On in-plane drill rotations for Cosserat surfaces
OOn in-plane drill rotations for Cosserat surfaces
Maryam Mohammadi Saem , ∗ , Peter Lewintan and Patrizio Neff February 23, 2021
Abstract
We show under some natural smoothness assumptions that pure in-plane drill rotations as deformationmappings of a C -smooth regular shell surface to another one parametrized over the same domain areimpossible provided that the rotations are fixed at a portion of the boundary. Put otherwise, if the tangentvectors of the new surface are obtained locally by only rotating the given tangent vectors, and if theserotations have a rotation axis which coincides everywhere with the normal of the initial surface, then thetwo surfaces are equal provided they coincide at a portion of the boundary. In the language of differentialgeometry of surfaces we show that any isometry which leaves normals invariant and which coincides withthe given surface at a portion of the boundary, is the identity mapping. Keywords: compatibility, integrability, rigidity, in-plane drill rotations, isometry, invariant normal field,minimal surfaces, associate family of minimal surfaces, mean curvature, Rodrigues formula.
AMS 2020 subject classification:
In this contribution we consider the apparently novel question whether nontrivial pure in-plane drill rotationsmay appear in the deformation of shells if boundary conditions are prescribed that fix the Cosserat drillrotations at a portion of the boundary.
In this context, we prove the following main improved rigidity result for surfaces:
Proposition 1.1.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces, Q ∈ C ( ω, SO(3)) and D m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , x ∈ ω ,m | γ d = y | γ d , (1.1) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) and γ d is a relatively open, non-empty subset ofthe boundary ∂ω . Then m ≡ y . Expressed in the language of classical differential geometry of surfaces we prove: Chair for Nonlinear Analysis and Modeling, Faculty of Mathematics, University of Duisburg-Essen, Thea-Leymann-Str. 9,45127 Essen, Germany ∗ Corresponding author, email: [email protected]. a r X i v : . [ m a t h . DG ] F e b orollary 1.2. Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are two regularsurfaces and I m ( x ) = [D m ( x )] T D m ( x ) = [D y ( x )] T D y ( x ) = I y ( x ) , n ( x ) = n ( x ) ∀ x ∈ ω ,m | γ d = y | γ d , (1.2) where n = ∂ m × ∂ m (cid:107) ∂ m × ∂ m (cid:107) and n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) are the respective normal fields and γ d is a relatively open,non-empty subset of the boundary ∂ω . Then m ≡ y . The results of this paper should perhaps not come as a surprise to experts in the field of differentialgeometry. Indeed, except for minimal surfaces, the Gauss map n already determines the surface essentially,cf. [21, Theorem 2.5]. On the contrary, minimal surfaces come with a family of ‘associate surfaces’, whichhave all the same Gauss map but are distinct to each other. Comparable results to Corollary 1.2 can befound in [1, 21, 16] where different methods of proof were used and any connection to applications weremissing. The aim of the present paper is to give a straight foreward proof without the techniques comingfrom differential geometry. We use the Rodrigues representation formula for rotations with given axis as wellas repeated properties of the cross-product.In order to set the stage, we recall some of the better known rigidity and integrability theorems, for3D − bulk materials and 2D − surfaces. For a warm up, we tackle first the small rotation problem whichalready discloses some of the necessary techniques. The paper is now structured as follows: after setting ournotations, in the next subsections 1.3 and 1.4, we will motivate the problem from an engineering point ofview. The reader only interested in the mathematical development can safely skip this part. Then we arein section 2 recapitulating some well-known rigidity and integrability results for bulk materials and surfacestogether with simple observations which turn up in our question. We complement this section with somepreliminaries on rotations and induced boundary conditions. Then we consider in section 3 the case wherethe rotations in Proposition 3.2 are assumed to be small. In the subsequent section 4 we give the proof ofProposition 1.1 for the large rotation problem. We end with a conclusion, in section 5, putting our findingsback in an engineering context.The complementary problem D m ( x ) = U ( x ) D y ( x ) , (1.3)of finding all “compatible” in-plane stretches U , characterized by U ( x ) n ( x ) = κ + ( x ) n ( x ) , U ( x ) ∈ Sym + (3) , κ + ( x ) > , (1.4)has been completely solved more than ten years ago by Szwabowicz [38]. For n ∈ N , we denote the set of all n × n second order tensors by R n × n . The identity tensor on R n × n isdenoted by n . We define GL( n ) = { X ∈ R n × n | det( X ) (cid:54) = 0 } , consequently SO( n ) = { X ∈ GL( n ) | X T X = n , det( X ) = 1 } , so ( n ) = { X ∈ R n × n | X T = − X } . The sets Sym( n ) and Sym + ( n ) denote the symmetric andpositive definite symmetric tensors respectively. For all X ∈ R n × n we have sym X = ( X + X T ) ∈ Sym( n )and skew X = ( X − X T ) ∈ so ( n ). For a, b ∈ R n we denote the scalar product on R n as (cid:104) a, b (cid:105) with associated(squared) norm (cid:107) a (cid:107) = (cid:104) a, a (cid:105) . We denote ( a | b | c ) as 3 × a, b, c ∈ R . For u ∈ C ( R , R )and x = ( x , x , x ) we denote the derivative (the Jacobi matrix) of u as D u = ( ∂ u | ∂ u | ∂ u ). We considerthe mapping m : ω ⊂ R → R as a deformation of the midsurface of a shell with Jacobi matrixD m = ∂ m ∂ m ∂ m ∂ m ∂ m ∂ m = ( ∂ m | ∂ m ) . (1.5) Szwabowicz uses a different notation, but his stretch tensor is basically the stretch tensor U satisfying (1.4). Note thatsince U ∈ Sym + (3) it can be orthogonally diagonalized, the stretch U ( x ) satisfying (1.4) leaves the tangent plane T y ( x ) y ( ω )invariant. Therefore, the Gauss map is preserved as well. so (3) and R is denoted by axl( · ). Note that with the standard vector productit is given by (axl A ) × ξ = A . ξ for all skew-symmetric matrices A ∈ so (3) and vectors ξ ∈ R , so thataxl − v v v − v − v v := v v v , Anti v v v := − v v v − v − v v , (1.6)where the inverse of axl( · ) is denoted by Anti( · ). A map y ∈ C ( ω, R ) is called a regular surface if theJacobi matrix D y = ( ∂ y | ∂ y ) has rank 2, i.e., if the surface y ( ω ) has a well-defined tangent space at all ofits (inner) points. Moreover, we make use of the first fundamental form I y on y ( ω ) in matrix representationI y := [D y ] T D y = (cid:18) (cid:107) ∂ y (cid:107) (cid:104) ∂ y , ∂ y (cid:105)(cid:104) ∂ y , ∂ y (cid:105) (cid:107) ∂ y (cid:107) (cid:19) ∈ Sym + (2) , (1.7)where the positive definitness follows from rank D y = 2. The matrix representation of the second fundamentalform on y ( ω ) is given byII y := − [D y ] T D n = − (cid:18) (cid:104) ∂ y , ∂ n (cid:105) (cid:104) ∂ y , ∂ n (cid:105)(cid:104) ∂ y , ∂ n (cid:105) (cid:104) ∂ y , ∂ n (cid:105) (cid:19) ∈ Sym(2) , (1.8)where the symmetry follows from the fact that (cid:104) n , ∂ i y (cid:105) = 0, for i = 1 , The elastic range of many engineering materials is restricted to small finite strains. Thin structures maytypically undergo large rotations (by bending) but are accompanied by small elastic strains.In [29] the following geometrically nonlinear (but small elastic strain) isotropic planar shell model hasbeen derived for such a situation: find the midsurface deformation m : ω ⊂ R → R and the independentrotation field R : ω ⊂ R → SO(3) minimizing the elastic energy I ( m, R ) = (cid:90) ω h (cid:104) µ (cid:107) sym(( R | R ) T D m − ) (cid:107) (cid:124) (cid:123)(cid:122) (cid:125) shear-stretch energy + µ c (cid:107) skew(( R | R ) T D m ) (cid:107) (cid:124) (cid:123)(cid:122) (cid:125) first order drill energy + ( µ + µ c )2 (cid:16) (cid:104) R , ∂ m (cid:105) + (cid:104) R , ∂ m (cid:105) (cid:17)(cid:124) (cid:123)(cid:122) (cid:125) transverse shear energy + µλ µ + λ tr(sym(( R | R ) T D m − )) (cid:124) (cid:123)(cid:122) (cid:125) stretch energy (cid:105) (1.9)+ h (cid:104) µL c (cid:107)K s (cid:107) + µL qc (cid:107)K s (cid:107) q (cid:105) + h (cid:104) µ (cid:107) sym K b (cid:107) + µ c (cid:107) skew K b (cid:107) + µλ µ + λ tr[ K b ] (cid:105) dω → min w.r.t ( m, R ) , where the Cosserat curvature tensor is given by K s = ( R T (D( R.e ) | , R T (D( R.e ) | , R T (D( R.e ) | , K b := K s, = R T (D( R.e ) | , (1.10)3nd the boundary condition of place for the midsurface deformation m on the Dirichlet part of the lateralboundary, m | γ d = g d ( x, y,
0) is imposed. This shell model is derived by dimensional descent from a three-dimensional bulk Cosserat model [12, 29] and the appearing parameters are the isotropic shear modulus µ >
0, the second Lam´e parameter λ (with 2 µ + λ >
0) and the so-called
Cosserat couple modulus µ c ≥ h > L c ≥ q ≥
0. This Cosserat shellmodel can be naturally related to the general 6-parameter theory of shells [6, 4, 5, 7, 9, 10, 23, 34, 35, 41, 39],see also [3, 8, 17, 22, 25, 40, 43]. One of the typical energy terms in these models is connected to so-called in-plane drill rotations [7, 42]. These in-plane drill rotations describe local rotations of the shell midsurfacewith rotation axis given by the local shell normal n of y . Typically, the constitutive coefficients whichgoverns this deformation mode are difficult to establish (and the ubiquitous Cosserat couple modulus µ c > µ c has been set to zero throughout and q = 2 has been adopted. In this case, the term in (1.9) denoted by ”first order drill energy” will drop out,while all other terms basically remain the same. It seems therefore mandatory to devote special attentionto this in-plane drill term in order to understand it’s physical and mathematical significance. This will beundertaken next. Let ω ⊂ R be a bounded Lipschitz domain and y : ω ⊂ R → R smooth and regular describing themid-surface of a shell. γ d ω ⊂ R y ∂ y ∂ y n e Figure 1: The midsurface y ∈ C ( ω, R ) of a shellis visualized, together with a tangent plane T y ( x ) y spanned by ∂ y ( x ) , ∂ y ( x ) and with unit normal n ( x ). Prescribed boundary conditions at γ d fix theshell mid-surface in space.Let us analyze the energy term corresponding to drill rotations shown in (1.9). In order to measurein-plane drill rotations of a shell in a continuum description one first needs to endow the shell with a givenorthonormal frame, tangent to the surface y , against which in-plane rotations can be seen. The role of thisframe will be taken here by Q ∈ SO(3), defined by Q := polar(D y | n ) (already used as such by Darboux,see [13]), where polar( F ) denotes the orthogonal part in the polar-decomposition of F ∈ GL + (3). First, it4olds that skew( Q T (D y | n )) = 0 for Q = polar(D y | n ) ∈ SO(3) , (1.11)due to the properties of the polar decomposition [31](D y | n ) = Q
000 0 1 (cid:112) [D y ] T D y (cid:124) (cid:123)(cid:122) (cid:125) ∈ Sym + (3) , Q e = n , ⇐⇒ (1.12)D y = ( Q | Q ) (cid:113) [D y ] T D y ⇐⇒ ( Q | Q ) T D y = (cid:113) [D y ] T D y ∈ Sym + (2) . Here, it can be seen that Q : ω ⊂ R → SO(3) is an orthonormal frame whose third column coincides withthe normal n of the surface such that there is also no induced transverse shear . The three dimensionalcondition (1.11) can be expressed equivalently as (see also (1.12) )skew( Q T (D y | n )) = skew (cid:104) ( Q | Q | Q ) T (cid:16) ∂ y | ∂ y | n (cid:17)(cid:105) = skew (cid:104) Q , ∂ y (cid:105) (cid:104) Q , ∂ y (cid:105) (cid:104) Q , ∂ y (cid:105) (cid:104) Q , ∂ y (cid:105)
