Classification of Quadratic Surfaces

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 − û) = (u − û) · K (u − û) (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. So we have a hyperbolic cylinder.

3.9 Case 9: K1 6= K2,K13 = 0 = K23 = 0 and either K1uˆ1 = 0 or K2uˆ2 = 0

Then MQ and MS have rank 2. The nonzero eigenvalues of MQ still have opposite signs so we have intersecting planes.

5 3.10 Case 10: K1 6= K2,K13 = 0 = K23 = 0 and K1uˆ1 = 0 = K2uˆ2

Same as case 9.

3.11 Non quadratic case : K1 = K2,K13 = 0 = K23 = 0

With these assumptions, the equation (5) becomes

K1uˆ1u2 − K2uˆ2u1 = 0. (14)

Provided at least one of K1uˆ1,K2uˆ2 is nonzero, this is a plane. Otherwise, this is not a surface.

4 Summarizing table

We set κ = K2K13uˆ2 − K1K23uˆ1. H1 stands for one-sheeted hyperboloid, HP for hyperbolic paraboloid, EC for elliptic cone, IP for intersecting planes and HC for hyperbolic cylinder.

κ K2 − K1 K13 K23 K1uˆ1 K2uˆ2 Surface u3-axis 6= 0 6= 0 6= 0 6= 0 H1 ∅ 6= 0 6= 0 6= 0 0 HP ∅ 6= 0 6= 0 0 6= 0 HP ∅ 6= 0 0 6= 0 6= 0 HP ∅ 6= 0 0 6= 0 0 HP ∅ 6= 0 0 0 6= 0 HP ∅ K1uˆ1 K2uˆ2 0 6= 0 6= 0 6= 0 EC 0, 0, +u ˆ3 = 0, 0, +u ˆ3 K13 K23 K1uˆ1 0 6= 0 6= 0 0 0 IP 0, 0, +u ˆ3 K13 K2uˆ2 0 6= 0 0 6= 0 0 IP 0, 0, +u ˆ3 K23 0 6= 0 0 0 6= 0 6= 0 HC ∅ 0 6= 0 0 0 6= 0 0 IP ∅ 0 6= 0 0 0 0 6= 0 IP ∅ 0 6= 0 0 0 0 0 IP Whole axis K1uˆ1 K2uˆ2 0 0 6= 0 6= 0 IP 0, 0, +u ˆ3 = 0, 0, +u ˆ3 K13 K23 K1uˆ1 0 0 6= 0 0 0 IP 0, 0, +u ˆ3 K13 K2uˆ2 0 0 0 6= 0 0 IP 0, 0, +u ˆ3 K23 0 0 0 0 6= 0 0 Plane ∅ 0 0 0 0 0 6= 0 Plane ∅

5 X-space

We set x = Pu + w with

1 1 −d − b P = 1 −1 d − b  0 0 1 and

−c − a w =  c − a  bc+ad 2bd

6 where a = K1uˆ1+K13uˆ3 , b = K13 , c = K2uˆ2+K23uˆ3 and d = − K23 . K2−K1 K2−K1 K2−K1 K2−K1 This transforms (5) into equation in the canonical form

1 1 (bc − ad)2 x2 − x2 − bdx2 + = 0. (15) 4 1 4 2 3 4bd

However, this equation is of not useful in the study limiting cases : K2 = K1 is not allowed because of factor bd multiplying x3, and K23 = 0 and K23 = 0 are forbidden by the division of the constant coeﬃcient by b and d.

We can still see what MQ and MS look like in this conﬁguration :

 1  4 0 0 1 MQ = 0 − 4 0  0 0 −bd and

 1  4 0 0 0 1 0 − 4 0 0  MS =   . 0 0 −bd 0  (bc−ad)2 0 0 0 4bd

1 2 We see immediately that det MS = 64 (bc − ad) > 0.

K13K2uˆ2−K23K1uˆ1 But bc − ad = 2 so if K1 6= K2, det MS in x-space is zero if and only if det MS (K2−K1) in u-space is zero.

