 : u 3 R. - If K 23 = 0: 9

10 u 2 ((K 2 K ) u K 3 u 3 + (K û + K 3 û 3 )) = 0 {( ) } K so the intersection line is 0, 3 K 2 K u 3 Kû+K3û3 K 2 K, u 3 : u 3 R. - If K 2 K = 0 and K 3 = 0: u (K 23 u 3 (K 2 û 2 + K 23 û 3 )) = 0 so the intersection line is {( ) 0, u 2, K2û2+K23û3 K 23 : u 2 R }. - If K 2 K = 0 and K 23 = 0: u 2 ( K 3 u 3 + (K û + K 3 û 3 )) = 0 so the intersection line is {( ) u, 0, Kû+K3û3 K 3 : u R }. - If K 3 = 0, K 23 = 0 and K û = 0 : u ((K 2 K ) u 2 K 2 û 2 ) = 0 {( ) } so the intersection line is 0, K2û2 K 2 K, u 3 : u 3 R. - If K 3 = 0, K 23 = 0 and K 2 û 2 = 0 : ((K 2 K ) u + K û ) u 2 = 0 {( ) } so the intersection line is Kû K 2 K, 0, u 3 : u 3 R. 9 u 3 -axis Remember that for going from (4) to (5), we made the assumption that u and u 2 are not zero at the same time, i.e. we are not on the u 3 -axis. This means that even if the u 3 axis is in all the surfaces we have seen, it may not satisfy equation (4). In fact, as we will see, the u 3 -axis satisfies (4) only in one case. Suppose the u 3 -axis satisfies (4) then all u 3 R satisfy K 3 u 3 = K û + K 3 û 3 K 23 u 3 = K 2 û 2 + K 23 û 3 (2) (K 3 µ ) u 3 = K 3 û + K 23 û 2 + K 3 û 3 µ 2 The third equation gives values for µ and µ 2 and the first two lead to K 3 = 0, K 23 = 0, K û = 0, K 2 û 2 = 0. Now see if there is points on the u 3 -axis satisfying (4) in other cases, i.e. if there is u 3 R satisfying (2). Note that in (2), K 3 = 0 K û = 0 and K 23 = 0 K 2 û 2 = 0. So all surfaces with either K 3 = 0 and K û 0 or K 23 = 0 and K 2 û 2 0 have no point on their u 3 -axis satisfying (4). There is a solution only if K 3 K 2 û 2 K 23 K û = 0. Provided( one of K 3, K 23 ) 0, ( there is a unique ) point of the u 3 -axis satisfying (4). This point is either 0, 0, Kû K 3 + û 3 or 0, 0, K2û2 K 23 + û 3 or both if K 3 0 K 23 and they are equal in this case. Note that all forms of (4) with at least one point of the u 3 -axis satisfying it become equations of intersecting planes when transformed to (5), except the first one which is an elliptic cone. Here is a summarizing table of these results : 0

