Classification of Quadratic Surfaces Pauline Rüegg-Reymond June 14, 2012 Part I Classification of Quadratic Surfaces 1 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 ∈ R3. The strains of the reference state helix are denoted by ˆu. The strain-energy density of a helix given by u is the quadratic function 1 W (u − ˆu) = (u − ˆu) · K (u − ˆu) (1) 2 where K ∈ R3×3 is assumed to be of the form K1 0 K13 K = 0 K2 K23 K13 K23 K3 with K1 6 K2. A helical rod also has stresses m ∈ R3 related to strains u through balance laws, which are equivalent to m = µ1u + µ2e3 (2) for some scalars µ1 and µ2 and e3 = (0, 0, 1), and constitutive relation m = K (u − ˆu) . (3) Every helix u at equilibrium, with reference state uˆ, is such that there is some scalars µ1, µ2 with µ1u + µ2e3 = K (u − ˆu) µ1u1 = K1 (u1 − uˆ1) + K13 (u3 − uˆ3) (4) ⇔ µ1u2 = K2 (u2 − uˆ2) + K23 (u3 − uˆ3) µ1u3 + µ2 = K13 (u1 − uˆ1) + K23 (u1 − uˆ1) + K3 (u3 − uˆ3) Assuming u1 and u2 are not zero at the same time, we can rewrite this surface (K2 − K1) u1u2 + K23u1u3 − K13u2u3 − (K2uˆ2 + K23uˆ3) u1 + (K1uˆ1 + K13uˆ3) u2 = 0. (5) 1 Since this is a quadratic surface, we will study further their properties. But before going to general cases, let us observe that the u3 axis is included in (5) for any values of ˆu and K components. 2 Quadratic Surfaces A quadratic surface is a set of points (x, y, z) ∈ R3 satisfying Ax2 + By2 + Cz2 + 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) = Ax2 + By2 + Cz2 + 2Dxy + 2Exz + 2F yz and by MQ its matrix ADE MQ = DBF . EFC MS will denote the matrix ADEG DBFH MS = . EFCI GHIJ The signature of a matrix M ∈ Rn×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 MQ. The rank of M is α + β. 2.1 Classification Tables [2] classifies quadratic surfaces in the following way : Rank of MQ Signature of MQ Surface 3 (3, 0, 0) or (0, 3, 0) Ellipsoid 3 (2, 1, 0) or (1, 2, 0) 1 or 2-sheeted hyperboloid or cone 2 (2, 0, 1) or (0, 2, 1) Elliptic paraboloid or elliptic cylinder 2 (1, 1, 1) Hyperbolic paraboloid or hyperbolic cylinder 1 (1, 0, 2) or (0, 1, 2) Parabolic cylinder A proof for this classification is provided. [1] provides a more detailed classification without proof : 2 Rank of MQ Rank of MS Signature of MQ Sign of |MS| 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, 1, 0) or (1, 2, 0) − 2-sheeted hyperboloid 3 4 (2, 1, 0) or (1, 2, 0) + 1-sheeted hyperboloid 3 3 (2, 1, 0) or (1, 2, 0) Real elliptic cone 2 4 (2, 0, 1) or (0, 2, 1) − Elliptic paraboloid 2 3 (2, 0, 1) or (0, 2, 1) Elliptic cylinder 2 2 (2, 0, 1) or (0, 2, 1) Imaginary intersecting planes 2 4 (1, 1, 1) + Hyperbolic paraboloid 2 3 (1, 1, 1) Hyperbolic cylinder 2 2 (1, 1, 1) Real intersecting planes 1 3 Parabolic cylinder 1 2 Parallel planes 1 1 Coincident planes 3 Our surface 1 1 1 1 Our surface has A = B = C = 0, D = 2 (K2 − K1) ,E = 2 K23,F = − 2 K13, G = − 2 (K2uˆ2 + K23uˆ3), 1 H = 2 (K1uˆ1 + K13uˆ3) and I = J = 0. So 1 1 0 2 (K2 − K1) 2 K23 1 1 MQ = 2 (K2 − K1) 0 − 2 K13 (7) 1 1 2 K23 − 2 K13 0 and its characteristic polynomial is (K − K )2 + K2 + K2 (K − K ) K K P (λ) = −λ3 + 2 1 13 23 λ − 2 1 13 23 . (8) MQ 4 4 2 2 2 (K2−K1) +K13+K23 (K2−K1)K13K23 For further simplification we will denote α = 4 and β = 4 . So in order to study the eigenvalues of MQ, we will study the roots of the polynomial P (λ) = λ3 − αλ + β. (9) We can observe that α > 0. The case α = 0 implies