6 M-space

Remember that m = K (u − uˆ) ⇔ u = K−1m + ˆu. So equation (5) is

0 = −ξ · K (u − uˆ) = Au · m = A K−1m + uˆ · m = AK−1m · m + Aˆu · m (16) where

0 −1 0 A = 1 0 0 . 0 0 0

K K − K2 K K −K K  1 2 3 23 13 23 2 13 K−1 = K K K K − K2 −K K det K  13 23 1 3 13 1 23 −K13K2 −K1K23 K1K2 so (16) becomes

1 −K K m2 + K K m2 + K2 − K2 + (K − K ) K m m + K K m m − K K m m det K 13 23 1 13 23 2 13 23 2 1 3 1 2 1 23 1 3 2 13 2 3 − uˆ2m1 +u ˆ1m2 = 0.

2 2 Note that positive deﬁniteness of K ensures det K = K1K2K3 − K2K13 − K1K23 > 0. In particular, we cannot have K1 = 0 and K2 = 0 at the same time. We have in m-space

7  1 2 2 1  −K13K23 2 K13 − K23 + (K2 − K1) K3 2 K1K23 1 1 2 2 1 MQ =  2 K13 − K23 + (K2 − K1) K3 K13K23 − 2 K2K13 det K 1 1 2 K1K23 − 2 K2K13 0 and

 1 1 2 2 1 1  − det K K13K23 2 det K K13 − K23 + (K2 − K1) K3 2 det K K1K23 − 2 uˆ2 1 2 2 1 1 1  2 det K K13 − K23 + (K2 − K1) K3 det K K13K23 − 2 det K K2K13 2 uˆ1  MS =  1 1  .  2 det K K1K23 − 2 det K K2K13 0 0  1 1 − 2 uˆ2 2 uˆ1 0 0

1 2 So det MS = 16(det K)2 (K1K23uˆ1 − K13K2uˆ2) .

Determinant of MQ is

1 det MQ = − (K2 − K1) K13K23. 4 (det K)2

Therefore, the presence of K3 in equation (5) expressed in m-space has no inﬂuence on classiﬁ- cation of the surface.

7 Hyperbolic paraboloids

We saw that when K2K13uˆ2 − K1K23uˆ1 6= 0 and one or two of K2 − K1,K13 and K23 are zero (but not K13 and K23 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 pn. It also has a saddle point s at which the normal direction to the surface is pn. There are two canonical equations of a hyperbolic paraboloid:

y2 x2 z = − (17) b2 a2 and

z = xy. (18)

T In both these canonical forms, pn = (0, 0, 1) and s = (0, 0, 0).

In cases K1 = K2 and either K13 or K23 = 0, (5) becomes either

K23u1u3 − (K2uˆ2 + K23uˆ3) u1 + K1uˆ1u2 = 0

((K2uˆ2 + K23uˆ3) − K23u3) u1 = K1uˆ1u2 or

−K13u2u3 − K2uˆ2u1 + (K1uˆ1 + K13uˆ3) u2 = 0

⇔ ((K1uˆ1 + K13uˆ3) − K13u3) u2 = K2uˆ2u1.

8 K2uˆ2+K23uˆ3 So, in the ﬁrst case, pn is parallel to the u2-axis and s = 0, 0, and in the second K23 K1uˆ1+K13uˆ3 case, pn is parallel to the u1-axis and s = 0, 0, . K13

In cases with either K1 = K2 or K13 = 0 or K23 = 0, we have

K23u1u3 − K13u2u3 − (K2uˆ2 + K23uˆ3) u1 + (K1uˆ1 + K13uˆ3) u2 = 0

⇔ (K23u1 − K13u2) u3 = (K2uˆ2 + K23uˆ3) u1 − (K1uˆ1 + K13uˆ3) u2 or

(K2 − K1) u1u2 + K23u1u3 − (K2uˆ2 + K23uˆ3) u1 + K1uˆ1u2 = 0

⇔ ((K2 − K1) u2 + K23u3) u1 = (K2uˆ2 + K23uˆ3) u1 − K1uˆ1u2 or

(K2 − K1) u1u2 − K13u2u3 − K2uˆ2u1 + (K1uˆ1 + K13uˆ3) u2 = 0

⇔ ((K2 − K1) u1 − K13u3) u2 = K2uˆ2u1 − (K1uˆ1 + K13uˆ3) u2.