11 K 2 K 3 û 2 K K 23 û K 2 K K 3 K 23 K û K 2 û 2 ( u ) 3 -axis ( ) , 0, Kû K 3 + û 3 = 0, 0, K2û2 K 23 + û 3 ( ) ( ) , 0, Kû K 3 + û 3 = 0, 0, K2û2 K 23 + û 3 ( ) , 0, K2û2 K 23 + û 3 ( ) , 0, Kû K 3 + û 3 ( ) , 0, K2û2 K 23 + û 3 ( ) , 0, Kû K 3 + û Whole axis Part II Maximum Stability Length 0 Context The energy of a rod q(s) = (r(s), R(s)) where s [0, L] is with W given by (). E [q(s)] = L 0 W (u(s) û(s), v(s) ˆv(s))ds (22) Computing its first variation, we get equilibrium conditions that lead, for a given û and ˆv = v = (0, 0, ), to the surfaces we saw in part I. Computing the second variation of the energy, and finding the first conjugate point to 0 of its Jacobi equation allows to set a maximum stability length for the rod, i.e. the maximal length the rod can have remaining stable. In autumn 20, I did a semester project consisting in writing a matlab code to compute the maximum stability length of helices on a hyperboloid. The computation on each point of a 00 by 00 mesh took about forty-five minutes on my laptop. During the same semester, two other student made an application in C++ to visualize the hyperboloid corresponding to chosen parameters. So in order to integrate the computation of the maximum stability length into the helix viewer program, I translated the code from matlab to C++. To make it even faster, we decided to externalize the computations on a server. The translation in C++ really improves the computation time. Now, the computation of a whole mesh takes from eight seconds for a far from degenerate case to fifteen minutes for a case approaching a degenerate one, for a 00 by 00 points mesh. However, the externalization on the server is not yet useful. A computation taking eight seconds on my laptop takes one minute when sent on the server. This is probably due to transfer time. Case illustration The images shown in the following represent the level sets of the maximum stability length on the surface. The meshes have been computed with the helix viewer application. This application doesn t yet include the degenerate cases, so all the images represent hyperboloid. The curves represent levels, 2, 3, 4, 5, 6, 7, 8, 9, 0, 20, 30, 40, 50, 60, 70, 80, 90 and 00 with blue for low values and red for high values. This implies solving an ordinary differential equation y = Ay where A is a twelve by twelve matrix with six different initial conditions.

12 Each degenerate case is approached by modifying some parameters. There is a symmetry between K 3 and K 23 and between K û and K 2 û 2 so there is only one set of pictures for symmetric cases. It will be easier to analyze these pictures when we include the maximum stability length in the helix viewer, so I will only make as few descriptive comments.. Hyperbolic Paraboloid with K 3 = 0 (a) K 3 = 0.5 (b) K 3 = 0.4 (c) K 3 = 0.3 (d) K 3 = 0.2 (e) K 3 = 0. (f) K 3 = 0.08 (g) K 3 = 0.06 (h) K 3 = 0.04 (i) K 3 = 0.02 Figure : K 3 0 K =, K 2 =.5, K 3 =.2, K 23 = 0.6, û = (0.7, 0.4, 0.6) This set of image shows evolution of a non degenerate case to a hyperbolic paraboloid by decreasing the value of K 3 up to It would have been interesting to go even lower to see what happens with these curves appearing in the middle of the image, but it already took a lot of time to compute the stability lengths for K 3 =

13 (j) K 3 = 0.0 (k) K 3 = (l) K 3 = (m) K 3 = (n) K 3 = (o) K 3 = 0.00 Figure 3

14 .2 Hyperbolic Paraboloid with K = K 2 (a) K 2 =.5 (b) K 2 =.4 (c) K 2 =.3 (d) K 2 =.2 (e) K 2 =. (f) K 2 =.09 (g) K 2 =.08 (h) K 2 =.07 (i) K 2 =.06 Figure 2: K 2 K 0 K =, K 3 =.2, K 3 = 0.5, K 23 = 0.6, û = (0.7, 0.4, 0.6) Here we are again approaching a hyperbolic paraboloid but by decreasing the value of K 2 to K =. Here high values first gather near one line (the hyperboloid closes on itself) and then a second line appears nd the first one disappears. 4

15 (j) K 2 =.05 (k) K 2 =.04 (l) K 2 =.03 (m) K 2 =.02 (n) K 2 =.0 (o) K 2 =.009 (p) K 2 =.008 (q) K 2 =.007 (r) K 2 =.006 (s) K 2 =.005 (t) K 2 =.004 (u) K 2 =.003 Figure 2 5