00 0 1 = 0 ⇐⇒ skew[( Q | Q ) T D y ] = 0 , (1.13)which is the ”drill energy” argument from equation (1.9) and in this special situation we have an initially”drill-free” setting.Now, what happens if we only locally rotate (drill) the given tangent vectors ∂ y , ∂ y about therotation axis n ? For this, we take a drill rotation Q ( α ) n = n , Q ( α ) = Q ( α ( x )) ∈ SO(3), where α = α ( x )is the rotation angle and n = n ( x ) is the prescribed axis of rotation normal to the surface y and we considerlocally the mapping D y → Q ( α ) D y , (1.14)which leaves the first fundamental form I y = D y T D y = (cid:0) Q ( α ) D y (cid:1) T (cid:0) Q ( α ) D y (cid:1) invariant. This impliesthat the surface y is locally changed isometrically. For simplicity, taking into account the subsequentrepresentation (2.12) of rotations with given axis of rotations, we consider presently only small drill rotationangles α so that we can duly approximate Q ( α ) ≈ + α Anti( n ) . (1.15)Inserting then ( + α Anti( n )) D y instead of D y into the drill term from (1.13) we obtainskew (cid:104) ( Q | Q ) T ( + α Anti( n )) (cid:16) ∂ y | ∂ y (cid:17)(cid:105) = skew(( Q | Q ) T D y ) (cid:124) (cid:123)(cid:122) (cid:125) =0 + skew (cid:104) ( Q | Q ) T α (cid:16) n × ∂ y (cid:12)(cid:12)(cid:12) n × ∂ y (cid:17)(cid:105) = α skew (cid:104) (cid:18) [ Q ] T [ Q ] T (cid:19) (cid:16) n × ∂ y (cid:12)(cid:12)(cid:12) n × ∂ y (cid:17)(cid:105) = α (cid:18) − [ (cid:104) Q , n × ∂ y (cid:105) − (cid:104) Q , n × ∂ y (cid:105) ][ (cid:104) Q , n × ∂ y (cid:105) − (cid:104) Q , n × ∂ y (cid:105) ] 0 (cid:19) . (1.16)We will now show that for any non-zero small angle of rotation α , expression (1.16) is non-zero implying thatthe related drill energy term hµ c (cid:107) skew(( Q | Q ) T D y ) (cid:107) serves to introduce a linear torsional spring tiffness against superposed in-plane rotations (with spring constant hµ c , where µ c ≥ ∂ y , ∂ y , Q , Q lie in the same tangent plane, therefore,there exist a = a ( x ) , b = b ( x ) , c = c ( x ) , d = d ( x ) ∈ R such that ∂ y = a Q + c Q , and ∂ y = b Q + d Q (1.17)or, respectively, D y = ( Q (cid:12)(cid:12) Q ) (cid:18) a bc d (cid:19) . (1.18)We also observe that the orientation of ( ∂ y , ∂ y ) and ( Q , Q ) must coincide, due to (1.11) . Here, thismeans ∂ y × ∂ y = β + Q × Q for some β + >
0. Using (1.17), it holds ∂ y × ∂ y = ( a Q + c Q ) × ( b Q + d Q ) = ( ad − bc ) Q × Q . (1.19)Thus, the orientation preservation condition in our setting is ad − bc >
0. The symmetry condition (1.13)implies that (cid:104) Q , ∂ y (cid:105) = (cid:104) Q , ∂ y (cid:105) , and insertion into (1.17) shows that b = c by orthogonality of Q and Q . Observe now that[ (cid:104) Q , n × ∂ y (cid:105) − (cid:104) Q , n × ∂ y (cid:105) ] = [ (cid:104) Q , n × ( a Q + c Q ) (cid:105) − (cid:104) Q , n × ( b Q + d Q ) (cid:105) ]= a (cid:104) Q , n × Q (cid:105) − d (cid:104) Q , n × Q (cid:105) = ( a + d ) (cid:104) Q , n × Q (cid:105) = a + d . (1.20)Orientation preservation ad − bc > b = c imply that ad > b , which is sufficient to conclude that a + d (cid:54) = 0 for otherwise a + d = 0 would lead to − d > b , a contradiction. This proves our claim.Note that at present, the discussion is purely local: at no place did we require that Q ( α ( x )) D y ( x ) can bedetermined as the gradient of a mapping.The global question whether Q ( α ( x )) D y ( x ) can be the gradient of a regular embedding m : ω ⊂ R → R will be considered next. Consider a given initial curved shell surface parametrized locally by y : ω ⊂ R → R , where we assume that y is sufficiently smooth and regular (rank(D y ) = 2). Let m : ω ⊂ R → R be any smooth deformation ofthe given surface y parametrized over the same domain and consider a smooth in-plane drill rotation field Q : ω ⊂ R → SO(3) , Q ( x ) n ( x ) = n ( x ) , (2.1)where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) is the unit normal vector field on the initial surface y . We will assume furtheron that Q | γ d = , (2.2)where γ d is a relatively open, non-empty subset of the boundary ∂ω . The motivation for this boundarycondition will be given in Lemma 2.7. Problem 2.1.
Let ω ⊂ R be a bounded Lipschitz domain and assume that m, y ∈ C ( ω, R ) are regularsurfaces. Does there exist a nontrivial in-plane drill rotation field Q ∈ C ( ω, SO(3)) such that D m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , x ∈ ω,Q | γ d = . (2.3)6 y γ d ω ⊂ R n Figure 2: At any point of the surface y tan-gent planes are rotated in their own plane leav-ing the orientation of the surface invariant toobtain a new surface m . How can m look like ifat a part of the boundary γ d ⊂ ∂ω the surfaces m and y coincide? Remark 2.2.
For a given shell surface y any pure bending (flexure) deformation m satisfies locallyD m ( x ) = Q ( x ) D y ( x ) , Q ( x ) ∈ SO(3) , (2.4)such that the first fundamental forms coincide I m = [D m ] T D m = [D y ] T D y = I y . Considering a flat pieceof paper assumed to be made of unstretchable material, the rotations can even be fixed at one side of thepaper still allowing for nontrivial bending deformations of the paper. However, the appearing local rotationfield Q ( x ) ∈ SO(3) in this case is not of in-plane drill type, i.e., the rotation axis of Q is not everywhere givenby n . ω ⊂ R γ d my Figure 3: A pure bending deformation of asurface leaves length invariant, the first fun-damental form is unchanged, but there is lo-cal rotation. We have D m = Q ( x ) D y forsome non-constant Q ∈ SO(3), but the ro-tation in this example does not have an in-plane rotation axis.
Remark 2.3 (Rigidity in 3D) . Assume that M : Ω ⊂ R → R and Y : Ω ⊂ R → R are two diffeomor-phisms, the formally similar to (2.3) looking conditionD M ( x ) = Q ( x ) D Y ( x ) , Q ( x ) ∈ SO(3) , x ∈ Ω , (2.5)implies that Q ( x ) ≡ const by rigidity [32, 2], as can easily be seen as follows.By the chain rule we haveD( M ( Y − ( ξ ))) = D M ( Y − ( ξ )) D[ Y − ( ξ )] , where D[ Y − ( ξ )] = [D Y ( x )] − , therefore, D( M ( Y − ( ξ ))) = Q ( x )[D Y ( x )] [D Y ( x )] − , ⇒ D( M ( Y − ( ξ ))) = Q ( x ) ∈ SO(3) rigidity ===== ⇒ Q ≡ const. (2.6)Note that the smoothness of Q : Ω → SO(3) can be a priori controlled by the smoothness of M and Y .7 ξY M ( Y − ( ξ )) M Ω Figure 4: Multiplicative composition of mappings.