T So in all these cases, s = (0, 0, 0) and pn = ((K2uˆ2 + K23uˆ3) , − (K1uˆ1 + K13uˆ3) , 0) (with K13 or K23 = 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 K2K13uˆ2 − K1K23uˆ1 = 0 and one or two of K2 − K1,K13 and K23 = 0, except when K13 = K23 = 0 and K1uˆ1 6= 0 6= K2uˆ2. As for hyperbolic paraboloid, there are two canonical equations for intersecting planes:

y2 x2 − = 0 (19) b2 a2 and

xy = 0. (20)

In these forms, the intersection line is the z-axis, i.e. points (x, y, z) ∈ R3 with x = 0 and y = 0.

Suppose now K2K13uˆ2 − K1K23uˆ1 = 0. Then (5) becomes

- if K2 − K1 = 0 :

K1uˆ1 (K23u1 − K13u2) u3 − +u ˆ3 = 0 K13 n o K23 K1uˆ1 so the intersection line is u1, u1, +u ˆ3 : u1 ∈ . K13 K13 R

- If K13 = 0:

u1 ((K2 − K1) u2 + K23u3 − (K2uˆ2 + K23uˆ3)) = 0 n o K2uˆ2+K23uˆ3 K23 so the intersection line is 0, − u3, u3 : u3 ∈ . K2−K1 K2−K1 R

- If K23 = 0:

9 u2 ((K2 − K1) u1 − K13u3 + (K1uˆ1 + K13uˆ3)) = 0 n o K13 K1uˆ1+K13uˆ3 so the intersection line is 0, u3 − , u3 : u3 ∈ . K2−K1 K2−K1 R

- If K2 − K1 = 0 and K13 = 0:

u1 (K23u3 − (K2uˆ2 + K23uˆ3)) = 0 n o K2uˆ2+K23uˆ3 so the intersection line is 0, u2, : u2 ∈ . K23 R

- If K2 − K1 = 0 and K23 = 0:

u2 (−K13u3 + (K1uˆ1 + K13uˆ3)) = 0 n o K1uˆ1+K13uˆ3 so the intersection line is u1, 0, : u1 ∈ . K13 R

- If K13 = 0,K23 = 0 and K1uˆ1 = 0 :

u1 ((K2 − K1) u2 − K2uˆ2) = 0 n o K2uˆ2 so the intersection line is 0, − , u3 : u3 ∈ . K2−K1 R

- If K13 = 0,K23 = 0 and K2uˆ2 = 0 :

((K2 − K1) u1 + K1uˆ1) u2 = 0 n o K1uˆ1 so the intersection line is − , 0, u3 : u3 ∈ . K2−K1 R

9 u3-axis

Remember that for going from (4) to (5), we made the assumption that u1 and u2 are not zero at the same time, i.e. we are not on the u3-axis. This means that even if the u3 axis is in all the surfaces we have seen, it may not satisfy equation (4). In fact, as we will see, the u3-axis satisﬁes (4) only in one case.

Suppose the u3-axis satisﬁes (4) then all u3 ∈ R satisfy   K13u3 = K1uˆ1 + K13uˆ3 K23u3 = K2uˆ2 + K23uˆ3 (21)  (K3 − µ1) u3 = K13uˆ1 + K23uˆ2 + K3uˆ3 − µ2

The third equation gives values for µ1 and µ2 and the ﬁrst two lead to K13 = 0,K23 = 0,K1uˆ1 = 0,K2uˆ2 = 0.

Now see if there is points on the u3-axis satisfying (4) in other cases, i.e. if there is u3 ∈ R satisfying (21).

Note that in (21), K13 = 0 ⇒ K1uˆ1 = 0 and K23 = 0 ⇒ K2uˆ2 = 0. So all surfaces with either K13 = 0 and K1uˆ1 6= 0 or K23 = 0 and K2uˆ2 6= 0 have no point on their u3-axis satisfying (4).

There is a solution only if K13K2uˆ2 − K23K1uˆ1 = 0.

Provided one of K13,K23 6= 0, there is a unique point of the u3-axis satisfying (4). This point K1uˆ1 K2uˆ2 is either 0, 0, +u ˆ3 or 0, 0, +u ˆ3 or both if K13 6= 0 6= K23 and they are equal in K13 K23 this case.