16 .3 Hyperbolic Paraboloid with K = K 2 and K 3 = 0 (a) K 2 =.5, K 3 = 0.5 (b) K 2 =.4, K 3 = 0.4 (c) K 2 =.3, K 3 = 0.3 (d) K 2 =.2, K 3 = 0.2 (e) K 2 =., K 3 = 0. (f) K 2 =.09, K 3 = 0.09 (g) K 2 =.08, K 3 = 0.08 (h) K 2 =.07, K 3 = 0.07 (i) K 2 =.06, K 3 = 0.06 Figure 3: K 2 K 0 and K 3 0 K =, K 3 =.2, K 23 = 0.6, û = (0.7, 0.4, 0.6) This time, we tend to a hyperbolic paraboloid by letting both K 3 0 and K 2 K =. It looks more like the second case than the first one. The computation time began to be to long for higher values of K 3 and K 2 than in previous cases. 6

17 (j) K 2 =.05, K 3 = 0.05 (k) K 2 =.04, K 3 = 0.04 (l) K 2 =.03, K 3 = 0.03 (m) K 2 =.02, K 3 = 0.02 Figure 3 7

18 .4 Elliptic cone with K 3 K 2 û 2 K 23 K û = 0 (a) K 3 = 0.5, K 23 = (b) K 3 = 0.6, K 23 =. (c) K 3 = 0.7, K 23 =.2 (d) K 3 = 0.8, K 23 =.3 (e) K 3 = 0.9, K 23 =.4 (f) K 3 = 0.92, K 23 =.42 (g) K 3 = 0.94, K 23 =.44 (h) K 3 = 0.96, K 23 =.46 (i) K 3 = 0.98, K 23 =.48 Figure 4: K 3 K 2 û 2 K 23 K û 0 K =, K 2 =.5, K 3 = 2.5, û = (0.7, 0.7, 0.6) In this case, we let K 3 K 2 û 2 K 23 K û 0 in order to approach an elliptic cone. This is done by having û = û 2 = 0.7, K 3 K 2 =.5 and K 23 K =. In order to have K positive definite, we set K 3 = 2.5. Furthermore, û has been modified, so the starting picture is not the same as in previous cases. The pattern appearing here is really different from all the other cases. 8

19 (j) K 3 = 0.99, K 23 =.49 (k) K 3 = 0.992, K 23 =.492 (l) K 3 = 0.994, K 23 =.494 (m) K 3 = 0.996, K 23 =.496 (n) K 3 = 0.998, K 23 =.498 (o) K 3 = 0.999, K 23 =.499 Figure 4 9

20 .5 Intersecting planes with K 3 K 2 û 2 K 23 K û = 0 and K 3 = 0 (a) K 3 = 0.5, û = 0.5 (b) K 3 = 0.4, û = 0.4 (c) K 3 = 0.3, û = 0.3 (d) K 3 = 0.2, û = 0.2 (e) K 3 = 0., û = 0. (f) K 3 = 0.08, û = 0.08 (g) K 3 = 0.06, û = 0.06 (h) K 3 = 0.04, û = 0.04 (i) K 3 = 0.02, û = 0.02 Figure 5: K 3 0 and K 3 K 2 û 2 K 23 K û 0 K =, K 2 =.5K 3 =.2, K 23 = 0.6, û 2 0.7, û 3 = 0.6 This is the first of five cases of intersecting plane. This one is obtained by letting K 2 K = and K 3, K 23 0 which implies K 3 K 2 û 2 K 23 K û 0. The pattern obtained is especially interesting. 20

21 (j) K 3 = 0.0, û = 0.0 (k) K 3 = 0.008, û = (l) K 3 = 0.006, û = (m) K 3 = 0.004, û = (n) K 3 = 0.002, û = (o) K 3 = 0.00, û = 0.00 (p) K 3 = , û = (q) K 3 = , û = (r) K 3 = , û = (s) K 3 = , û = Figure 5 2