Remark 2.4 (Rigidity in 3D) . In the 3D − case we have another condition which turns out to yield homoge-neous rotations as well. Assume again that M, Y : Ω ⊂ R → R are two diffeomorphisms. Then ∀ x ∈ Ω : [D M ( x )] T D M ( x ) = [D Y ( x )] T D Y ( x ) ⇐⇒ M ( x ) = Q Y ( x ) , (2.7)where Q ∈ SO(3) is a constant rotation, as is shown, e.g. in [11]. However, as already seen, 2D − structuresare much more flexible in the sense that for m, y : ω ⊂ R → R (smooth embeddings)I m = [D m ( x )] T D m ( x ) = [D y ( x )] T D y ( x ) = I y (cid:54) = ⇒ m ( x ) = Q y ( x ) , (2.8)as any pure bending deformation shows. Example 2.5 (Flat case) . Consider y : ω ⊂ R → R with y ( x ) = ( x , x , T . Then D y ( x ) = and n ≡ e . Thus, the conditions Q ( x ) n = n , m ∈ C ( ω, R ) and D m ( x ) = Q ( x ) D y ( x ) for Q ( x ) ∈ SO(3),together with Q | γ d = , implyD m ( x ) = Q ( x ) =
000 0 1 (cid:98) Q ( x ) = (cid:98) Q ( x ) , (2.9)with (cid:98) Q ( x ) ∈ SO(2). Hence, D m ( x ) ≡ (cid:18) m ( x ) m ( x ) (cid:19) = (cid:98) Q ( x ) ∈ SO(2) . (2.10)Then, again by rigidity [32] we obtain that (cid:98) Q ≡ const. Applying the boundary conditions we have (cid:98) Q | γ d = and finally Q ≡ . Thus m − y ≡ const. SO(3) and the Euler-Rodrigues formula
We need to consider the matrix exponential functionexp : so (3) → SO(3) , exp( A ) = ∞ (cid:88) k =0 k ! A k , A ∈ so (3) . (2.11)According to Euler, any rotation can be realized by a rotation around one axis with a certain rotation angle.However, any rotation Q ∈ SO(3) with prescribed rotation axis n and angle of rotation α can be written8ith the Euler-Rodrigues representation in matrix notation as [33] Q ( α ) = exp(Anti( α n )) = + sin α Anti( n ) + (1 − cos α ) (Anti( n )) = (1 − cos α ) n ⊗ n + cos α + sin α Anti( n ) . (2.12)For small rotation angle | α | (cid:28) Q ≈ + α Anti( n ) , (2.13)since sin α → α and 1 − cos α → | α | →
0. For later use, note that any nontrivial rotation (cid:54) = Q ∈ SO(3) has (only) one rotation axis η ∈ R such that Qη = η where η is the eigenvector to the realeigenvalue 1.Taking the trace in (2.12), we also see thattr( Q ( α ( x ))) = 2 cos α ( x ) + 1 ⇐⇒ cos( α ( x )) = tr( Q ( α ( x ))) − . (2.14)The inverse cosine is a multivalued function and each branch is differentiable only on ( − , Q = or Q = diag( − , − ,
1) we have tr( Q ) − ∈ {− , } , i.e., in the neighborhood of both these rotationsthe simple formula (2.14) is not meaningful for extracting a smooth rotation angle. In order to solve thisproblem for small rotation angle α (for Q near to ) we proceed as follows (the simple idea is taken from[26]). Multiplying (2.12) on both sides with Anti( n ) from the left givesAnti( n ) Q ( α ) = Anti( n ) (cid:124) (cid:123)(cid:122) (cid:125) ∈ so (3) + sin α (Anti( n )) + (1 − cos α )(Anti( n )) (cid:124) (cid:123)(cid:122) (cid:125) ∈ so (3) . (2.15)Tanking the trace gives tr (cid:0) Anti( n ) Q ( α ) (cid:1) = − sin α (cid:107) Anti( n ) (cid:107) = − α , (2.16)since Anti( n ) ∈ so (3) and (cid:107) n (cid:107) = 1. Thus with (2.14) we arrive atsin α = − tr (cid:16) Anti( n ) Q (cid:17) , cos α = tr( Q ) − tr( Q ) (cid:54) =1 = ⇒ tan α = − tr (cid:16) Anti( n ) Q (cid:17) tr( Q ) − , (2.17)whereby any branch of the inverse tangent is smooth on R . This shows that for tr( Q ) − >
0, i.e., in a largeneighborhood of Q = , the extraction of the rotation angle α from the rotation Q is as smooth as Q andthe surface allows. Lemma 2.6.
Assume y ∈ C ( ω, R ) is a regular surface and let Q ∈ C ( ω, SO(3)) be given. Assume thatfor a point x ∈ ω and α ∈ R it holds Q ( x ) = (1 − cos α ) n ( x ) ⊗ n ( x ) + cos α + sin α Anti( n ( x )) , (2.18) where n is the normal field on y . Then there exists a neighborhood U ( x ) ⊂ ω and a continuously-differentiable function α : U ( x ) → R satisfying α ( x ) = α , (2.19) such that for all x ∈ U ( x ) Q ( x ) = (1 − cos α ( x )) n ( x ) ⊗ n ( x ) + cos α ( x ) + sin α ( x ) Anti( n ( x )) . (2.20) Proof.
The C -regularity of α in a sufficiently small neighborhood of x follows from one of the expressionscontained in (2.17). Indeed, for tr( Q ( x )) (cid:54) = 1 consider in (2.17) the branch of the inverse tangent whichcontains α . Otherwise, for tr( Q ( x )) = 1 take in (2.17) the branch of the inverse cosine which contains α ,cf. Figure 5. (cid:4) π π π π − π − π − π ϕ bbbb b α tan( α ) tan ϕ π π π π π − π − π − π ϕ bbbb b α cos ϕ Figure 5: The range for the corresponding branch of the inverse trigonometric functions are indicated in blue.
Lemma 2.7.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces, Q ∈ C ( ω, SO(3)) and D m ( x ) = Q ( x ) D y ( x ) , x ∈ ω , m | γ d = y | γ d , (2.21) where γ d is a relatively open, non-empty subset of the boundary ∂ω . If for all x ∈ γ d we have Q ( x ) n ( x ) = n ( x ) , where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) , then Q | γ d ≡ . Remark 2.8.
For the conclusion of this lemma we only need the assumptions on γ d . However, the requiredconditions in the interior of ω are those which will be considered later. Proof of Lemma 2.7.
Consider a C , -parametrization γ : (0 , → R , γ (0 , ⊂ γ d ⊂ ∂ω . Then, m ( γ ( s )) = y ( γ ( s )) on (0 ,
1) implies for ˙ γ ( s ) ∈ R dds m ( γ ( s )) = dds y ( γ ( s )) ⇒ D m ( γ ( s )) ˙ γ ( s ) = D y ( γ ( s )) ˙ γ ( s ) ∈ R a.e. on (0 , m ( γ ( s )) ˙ γ ( s ) (2.21) = Q ( γ ( s )) D y ( γ ( s )) ˙ γ ( s ) (2.22) = Q ( γ ( s )) D m ( γ ( s )) ˙ γ ( s ) (cid:124) (cid:123)(cid:122) (cid:125) =: q ( s ) ∈ R a.e. on (0 , q ( s ) = Q ( γ ( s )) q ( s ), for almost all s ∈ (0 , q ( s ) is a tangent vector to y at y ( γ ( s )). Together with the assumption Q ( γ ( s )) n ( γ ( s )) = n ( γ ( s )) it follows that Q ( γ ( s )) has two linearindependent eigenvectors q ( s ) and n ( γ ( s )). Since the axis of rotation is unique for any nontrivial rotation,it follows Q ( γ ( · )) = a.e. on (0 ,
1) and by continuity Q | γ d ≡ . (cid:4) We repeat a similar reasoning for the small rotation case.