Note that all forms of (4) with at least one point of the u3-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 :

10 K2K13uˆ2 − K1K23uˆ1 K2 − K1 K13 K23 K1uˆ1 K2uˆ2 u3-axis K1uˆ1 K2uˆ2 0 6= 0 6= 0 6= 0 0, 0, +u ˆ3 = 0, 0, +u ˆ3 K13 K23 K1uˆ1 K2uˆ2 0 0 6= 0 6= 0 0, 0, +u ˆ3 = 0, 0, +u ˆ3 K13 K23 K2uˆ2 0 6= 0 0 6= 0 0 0, 0, +u ˆ3 K23 K1uˆ1 0 6= 0 6= 0 0 0 0, 0, +u ˆ3 K13 K2uˆ2 0 0 0 6= 0 0 0, 0, +u ˆ3 K23 K1uˆ1 0 0 6= 0 0 0 0, 0, +u ˆ3 K13 0 6= 0 0 0 0 0 Whole axis

Part II Maximum Stability Length

10 Context

The energy of a rod q(s) = (r(s), R(s)) where s ∈ [0,L] is

Z L E [q(s)] = W (u(s) − uˆ(s), v(s) − vˆ(s))ds (22) 0 with W given by (1). Computing its first variation, we get equilibrium conditions that lead, for a given û and ˆv = v = (0, 0, 1), 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 equation1 allows to set a maximum stability length for the rod, i.e. the maximal length the rod can have remaining stable. In autumn 2011, 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 100 by 100 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 100 by 100 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.

11 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 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100 with blue for low values and red for high values.

1This implies solving an ordinary diﬀerential equation y0 = Ay where A is a twelve by twelve matrix with six diﬀerent initial conditions.

11 Each degenerate case is approached by modifying some parameters. There is a symmetry between K13 and K23 and between K1uˆ1 and K2uˆ2 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.

11.1 Hyperbolic Paraboloid with K13 = 0

(a) K13 = 0.5 (b) K13 = 0.4 (c) K13 = 0.3

(d) K13 = 0.2 (e) K13 = 0.1 (f) K13 = 0.08

(g) K13 = 0.06 (h) K13 = 0.04 (i) K13 = 0.02

Figure 1: K13 → 0 K1 = 1, K2 = 1.5, K3 = 1.2, K23 = 0.6, ˆu = (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 K13 up to 0.001. 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 K13 = 0.001.

12 (j) K13 = 0.01 (k) K13 = 0.008 (l) K13 = 0.006

(m) K13 = 0.004 (n) K13 = 0.002 (o) K13 = 0.001

Figure 1

13 11.2 Hyperbolic Paraboloid with K1 = K2

(a) K2 = 1.5 (b) K2 = 1.4 (c) K2 = 1.3

(d) K2 = 1.2 (e) K2 = 1.1 (f) K2 = 1.09

(g) K2 = 1.08 (h) K2 = 1.07 (i) K2 = 1.06

Figure 2: K2 − K1 → 0 K1 = 1, K3 = 1.2, K13 = 0.5, K23 = 0.6, ˆu = (0.7, −0.4, 0.6)

Here we are again approaching a hyperbolic paraboloid but by decreasing the value of K2 to K1 = 1. Here high values ﬁrst gather near one line (the hyperboloid closes on itself) and then a second line appears nd the ﬁrst one disappears.

14 (j) K2 = 1.05 (k) K2 = 1.04 (l) K2 = 1.03

(m) K2 = 1.02 (n) K2 = 1.01 (o) K2 = 1.009

(p) K2 = 1.008 (q) K2 = 1.007 (r) K2 = 1.006

(s) K2 = 1.005 (t) K2 = 1.004 (u) K2 = 1.003

Figure 2

15 11.3 Hyperbolic Paraboloid with K1 = K2 and K13 = 0

(a) K2 = 1.5, K13 = 0.5 (b) K2 = 1.4, K13 = 0.4 (c) K2 = 1.3, K13 = 0.3

(d) K2 = 1.2, K13 = 0.2 (e) K2 = 1.1, K13 = 0.1 (f) K2 = 1.09, K13 = 0.09

(g) K2 = 1.08, K13 = 0.08 (h) K2 = 1.07, K13 = 0.07 (i) K2 = 1.06, K13 = 0.06

Figure 3: K2 − K1 → 0 and K13 → 0 K1 = 1, K3 = 1.2, K23 = 0.6, ˆu = (0.7, −0.4, 0.6)