K2 − K1 = K13 = K23 = 0 and therefore MQ = 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 λ1, λ2 and λ3 are the eigenvalues of MQ, we have P (λ) = (λ − λ1)(λ − λ2)(λ − λ2) 3 2 (10) = λ − (λ1 + λ2 + λ3) λ + (λ1λ2 + λ1λ3 + λ2λ3) λ − λ1λ2λ3. So λ1λ2 + λ1λ3 + λ2λ3 < 0 which implies there is one positive and one negative eigenvalue. The third one can be either positive, negative or zero. 1 1 1 0 2 (K2 − K1) 2 K23 − 2 (K2uˆ2 + K23uˆ3) 1 1 1 2 (K2 − K1) 0 − 2 K13 2 (K1uˆ1 + K13uˆ3) MS = 1 1 (11) 2 K23 − 2 K13 0 0 1 1 − 2 (K2uˆ2 + K23uˆ3) 2 (K1uˆ1 + K13uˆ3) 0 0 3 and its determinant is K K uˆ − K K uˆ 2 |M | = 2 13 2 1 23 1 (12) S 4 We have |MS| > 0 for any values of the components of K and ˆu. 3.1 Case 1: K1 6= K2,K13 6= 0 6= K23 and K2K13uˆ2 6= K1K23uˆ1 Under the above assumptions, MQ has rank 3, MS has rank 4 and its determinant is strictly positive. We remarked before that MQ has one positive and one negative eigenvalue so its signature is either (2, 1, 0) or (1, 2, 0). So in this case, the surface is a 1-sheeted hyperboloid. 3.2 Case 2: K1 6= K2,K13 6= 0 6= K23 and K2K13uˆ2 = K1K23uˆ1 This case is similar to case 1, but MS has rank 3 and its determinant is 0. The surface is then a real elliptic cone. 3.3 Case 3: Either K1 = K2 or K13 = 0 or K23 = 0 and K2K13uˆ2 6= K1K23uˆ1 In this case, MQ has rank 2 and MS has rank 4 and its determinant is strictly positive. So the surface is a hyperbolic paraboloid. 3.4 Case 4: K1 = K2,K13 6= 0 6= K23 and K2K13uˆ2 = K1K23uˆ1 K13 MQ has rank 2 and K13uˆ2 = K23uˆ1 so uˆ1 = uˆ2. So K23 1 1 0 0 2 K23 − 2 (K1uˆ2 + K23uˆ3) 1 1 K13 0 0 − K13 (K1uˆ2 + K23uˆ3) 2 2 K23 MS = 1 1 2 K23 − 2 K13 0 0 1 1 K13 − (K1uˆ2 + K23uˆ3) (K1uˆ2 + K23uˆ3) 0 0 2 2 K23 so it has rank 2. 2 2 K13+K23 In (9) we have α = 4 and β = 0. So √ √ P (λ) = λ3 − αλ = λ λ + α λ − α . (13) √ √ So the two nonzero eigenvalue of MQ are α and − α. The signature of MQ is therefore (1, 1, 1). So the surface has shape of real intersecting planes. 3.5 Case 5: K1 6= K2 and either K13 = 0 or K23 = 0 and K2K13uˆ2 = K1K23uˆ1 In this case, MQ has rank 2. We have either K13 = 0 = K1uˆ1 and 1 1 1 0 2 (K2 − K1) 2 K23 − 2 (K2uˆ2 + K23uˆ3) 1 2 (K2 − K1) 0 0 0 MS = 1 2 K23 0 0 0 1 − 2 (K2uˆ2 + K23uˆ3) 0 0 0 4 or K23 = 0 = K2uˆ2 and 1 0 2 (K2 − K1) 0 0 1 1 1 2 (K2 − K1) 0 − 2 K13 2 (K1uˆ1 + K13uˆ3) MS = 1 . 0 − 2 K13 0 0 1 0 2 (K1uˆ1 + K13uˆ3) 0 0 So the rank of MS is 2. As in previous case, The signature of MQ is (1, 1, 1) so we have again intersecting planes. 3.6 Case 6: K1 = K2 and either K13 = 0 or K23 = 0 and K2K13uˆ2 6= K1K23uˆ1 Same as case 3. 3.7 Case 7: K1 = K2 and either K13 = 0 or K23 = 0 and K2K13uˆ2 = K1K23uˆ1 MQ has rank 2. Since K2K13uˆ2 = K1K23uˆ1, we have either K13 = 0 = K1uˆ1 and 1 1 0 0 2 K23 − 2 (K2uˆ2 + K23uˆ3) 0 0 0 0 MS = 1 2 K23 0 0 0 1 − 2 (K2uˆ2 + K23uˆ3) 0 0 0 or K23 = 0 = K2uˆ2 and 0 0 0 0 1 1 0 0 − 2 K13 2 (K1uˆ1 + K13uˆ3) MS = 1 0 − 2 K13 0 0 1 0 2 (K1uˆ1 + K13uˆ3) 0 0 So MS has also rank 2. √ √ As in case 4 nonzero eigenvalue of MQ are α and − α. So the surface has shape of real intersecting planes. 3.8 Case 8: K1 6= K2,K13 = 0 = K23 = 0 and K1uˆ1 6= 0 6= K2uˆ2 MQ has rank 2 and K2K13uˆ2 = K1K23uˆ1 so |MS| = 0. 1 1 0 2 (K2 − K1) 0 − 2 K2uˆ2 1 1 2 (K2 − K1) 0 0 2 K1uˆ1 MS = 0 0 0 0 1 1 − 2 K2uˆ2 2 K1uˆ1 0 0 It is easy to verify that MS has rank 3. √ √ K2−K1 As in case 4, the two nonzero eigenvalue of MQ are α = 2 and − α.