Lemma 2.9.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are two regularsurfaces, A ∈ C ( ω, so (3)) and D m ( x ) = ( + A ( x )) D y ( x ) , x ∈ ω , m | γ d = y | γ d , (2.24) where γ d is a relatively open, non-empty subset of the boundary ∂ω . If for all x ∈ γ d ⊂ ∂ω we have A ( x ) n ( x ) = 0 , where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) , then A | γ d ≡ . roof. We have again D m ( γ ( s )) ˙ γ ( s ) = D y ( γ ( s )) ˙ γ ( s ) ∈ R a.e. on (0 , , (2.25)for a C , -parametrization γ : (0 , → R , γ (0 , ⊂ γ d ⊂ ∂ω . Hence, we obtain a.e. on (0 , y ( γ ( s )) ˙ γ ( s ) (2.25) = D m ( γ ( s )) ˙ γ ( s ) (2.24) = D y ( γ ( s )) ˙ γ ( s ) + A ( γ ( s )) D y ( γ ( s )) ˙ γ ( s ) ⇐⇒ A ( γ ( s )) D y ( γ ( s )) ˙ γ ( s ) (cid:124) (cid:123)(cid:122) (cid:125) =: q ( s ) = A ( γ ( s )) q ( s ) , where q ( s ) ⊥ n ( γ ( s )) . (2.26)From the assumption we moreover have A ( γ ( s )) n ( γ ( s )) = 0 so that with A ( γ ( s )) q ( s ) = 0 along γ d weobtain A ( γ ( s )) = 0, since any non-zero skew-symmetric 3 × A has rank two. (cid:4) Proposition 2.10. [ Compatbility for surfaces ] Let v, w ∈ C ( ω, R ) be given vector fields with rank( v, w ) =2 everywhere and assume that ω is a bounded, simply connected domain. Then there exists a regular surface m ∈ C ( ω, R ) such that D m ( x ) = (cid:16) v ( x ) (cid:12)(cid:12) w ( x ) (cid:17) , x ∈ ω , (2.27) if and only if the compatibility condition ∂ v ( x , x ) = ∂ w ( x , x ) , (2.28) holds. The surface m is unique up to translation.Proof. We only need to observe that (2.27) reads m ,x m ,x m ,x m ,x m ,x m ,x = v w v w v w , (2.29)if and only if (cid:18) m ,x m ,x (cid:19) = (cid:18) v w (cid:19) , (cid:18) m ,x m ,x (cid:19) = (cid:18) v w (cid:19) , (cid:18) m ,x m ,x (cid:19) = (cid:18) v w (cid:19) , (2.30)and the existence of potentials m i ∈ C ( ω, R ) is guaranteed if and only if ∂ v = ∂ w , ∂ v = ∂ w , ∂ v = ∂ w , (2.31)which is equivalent to ∂ v = ∂ w . (2.32)The potentials m i , i = 1 , ,
3, are unique up to constants. (cid:4) A ∈ so (3) For m, y : ω ⊂ R → R we discussD m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , n unit normal vector field on y ( ω ) , (3.1)where Q ∈ C ( ω, SO(3)). Consider a linear approximation of this situation for small rotation angle α , | α | (cid:28) m ( x ) = D y ( x ) + D v ( x ) , Q ( x ) = + A ( x ) + h.o.t. , A ∈ C ( ω, so (3)) . (3.2)11ote that we do not assume that D v is small. We only assume that the rotations are close to . Then,D m ( x ) = D y ( x ) + D v ( x ) = ( + A ( x ) + . . . ) D y ( x ) ⇒ D v ( x ) = A ( x ) D y ( x ) + . . . , (3.3)hence we may consider the new problemD v ( x ) = A ( x ) D y ( x ) , A ∈ C ( ω, so (3)) , A ( x ) n ( x ) = 0 . (3.4)Therefore, we are led to study the problem Lemma 3.1.
Let ω ⊂ R be a bounded Lipschitz domain. Assume y ∈ C ( ω, R ) is a regular surface andlet v ∈ C ( ω, R ) . Moreover, assume α ∈ C ( ω, R ) and consider the system D v ( x ) = α ( x ) Anti( n ( x )) D y ( x ) , x ∈ ω , (3.5) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) . Then α ≡ const .Proof. We write ( ∂ v | ∂ v ) = α Anti( n )( ∂ y | ∂ y ) = α (cid:16) Anti( n ) ∂ y (cid:12)(cid:12)(cid:12) Anti( n ) ∂ y (cid:17) , (3.6) ⇐⇒ ∂ v = α [ n × ∂ y ] , ∂ v = α [ n × ∂ y ] , where we have used that Anti( n ) η = n × η . We proceed by taking the mixed derivatives ∂ ∂ v = ∂ α [ n × ∂ y ] + α [ ∂ n × ∂ y + n × ∂ ∂ y ] ,∂ ∂ v = ∂ α [ n × ∂ y ] + α [ ∂ n × ∂ y + n × ∂ ∂ y ] . (3.7)Hence, by equality of the mixed derivatives in (3.7) for y , v ∈ C ( ω, R ) we must have ∂ α [ n × ∂ y ] (cid:124) (cid:123)(cid:122) (cid:125) =: −→ Y + α [ ∂ n × ∂ y ] (cid:124) (cid:123)(cid:122) (cid:125) =: −→ B = ∂ α [ n × ∂ y ] (cid:124) (cid:123)(cid:122) (cid:125) =: −−→ X + α [ ∂ n × ∂ y ] (cid:124) (cid:123)(cid:122) (cid:125) =: −→ A ∈ R . (3.8)Especially we have that −→ A and −→ B are normal vectors whereas −→ X and −→ Y are linear independent tangentvectors, since (cid:104)−→ X , n (cid:105) = 0, (cid:104)−→ Y , n (cid:105) = 0 and −→ X × −→ Y = − ( n × ∂ y ) × ( n × ∂ y ) = −(cid:104) n , ∂ y × ∂ y (cid:105) n = (cid:104) n , (cid:107) ∂ y × ∂ y (cid:107) · n (cid:105) n = (cid:107) ∂ y × ∂ y (cid:107) · n = ∂ y × ∂ y , (3.9)where we have used that n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) . Thus, the vector fields −→ X , −→ Y and n form a 3 − frame on thesurface y ( ω ). However, (3.8) reads, ∂ α · −→ X + ∂ α · −→ Y = α · ( −→ A − −→ B ) = δ · n , (3.10)with a scalar field δ , so that by the linear independence of the vector fields −→ X , −→ Y and n we must alwayshave ∂ α = ∂ α = δ = 0, which gives α ≡ const. (cid:4) Thus, adding sufficient boundary conditions we arrive at
Proposition 3.2.
Let ω ⊂ R be a bounded Lipschitz domain. Assume y ∈ C ( ω, R ) is a regular surfaceand let v ∈ C ( ω, R ) . Moreover assume α ∈ C ( ω, R ) ∩ C ( ω, R ) and consider the system D v ( x ) = α ( x ) Anti( n ( x )) D y ( x ) , x ∈ ω , α | γ d = 0 , (3.11) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) and γ d is a relatively open, non-empty, subset ofthe boundary ∂ω . Then α ≡ . roof. By Lemma 3.1 it holds α ≡ const, so that due to the vanishing boundary condition α | γ d = 0 and thecontinuity of α we obtain α ≡ (cid:4) Corollary 3.3.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces and α ∈ C ( ω, R ) ∩ C ( ω, R ) is given with D m ( x ) = ( + α ( x ) Anti( n ( x ))) D y ( x ) , x ∈ ω , m | γ d = y | γ d , (3.12) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) and γ d is a relatively open, non-empty, subset ofthe boundary ∂ω . Then m ≡ y .Proof. We invoke Lemma 2.9 to see that m | γ d = y | γ d implies α | γ d ≡