This time, we tend to a hyperbolic paraboloid by letting both K13 → 0 and K2 → K1 = 1. It looks more like the second case than the ﬁrst one. The computation time began to be to long for higher values of K13 and K2 than in previous cases.

16 (j) K2 = 1.05, K13 = 0.05 (k) K2 = 1.04, K13 = 0.04 (l) K2 = 1.03, K13 = 0.03

(m) K2 = 1.02, K13 = 0.02

Figure 3

17 11.4 Elliptic cone with K13K2uˆ2 − K23K1uˆ1 = 0

(a) K13 = 0.5, K23 = 1 (b) K13 = 0.6, K23 = 1.1 (c) K13 = 0.7, K23 = 1.2

(d) K13 = 0.8, K23 = 1.3 (e) K13 = 0.9, K23 = 1.4 (f) K13 = 0.92, K23 = 1.42

(g) K13 = 0.94, K23 = 1.44 (h) K13 = 0.96, K23 = 1.46 (i) K13 = 0.98, K23 = 1.48

Figure 4: K13K2uˆ2 − K23K1uˆ1 → 0 K1 = 1, K2 = 1.5,K3 = 2.5, ˆu = (0.7, 0.7, 0.6)

In this case, we let K13K2uˆ2 − K23K1uˆ1 → 0 in order to approach an elliptic cone. This is done by having uˆ1 =u ˆ2 = 0.7, K13 → K2 = 1.5 and K23 → K1 = 1. In order to have K positive definite, we set K3 = 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.

18 (j) K13 = 0.99, K23 = 1.49 (k) K13 = 0.992, K23 = 1.492 (l) K13 = 0.994, K23 = 1.494

(m) K13 = 0.996, K23 = 1.496 (n) K13 = 0.998, K23 = 1.498 (o) K13 = 0.999, K23 = 1.499

Figure 4

19 11.5 Intersecting planes with K13K2uˆ2 − K23K1uˆ1 = 0 and K13 = 0

(a) K13 = 0.5, uˆ1 = 0.5 (b) K13 = 0.4, uˆ1 = 0.4 (c) K13 = 0.3, uˆ1 = 0.3

(d) K13 = 0.2, uˆ1 = 0.2 (e) K13 = 0.1, uˆ1 = 0.1 (f) K13 = 0.08, uˆ1 = 0.08

(g) K13 = 0.06, uˆ1 = 0.06 (h) K13 = 0.04, uˆ1 = 0.04 (i) K13 = 0.02, uˆ1 = 0.02

Figure 5: K13 → 0 and K13K2uˆ2 − K23K1uˆ1 → 0 K1 = 1,K2 = 1.5K3 = 1.2,K23 = 0.6, uˆ2 − 0.7, uˆ3 = 0.6

This is the ﬁrst of ﬁve cases of intersecting plane. This one is obtained by letting K2 → K1 = 1 and K13,K23 → 0 which implies K13K2uˆ2 − K23K1uˆ1 → 0. The pattern obtained is especially interesting.

20 (j) K13 = 0.01, uˆ1 = 0.01 (k) K13 = 0.008, uˆ1 = 0.008 (l) K13 = 0.006, uˆ1 = 0.006

(m) K13 = 0.004, uˆ1 = 0.004 (n) K13 = 0.002, uˆ1 = 0.002 (o) K13 = 0.001, uˆ1 = 0.001

(p) K13 = 0.0008, uˆ1 = 0.0008 (q) K13 = 0.0006, uˆ1 = 0.0006 (r) K13 = 0.0004, uˆ1 = 0.0004

(s) K13 = 0.0002, uˆ1 = 0.0002

Figure 5

21 11.6 Hyperbolic cylinder with K13K2uˆ2 −K23K1uˆ1 = 0,K13 = 0 = K23 and K1uˆ1 6= 0 6= K2uˆ2