22 .6 Hyperbolic cylinder with K 3 K 2 û 2 K 23 K û = 0, K 3 = 0 = K 23 and K û 0 K 2 û 2 (a) K 3 = 0.5, K 23 = 0.5 (b) K 3 = 0.4, K 23 = 0.4 (c) K 3 = 0.3, K 23 = 0.3 (d) K 3 = 0.2, K 23 = 0.2 (e) K 3 = 0.08, K 23 = 0.08 (f) K 3 = 0.06, K 23 = 0.06 (g) K 3 = 0.04, K 23 = 0.04 (h) K 3 = 0.02, K 23 = 0.02 (i) K 3 = 0.0, K 23 = 0.0 Figure 6: K 3, K 23 0 and K 3 K 2 û 2 K 23 K û 0 K =, K 2 =.5, K 3 =.2, û = (0.7, 0.4, 0.6) This is the only case of hyperbolic cylinder. It is obtained by decreasing K 3 and K 23 to 0. 22

23 (j) K 3 = 0.008, K 23 = (k) K 3 = 0.004, K 23 = (l) K 3 = 0.002, K 23 = Figure 6 23

24 .7 Intersecting planes with K 3 K 2 û 2 K 23 K û = 0, K 3 = 0 = K 23, K û = 0 and K 2 û 2 0 (a) K 3 = 0.5, K 23 = 0.5 (b) K 3 = 0.4, K 23 = 0.4 (c) K 3 = 0.3, K 23 = 0.3 (d) K 3 = 0.2, K 23 = 0.2 (e) K 3 = 0., K 23 = 0. (f) K 3 = 0.08, K 23 = 0.08 (g) K 3 = 0.06, K 23 = 0.06 (h) K 3 = 0.04, K 23 = 0.04 (i) K 3 = 0.02, K 23 = 0.02 Figure 7: K 3, K 23 0, K 3 K 2 û 2 K 23 K û 0 and K û = 0 K =, K 2 =.5, K 3 =.2, û = (0, 0.4, 0.6) This case of intersecting planes, obtained by letting K 3, K 23 0 with û = 0 so K û = 0 looks somewhat like the hyperbolic cylinder. 24

25 (j) K 3 = 0.0, K 23 = 0.0 (k) K 3 = 0.008, K 23 = (l) K 3 = 0.006, K 23 = (m) K 3 = 0.004, K 23 = Figure 7 25

26 .8 Intersecting planes with K 3 K 2 û 2 K 23 K û = 0, K 3 = 0 = K 23 and K û = 0 = K 2 û 2 (a) K 3 = 0.5, K 23 = 0.5, û 2 = 0.5 (b) K 3 = 0.4, K 23 = 0.4, û 2 = 0.4 (c) K 3 = 0.3, K 23 = 0.3, û 2 = 0.3 (d) K 3 = 0.2, K 23 = 0.2, û 2 = 0.2 (e) K 3 = 0., K 23 = 0., û 2 = 0. (f) K 3 = 0.08, K 23 = 0.08, û 2 = 0.08 (g) K 3 = 0.06, K 23 = 0.06, û 2 = (h) K 3 = 0.04, K 23 = 0.04, û 2 = (i) K 3 = 0.02, K 23 = 0.02, û 2 = Figure 8: K 3, K 23 0, K 3 K 2 û 2 K 23 K û 0, K 2 û 2 0 and K û = 0 K =, K 2 =.5, K 3 =.2, û = 0, û 2 = 0.6 We are here approaching the only surface having the whole u 3 -axis in it. K 3, K 23 0 and û 2 0 with û = 0. For this we let 26

27 (j) K 3 = 0.0, K 23 = 0.0, û 2 = (k) K 3 = 0.008, K 23 = 0.008, û 2 = (l) K 3 = 0.006, K 23 = 0.006, û 2 = (m) K 3 = 0.004, K 23 = 0.004, û 2 = (n) K 3 = 0.002, K 23 = 0.002, û 2 = (o) K 3 = 0.00, K 23 = 0.00, û 2 = Figure 8 27