0. Thus, for v = m − y we can applyProposition 3.2 to conclude that D v ≡ (cid:4) Proposition 3.4.
Let ω ⊂ R be a bounded Lipschitz domain. Assume m, y ∈ C ( ω, R ) are regular surfacesand α ∈ C ( ω, R ) with D m ( x ) = ( + α ( x ) Anti( n ( x ))) D y ( x ) , x ∈ ω , (3.13) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) . Then ∀ x ∈ ω : α ( x ) = 0 or H ( x ) = 0 , (3.14) where H denotes the mean curvature on the surface y .Proof. Recall that it holds for the vector field −→ A − −→ B = ∂ n × ∂ y − ∂ n × ∂ y = − H (cid:107) ∂ y × ∂ y (cid:107) n = − H ∂ y × ∂ y , (3.15)cf. [14, Section 2.5, Theorem 2]. Thus, for v = m − y the validity of (3.10) implies that we have pointwiseeither a vanishing angle α or a vanishing mean curvature H since0 (3.10) ≡ α ≡ const α · ( −→ A − −→ B ) (3.15) = − α · H · ∂ y × ∂ y . (cid:4) Remark 3.5 (Symmetry of the second fundamental form) . It is interesting to note that conclusion (3.14) canalso be obtained from the symmetry property of the second fundamental form on the surface m ( ω ). Indeed,the normal vector field on m ( ω ) coincides with n since ∂ m × ∂ m (3.13) = ( ∂ y + α Anti( n ) ∂ y ) × ( ∂ y + α Anti( n ) ∂ y )= ( ∂ y + α n × ∂ y ) × ( ∂ y + α n × ∂ y )= ∂ y × ∂ y + α ( n × ∂ y ) × ( n × ∂ y ) (3.9) = (1 + α ) ∂ y × ∂ y . (3.16)Thus, for the second fundamental form on m ( ω ) we obtainSym(2) (cid:51) II m = − [D m ] T D n = − [D m ] T D n = − [( + α Anti( n )) D y ] T D n = − [D y ] T D n − α [D y ] T Anti( n ) T D n = II v + α [D y ] T Anti( n ) D n . (3.17)13ince II v ∈ Sym(2), we are left with the single condition α [D y ] T Anti( n ) D n ∈ Sym(2) ⇐⇒ α (cid:18) ( ∂ y ) T ( ∂ y ) T (cid:19) ( n × ∂ n (cid:12)(cid:12)(cid:12) n × ∂ n ) ∈ Sym(2) ⇐⇒ α (cid:18) ∗ (cid:104) ∂ y , n × ∂ n (cid:105)(cid:104) ∂ y , n × ∂ n (cid:105) ∗ (cid:19) ∈ Sym(2) ⇐⇒ { α = 0 or (cid:104) ∂ y , n × ∂ n (cid:105) = (cid:104) ∂ y , n × ∂ n (cid:105)}⇐⇒ { α = 0 or (cid:104) n , ∂ n × ∂ y (cid:105) = (cid:104) n , ∂ n × ∂ y (cid:105)}⇐⇒ { α = 0 or 0 = (cid:104) n , ∂ n × ∂ y − ∂ n × ∂ y (cid:105)} (3.15) ⇐⇒ { α = 0 or 0 = (cid:104) n , − H ∂ y × ∂ y (cid:105)}⇐⇒ { α = 0 or H (cid:107) ∂ y × ∂ y (cid:107) = 0 }⇐⇒ { α = 0 or H = 0 } . (3.18) Corollary 3.6.
Let ω ⊂ R be a bounded Lipschitz domain. Assume y ∈ C ( ω, R ) is a regular surface andlet v ∈ C ( ω, R ) . Moreover assume α ∈ C ( ω, R ) and consider the system D v ( x ) = α ( x ) Anti( n ( x )) D y ( x ) , x ∈ ω , (3.19) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) . If the mean curvature H of y does not vanish atone point y ( x ) , then α ≡ .Proof. It follows from the previous Proposition 3.4, that if the mean curvature H does not vanish at somepoint y ( x ), we must have α ( x ) = 0 and the conclusion follows, since α ≡ const by Lemma 3.1. (cid:4) Overview large rotations : Q ∈ SO(3) infinitesimal rotations: A ∈ so (3)D m ( x ) = Q ( x ) D y ( x ) D m ( x ) = ( + A ( x )) D y ( x ) m | γ d = y | γ d m | γ d = y | γ d Q n = n A n = 0 ⇒ Q ( x ) q = q, q ⊥ n at γ d ⇒ A ( x ) q = 0 , q ⊥ n at γ d ⇒ Q | γ d ≡ ⇒ A | γ d ≡ m = I y (in-extensional) I m = I y + D y T A T A D y (cid:124) (cid:123)(cid:122) (cid:125) ∈ Sym + (2) ”expansive”isometry m ≡ y Q ≡ Proposition 4 . m ≡ y A ≡ Proposition 3 . . Q ∈ SO(3)
Lemma 4.1.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces, Q ∈ C ( ω, SO(3)) and D m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , x ∈ ω , (4.1)14 here n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) . Then α ≡ const , where α : ω → R denotes therotations angle in the Euler-Rodrigues representation of Q .Proof. By the Euler-Rodrigues representation there exists a function α : ω → R which fulfills Q ( x ) = (1 − cos α ( x )) n ( x ) ⊗ n ( x ) + cos α ( x ) + sin α ( x ) Anti( n ( x )) , (4.2)so that due to the expressions from (2.17) for all x ∈ ω there exists a neighborhood U of x such that α ∈ C ( U ∩ ω, R ).In view of (4.2), the problem (4.1) can be recast (at least locally) as followsD m = (cid:16) (1 − cos α ) n ⊗ n + cos α + sin α Anti( n ) (cid:17) D y . (4.3)Since ( n ⊗ n ) D y = n ⊗ ([D y ] T n ) = 0, the latter formula simplifies toD m = cos α D y + sin α Anti( n ) D y . (4.4)Obviously we have ∂ m = cos α ∂ y + sin α ( n × ∂ y ) and ∂ m = cos α ∂ y + sin α ( n × ∂ y ) , where for i = 1 ,
2, we used that Anti( n ) ∂ i y = n × ∂ i y . By taking the mixed derivatives, we arrive at ∂ ∂ m = ∂ (cos α ∂ y ) + ∂ (sin α ( n × ∂ y )) (4.5)= − sin α ∂ α ∂ y + cos α ∂ ∂ y + cos α ∂ α ( n × ∂ y ) + sin α ∂ n × ∂ y + sin α n × ∂ ∂ y , as well as ∂ ∂ m = ∂ (cos α ∂ y ) + ∂ (sin α ( n × ∂ y )) (4.6)= − sin α ∂ α ∂ y + cos α ∂ ∂ y + cos α ∂ α ( n × ∂ y ) + sin α ∂ n × ∂ y + sin α n × ∂ ∂ y . By using the equality of mixed derivatives for m, y ∈ C ( ω, R ) we must have ∂ α (cid:16) − sin α ∂ y + cos α ( n × ∂ y ) (cid:17)(cid:124) (cid:123)(cid:122) (cid:125) =: −→ Y α + sin α ∂ n × ∂ y (cid:124) (cid:123)(cid:122) (cid:125) = −→ B (4.7)= ∂ α (cid:16) − sin α ∂ y +cos α ( n × ∂ y ) (cid:17)(cid:124) (cid:123)(cid:122) (cid:125) =: −−→ X α +sin α ∂ n × ∂ y (cid:124) (cid:123)(cid:122) (cid:125) −→ A . As in the linear case we obtain that −→ A and −→ B are normal vectors whereas −→ X α and −→ Y α are linear independenttangent vectors, since (cid:104)−→ X α , n (cid:105) = 0, (cid:104)−→ Y α , n (cid:105) = 0 and −→ X α × −→ Y α = (sin α ∂ y + cos α −→ X ) × ( − sin α ∂ y + cos α −→ Y )= sin α ∂ y × ∂ y + cos α −→ X × −→ Y + sin α cos α ( ∂ y × −→ Y − −→ X × ∂ y ) (3.9) = ∗ sin α ∂ y × ∂ y + cos α ∂ y × ∂ y = ∂ y × ∂ y , (4.8)where in ∗ we also used that ∂ y × −→ Y − −→ X × ∂ y = ∂ y × ( n × ∂ y ) + ( n × ∂ y ) × ∂ y = 0.Thus, the vector fields −→ X α , −→ Y α and n form a 3 − frame on the surface y ( ω ). However, (4.7) reads, ∂ α · −→ X α + ∂ α · −→ Y α = sin α · ( −→ A − −→ B ) = (cid:101) δ · n , (4.9)with a scalar field (cid:101) δ , so that by the linear independence of the vector fields −→ X α , −→ Y α and n we must alwayshave ∂ α = ∂ α = (cid:101) δ = 0, which gives α ≡ const (at least locally, but the constant has to be always the samesince the neighborhoods overlap). (cid:4) ( x i ) ω ⊂ R Figure 6: Although the rotation angle α in Q can be chosen in each point, its C -regularityis a priori not clear. However, it is given in asufficient neighborhood of each point. We haveseen, that the angle should be constant there,and due to the overlapping of the neighbor-hoods it has to be always the same constant. Remark 4.2.