(a) K13 = 0.5,K23 = 0.5 (b) K13 = 0.4,K23 = 0.4 (c) K13 = 0.3,K23 = 0.3

(d) K13 = 0.2,K23 = 0.2 (e) K13 = 0.08,K23 = 0.08 (f) K13 = 0.06,K23 = 0.06

(g) K13 = 0.04,K23 = 0.04 (h) K13 = 0.02,K23 = 0.02 (i) K13 = 0.01,K23 = 0.01

Figure 6: K13,K23 → 0 and K13K2uˆ2 − K23K1uˆ1 → 0 K1 = 1,K2 = 1.5,K3 = 1.2, ˆu = (0.7, −0.4, 0.6)

This is the only case of hyperbolic cylinder. It is obtained by decreasing K13 and K23 to 0.

22 (j) K13 = 0.008,K23 = 0.008 (k) K13 = 0.004,K23 = 0.004 (l) K13 = 0.002,K23 = 0.002

Figure 6

23 11.7 Intersecting planes with K13K2uˆ2−K23K1uˆ1 = 0,K13 = 0 = K23,K1uˆ1 = 0 and K2uˆ2 6= 0

(a) K13 = 0.5,K23 = 0.5 (b) K13 = 0.4,K23 = 0.4 (c) K13 = 0.3,K23 = 0.3

(d) K13 = 0.2,K23 = 0.2 (e) K13 = 0.1,K23 = 0.1 (f) K13 = 0.08,K23 = 0.08

(g) K13 = 0.06,K23 = 0.06 (h) K13 = 0.04,K23 = 0.04 (i) K13 = 0.02,K23 = 0.02

Figure 7: K13,K23 → 0,K13K2uˆ2 − K23K1uˆ1 → 0 and K1uˆ1 = 0 K1 = 1,K2 = 1.5,K3 = 1.2, ˆu = (0, −0.4, 0.6)

This case of intersecting planes, obtained by letting K13,K23 → 0 with uˆ1 = 0 so K1uˆ1 = 0 looks somewhat like the hyperbolic cylinder.

24 (j) K13 = 0.01,K23 = 0.01 (k) K13 = 0.008,K23 = 0.008 (l) K13 = 0.006,K23 = 0.006

(m) K13 = 0.004,K23 = 0.004

Figure 7

25 11.8 Intersecting planes with K13K2uˆ2 − K23K1uˆ1 = 0,K13 = 0 = K23 and K1uˆ1 = 0 = K2uˆ2

(a) K13 = 0.5,K23 = 0.5, uˆ2 = −0.5 (b) K13 = 0.4,K23 = 0.4, uˆ2 = −0.4 (c) K13 = 0.3,K23 = 0.3, uˆ2 = −0.3

(d) K13 = 0.2,K23 = 0.2, uˆ2 = −0.2 (e) K13 = 0.1,K23 = 0.1, uˆ2 = −0.1 (f) K13 = 0.08,K23 = 0.08, uˆ2 = −0.08

(g) K13 = 0.06,K23 = 0.06, uˆ2 =(h) K13 = 0.04,K23 = 0.04, uˆ2 =(i) K13 = 0.02,K23 = 0.02, uˆ2 = −0.06 −0.04 −0.02

Figure 8: K13,K23 → 0,K13K2uˆ2 − K23K1uˆ1 → 0,K2uˆ2 → 0 and K1uˆ1 = 0 K1 = 1,K2 = 1.5,K3 = 1.2, uˆ1 = 0, uˆ2 = 0.6

We are here approaching the only surface having the whole u3-axis in it. For this we let K13,K23 → 0 and uˆ2 → 0 with uˆ1 = 0.