28 .9 Intersecting planes with K 3 K 2 û 2 K 23 K û = 0 and K = K 2 (a) K 2 =.5 (b) K 2 =.4 (c) K 2 =.3 (d) K 2 =.2 (e) K 2 =. (f) K 2 =.08 (g) K 2 =.06 (h) K 2 =.04 (i) K 2 =.02 Figure 9: K 2 K 0 and K 3 K 2 û 2 K 23 K û 0 K =, K 3 =.2, K 3 = 0.5, K 23 = 0.5, û = (0.7, 0.7, 0.6) Another case of intersecting planes, with K 2 K =. To have K 3 K 2 û 2 K 23 K û 0, we set K 3 = K 23 = 0.5 and û = û 2 = 0.7. Here, the values of the maximum stability length seem to grow bigger and bigger. In the last picture, there is no level curves for values under

29 (j) K 2 =.0 (k) K 2 =.008 (l) K 2 =.006 (m) K 2 =.004 Figure 9 29

30 .0 Intersecting planes with K 3 K 2 û 2 K 23 K û = 0, K = K 2, K 3 = 0 and K û = 0 (a) K 2 =.5, K 3 = 0.5 (b) K 2 =.4, K 3 = 0.4 (c) K 2 =.3, K 3 = 0.3 (d) K 2 =.2, K 3 = 0.2 (e) K 2 =., K 3 = 0. (f) K 2 =.08, K 3 = 0.08 (g) K 2 =.06, K 3 = 0.06 (h) K 2 =.04, K 3 = 0.04 (i) K 2 =.02, K 3 = 0.02 Figure 0: K 2 K 0, K 3 0, K 3 K 2 û 2 K 23 K û 0 and K û = 0 K =, K 3 =.2, K 23 = 0.5, û = (0, 0.7, 0.6) The last case of intersecting planes. We set û = 0 to have K û = 0 and we let K 2 K = and K 3 0. The evolution looks rather like the three cases of hyperbolic paraboloid. 30

31 (j) K 2 =.0, K 3 = 0.0 (k) K 2 =.008, K 3 = (l) K 2 =.006, K 3 = (m) K 2 =.004, K 3 = (n) K 2 =.002, K 3 = (o) K 2 =.00, K 3 = 0.00 (p) K 2 =.0008, K 3 = (q) K 2 =.0006, K 3 = Figure 0 3

32 . Plane with K 3 K 2 û 2 K 23 K û = 0, K = K, K 3 = 0, K 23 = 0 and K û = 0 (a) K 2 =.5, K 3 = 0.5,K 23 = 0.5 (b) K 2 =.4, K 3 = 0.4,K 23 = 0.4 (c) K 2 =.3, K 3 = 0.3,K 23 = 0.3 (d) K 2 =.2, K 3 = 0.2,K 23 = 0.2 (e) K 2 =., K 3 = 0.,K 23 = 0. (f) K 2 =.08, K 3 = 0.08,K 23 = 0.08 (g) K 2 =.06, K 3 = 0.06,K 23 = (h) K 2 =.04, K 3 = 0.04,K 23 = (i) K 2 =.02, K 3 = 0.02,K 23 = Figure : K 2 K 0, K 3, K 23 0, K 3 K 2 û 2 K 23 K û 0 and K û = 0 K =, K 3 =.2, û = (0, 0.7, 0.6) Letting K 2 K = and K 3, K 23 0 with û = 0, we approach a plane. The high values gather on a line but only on one side of the u 3 -axis. 32

33 (j) K 2 =.0, K 3 = 0.0,K 23 = 0.0 Figure 33

34 2 Conclusion Both study of degenerate cases and improvement of the computation time for maximum stability lengths open new perspectives for the helix viewer application. As I said, the externalization of the maximum stability length is not optimal, but I have a few ideas to improve it. We also have talked a little about how to include visualization of degenerate cases in the helix viewer. References [] Quadratic surface. [2] Quadriques, march