Note that the rotation angle α has always to be constant but unconstrained, if, e.g., nofurther boundary conditions are appended. This is, e.g., provided by members of the same associate family of minimal surfaces (i.e., it holds H ≡ Proposition 4.3.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces, Q ∈ C ( ω, SO(3)) and D m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , x ∈ ω , m | γ d = y | γ d , (4.10) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) and γ d is a relatively open, non-empty, subset ofthe boundary ∂ω . Then m ≡ y .Proof. Using the boundary condition m | γ d = y | γ d , Lemma 2.7 allows to conclude that Q | γ d ≡ whichimplies sin α | γ d = 0 and cos α | γ d = 1. By Lemma 4.1 we have α ≡ const, so that sin α ≡ α ≡ ω , and consequently by the Euler-Rodrigues representation it follows Q ≡ . (cid:4) Corollary 4.4.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are two regularsurfaces and I m ( x ) = [D m ( x )] T D m ( x ) = [D y ( x )] T D y ( x ) = I y ( x ) , n ( x ) = n ( x ) , ∀ x ∈ ω ,m | γ d = y | γ d , (4.11) where n = ∂ m × ∂ m (cid:107) ∂ m × ∂ m (cid:107) and n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) are the respective normal fields and γ d is a relatively open,non-empty subset of the boundary ∂ω . Then m ≡ y .Proof. Consider the lifted quantities (D m | n ) versus (D y | n ). It holds(D m | n ) T (D m | n ) =
000 0 1[D m ] T D m , (D y | n ) T (D y | n ) =
000 0 1[D y ] T D y . (4.12)Since by assumption I m = D m T D m = D y T D y = I y , it follows from (4.12) that(D m | n ) T (D m | n ) = (D y | n ) T (D y | n ) . (4.13)Then for all x ∈ ω there exists Q ( x ) ∈ SO(3) such that (D m ( x ) | n ( x )) = Q ( x )(D y ( x ) | n ( x )). The assumption n ( x ) = n ( x ) gives (D m | n ) = Q (D y | n ) . (4.14)16ultiplying both sides with e , we obtain Q ( x ) n ( x ) = n ( x ) for all x ∈ ω and from (4.14) we obtain Q = (D m | n )(D y | n ) − = (D m | n ) 1det(D y | n ) Cof(D y | n ) . (4.15)Since by assumption m, y ∈ C ( ω, R ) and y is a regular surface with det(D y | n ) ≥ c + > Q ∈ C ( ω, SO(3)). Thus, we are again in the situation of Proposition 4.3 and the proof isfinished. (cid:4)
Proposition 4.5.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces, Q ∈ C ( ω, SO(3)) and D m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , x ∈ ω , (4.16) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) . Then ∀ x ∈ ω : sin α ( x ) = 0 or H ( x ) = 0 , (4.17) where α : ω → R denotes the rotations angle in the Euler-Rodrigues representation of Q .Proof. Again, in view of (3.15), the validity of (4.9) implies that we have pointwise either vanishing sin α ora vanishing mean curvature H since0 (4.9) ≡ α ≡ const sin α · ( −→ A − −→ B ) (3.15) = − α · H · ∂ y × ∂ y . (cid:4) Remark 4.6 (Symmetry of the second fundamental form) . Conclusion (4.17) can also be obtained from thesymmetry property of the second fundamental form on the surface m ( ω ). Indeed, the normal vector field on m ( ω ) coincides with n since ∂ m × ∂ m (4.16) = ( Q ∂ y ) × ( Q ∂ y ) Q ∈ SO(3) = Q ∂ y × ∂ y = ∂ y × ∂ y . (4.18)Thus, for the second fundamental form on m ( ω ) we obtainSym(2) (cid:51) II m = − [D m ] T D n = − [D m ] T D n = − [D y ] T Q T D n = − [D y ] T [(1 − cos α ) n ⊗ n + cos α + sin α Anti( n )] T D n = (1 − cos α )[D y ] T n ⊗ n (cid:124) (cid:123)(cid:122) (cid:125) =0 D n − cos α [D y ] T D n − sin α [D y ] T Anti( n ) T D n = cos α II v + sin α [D y ] T Anti( n ) D n . (4.19)Since II v ∈ Sym(2), we are again left with the single conditionsin α [D y ] T Anti( n ) D n ∈ Sym(2) (3.18) ⇐⇒ { sin α = 0 or H = 0 } . (4.20) Corollary 4.7.
Let ω ⊂ R be a bounded Lipschitz domain. Assume that m, y ∈ C ( ω, R ) are regularsurfaces, Q ∈ C ( ω, SO(3)) and D m ( x ) = Q ( x ) D y ( x ) , Q ( x ) n ( x ) = n ( x ) , x ∈ ω , (4.21) where n = ∂ y × ∂ y (cid:107) ∂ y × ∂ y (cid:107) denotes the normal field on y ( ω ) . If the mean curvature of y does not vanish atone point y ( x ) , then m ( x ) = y ( x ) + b or m ( x ) = − y ( x ) + b for some constant translation b ∈ R . roof. It follows from Propostion 4.5, that if the mean curvature H is distinct from 0 at some point y ( x ),we must have sin α ( x ) = 0. Thus, sin α ≡ α = ±
1, where the sign of cos α does not change sincethe rotation angle α is constant by Lemma 4.1. Hence, we obtain for Q ( x ) (4.2) = (1 − cos α ) n ( x ) ⊗ n ( x ) ± , (4.22)and a multiplication with D y givesD m ( x ) = Q ( x ) D y ( x ) (4.22) = (1 − cos α ) n ( x ) ⊗ n ( x ) D y ( x ) (cid:124) (cid:123)(cid:122) (cid:125) =0 ± D y ( x ) = ± D y ( x ) . (cid:4) Remark 4.8.
For a C ∞ -embedding y a comparable result, using other techniques, has been obtained in [1,16]. We have proved some improved rigidity results for C -smooth regular embedded surfaces. The underlyingmechanical problem, namely the possibility of a pure in-plane drill rotation field as deformation mode of asurface when boundary conditions of place are prescribed somewhere has been negatively answered, for bothsmall and large drill rotations, assuming some natural level of smoothness for the rotation fields.The implication for the application of shell models with independent rotation fields in the FEM-contextcan be summarized as follows: since the pure drilling degree of freedom is kinematically impossible bycompatibility, any ”torsional spring stiffness” attributed to this pseudo-deformation mode must be regardedwith extreme caution (here, the Cosserat couple modulus µ c ≥ γ d of the shell.Then, the mentioned stiffness is a boundary value problem dependent parameter which needs to bedetermined for each new problem again. Thus one should call it always a fictitious stiffness and treat itaccordingly.However, it is remarkable that in the planar Cosserat shell model [29] existence can be shown also forzero Cosserat couple modulus µ c ≡ q > Acknowledgment.