26 (j) K13 = 0.01,K23 = 0.01, uˆ2 =(k) K13 = 0.008,K23 = 0.008, uˆ2 =(l) K13 = 0.006,K23 = 0.006, uˆ2 = −0.01 −0.008 −0.006

(m) K13 = 0.004,K23 = 0.004, uˆ2 =(n) K13 = 0.002,K23 = 0.002, uˆ2 =(o) K13 = 0.001,K23 = 0.001, uˆ2 = −0.004 −0.002 −0.001

Figure 8

27 11.9 Intersecting planes with K13K2uˆ2 − K23K1uˆ1 = 0 and K1 = K2

(a) K2 = 1.5 (b) K2 = 1.4 (c) K2 = 1.3

(d) K2 = 1.2 (e) K2 = 1.1 (f) K2 = 1.08

(g) K2 = 1.06 (h) K2 = 1.04 (i) K2 = 1.02

Figure 9: K2 − K1 → 0 and K13K2uˆ2 − K23K1uˆ1 → 0 K1 = 1,K3 = 1.2,K13 = 0.5,K23 = 0.5, ˆu = (0.7, 0.7, 0.6)

Another case of intersecting planes, with K2 → K1 = 1. To have K13K2uˆ2 − K23K1uˆ1 → 0, we set K13 = K23 = 0.5 and uˆ1 =u ˆ2 = 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 30.

28 (j) K2 = 1.01 (k) K2 = 1.008 (l) K2 = 1.006

(m) K2 = 1.004

Figure 9

29 11.10 Intersecting planes with K13K2uˆ2 − K23K1uˆ1 = 0,K1 = K2,K13 = 0 and K1uˆ1 = 0

(a) K2 = 1.5,K13 = 0.5 (b) K2 = 1.4,K13 = 0.4 (c) K2 = 1.3,K13 = 0.3

(d) K2 = 1.2,K13 = 0.2 (e) K2 = 1.1,K13 = 0.1 (f) K2 = 1.08,K13 = 0.08

(g) K2 = 1.06,K13 = 0.06 (h) K2 = 1.04,K13 = 0.04 (i) K2 = 1.02,K13 = 0.02

Figure 10: K2 − K1 → 0,K13 → 0,K13K2uˆ2 − K23K1uˆ1 → 0 and K1uˆ1 = 0 K1 = 1,K3 = 1.2,K23 = 0.5, ˆu = (0, 0.7, 0.6)

The last case of intersecting planes. We set uˆ1 = 0 to have K1uˆ1 = 0 and we let K2 → K1 = 1 and K13 → 0. The evolution looks rather like the three cases of hyperbolic paraboloid.

30 (j) K2 = 1.01,K13 = 0.01 (k) K2 = 1.008,K13 = 0.008 (l) K2 = 1.006,K13 = 0.006

(m) K2 = 1.004,K13 = 0.004 (n) K2 = 1.002,K13 = 0.002 (o) K2 = 1.001,K13 = 0.001

(p) K2 = 1.0008,K13 = 0.0008 (q) K2 = 1.0006,K13 = 0.0006

Figure 10

31 11.11 Plane with K13K2uˆ2 − K23K1uˆ1 = 0,K1 = K1,K13 = 0,K23 = 0 and K1uˆ1 = 0

(a) K2 = 1.5,K13 = 0.5,K23 = 0.5 (b) K2 = 1.4,K13 = 0.4,K23 = 0.4 (c) K2 = 1.3,K13 = 0.3,K23 = 0.3

(d) K2 = 1.2,K13 = 0.2,K23 = 0.2 (e) K2 = 1.1,K13 = 0.1,K23 = 0.1 (f) K2 = 1.08,K13 = 0.08,K23 = 0.08

(g) K2 = 1.06,K13 = 0.06,K23 =(h) K2 = 1.04,K13 = 0.04,K23 =(i) K2 = 1.02,K13 = 0.02,K23 = 0.06 0.04 0.02

Figure 11: K2 − K1 → 0,K13,K23 → 0,K13K2uˆ2 − K23K1uˆ1 → 0 and K1uˆ1 = 0 K1 = 1,K3 = 1.2, ˆu = (0, −0.7, 0.6)

Letting K2 → K1 = 1 and K13,K23 → 0 with uˆ1 = 0, we approach a plane. The high values gather on a line but only on one side of the u3-axis.

32 (j) K2 = 1.01,K13 = 0.01,K23 = 0.01

Figure 11

33 12 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

[1] Quadratic surface. http://mathworld.wolfram.com/QuadraticSurface.html. [2] Quadriques, march 2012. http://fr.wikipedia.org/wiki/Quadrique.