The last author is indebted to Amit Acharya (Carnegie Mellon University) for clarifyingdiscussions regarding his notion of normality preserving shell deformations. We also thank Robert L. Bryant(Duke University, North-Carolina) for fruitful discussions. Discussions with Marek L. Szwabowicz (MaritimeUniversity, Gdansk, Poland) and Krzysztof Wi´sniewski (Institute of Fundamental Technological ResearchPolish Academy of Sciences, Warsaw, Poland) are also acknowledged. This research has been supported bythe Deutsche Forschungsgemeinschaft (DFG, German Research Foundation)-Project No. , Neff902/8-1 (P. Lewintan and P. Neff). Maryam Mohammadi Saem is grateful for a grant of the Faculty ofMathematics, University Duisburg-Essen. [1] K. Abe and J. Erbacher. “Isometric immersions with the same Gauss map”.
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A Appendix
A.1 Associate family of minimal surfaces: the catenoid and helicoid family
Meusnier in 1776 discovered that the helicoid and the catenoid are minimal surfaces, i.e., they have vanishingmean curvature. It was Bonnet who showed that these minimal surfaces belong to the same associate family,more precisely, the catenoid can be bend without stretching into a portion of a helicoid in such a way thatthe surface normals remain unchanged. For further examples of associate families of minimal surfaces werefer the reader to [14, Chapter 3].Let ω = R × [0 , π ) then the catenoid X cat ∈ C ∞ ( ω, R ) and the helicoid X hel ∈ C ∞ ( ω, R ) areparametrized by X cat ( x , x ) = cos x cosh x sin x cosh x x , X hel ( x , x ) = sin x sinh x − cos x sinh x x , (A.1)and the associate surfaces, parametrized by θ , are given by X θ ( x , x ) = cos θ · X cat ( x , x ) + sin θ · X hel ( x , x ) for θ ∈ [0 , π ] . (A.2)Their partial derivatives fulfill ∂ X θ = cos θ · ∂ X cat − sin θ · ∂ X cat and ∂ X θ = sin θ · ∂ X cat + cos θ · ∂ X cat , (A.3)20o that, the surface normals remain unchanged n X θ ( x , x ) = n X cat ( x , x ) = n X hel ( x , x ) for all ( x , x ) ∈ ω . (A.4)Further properties of the members of this associate family X θ can be found in [24] and [14, Chapter 3], aswell as the references cited therein. (a) θ = 0 (b) θ = π (c) θ = π (d) θ = π Figure 7: Four consecutive stepsof an isometric deformation ofa catenoid (left) into a helicoid(right). Such a transformationexists, since both are membersof the same associate family X θ .Note that every member of thedeformation family has vanish-ing mean curvature, i.e., is aminimal surface.To see (A.3) and (A.4) we take the partial derivatives of (A.2): ∂ j X θ = cos θ · ∂ j X cat + sin θ · ∂ j X hel for j = 1 ,
2. (A.5)Since the partial derivatives of the catenoid and the helicoid satisfy the Cauchy-Riemann equations ∂ X cat = ∂ X hel and ∂ X cat = − ∂ X hel , (A.6)we obtain (A.3), which in matrix notation readsD X θ = D X cat (cid:18) cos θ sin θ − sin θ cos θ (cid:19) . (A.7)Moreover, ∂ X θ × ∂ X θ (A.3) = cos θ · ∂ X cat × ∂ X cat − sin θ · ∂ X cat × ∂ X cat = ∂ X cat × ∂ X cat , (A.8)which gives (A.4). Furthermore, ∂ X cat = cos x sinh x sin x sinh x and ∂ X cat = − sin x cosh x cos x cosh x , (A.9)so that (cid:107) ∂ X cat (cid:107) = sinh x + 1 = cosh x = (cid:107) ∂ X cat (cid:107) and (cid:104) ∂ X cat , ∂ X cat (cid:105) = 0 . (A.10)In regard with (A.3) we obtain (cid:107) ∂ X θ (cid:107) = cos θ (cid:107) ∂ X cat (cid:107) + sin θ (cid:107) ∂ X cat (cid:107) = cosh x , (A.11a) (cid:107) ∂ X θ (cid:107) = sin θ (cid:107) ∂ X cat (cid:107) + cos θ (cid:107) ∂ X cat (cid:107) = cosh x = (cid:107) ∂ X θ (cid:107) , (A.11b) (cid:104) ∂ X θ , ∂ X θ (cid:105) (A.10) = cos θ sin θ (cid:107) ∂ X cat (cid:107) − sin θ cos θ (cid:107) ∂ X cat (cid:107) = 0 . (A.11c)21n other words, the first fundamental form of all members of the associate family remains unchanged and isgiven by I X θ ( x , x ) = [D X θ ] T D X θ = (cid:18) (cid:107) ∂ X θ (cid:107) (cid:104) ∂ X θ , ∂ X θ (cid:105)(cid:104) ∂ X θ , ∂ X θ (cid:105) (cid:107) ∂ X θ (cid:107) (cid:19) = cosh x · . (A.12)Thus, [D X cat ] T D X θ (A.7) = [D X cat ] T D X cat (cid:18) cos θ sin θ − sin θ cos θ (cid:19) (A.12) = cosh x (cid:18) cos θ sin θ − sin θ cos θ (cid:19) , (A.13)and [D X cat ] T D X θ / ∈ Sym + (2) is not a pure in-plane stretch.Moreover, as in (4.13), there exists an in-plane drill rotation Q θ ( x ) ∈ SO(3) which fulfillsD X θ ( x ) = Q θ ( x ) D X cat ( x ) and Q θ ( x ) n ( x ) = n ( x ) , where n ( x ) := n X θ ( x ) = n X cat ( x ) . (A.14)Next we show, that the (constant) rotation angle, cf. Lemma 4.1, extracted from Q θ ( x ) is already given by − θ (or differs from it by an integer multiple of 2 π ), so that we have the representation Q θ ( x ) = (1 − cos( − θ )) n ( x ) ⊗ n ( x ) + cos( − θ ) + sin( − θ ) Anti( n ( x ))= (1 − cos θ ) n ( x ) ⊗ n ( x ) + cos θ − sin θ Anti( n ( x )) . (A.15)For that purpose, note that ∂ X cat × ∂ X cat = − cos x cosh x − sin x cosh x sinh x cosh x ⇒ (cid:107) ∂ X cat × ∂ X cat (cid:107) = cosh x , (A.16)so that n = 1cosh x − cos x − sin x sinh x , and n × ∂ X cat = ∂ X cat , n × ∂ X cat = − ∂ X cat . (A.17)Hence, with n ⊗ n ∂ j X cat = n (cid:104) n , ∂ j X cat (cid:105) ≡
0, we obtain Q θ ∂ X cat (A.15) = cos θ ∂ X cat − sin θ Anti( n ) ∂ X cat (A.17) = cos θ ∂ X cat − sin θ ∂ X cat (A.3) = ∂ X θ , (A.18)as well as Q θ ∂ X cat (A.15) = cos θ ∂ X cat − sin θ Anti( n ) ∂ X cat (A.17) = cos θ ∂ X cat + sin θ ∂ X cat (A.3) = ∂ X θ , (A.19)and we have shown that Q θ D X cat = D X θ , (A.20)where Q θ has the expression (A.15) and its columns read with the representation of the normal (A.17) Q θ e = − ( cos x − cosh x ) cos θ − cos x cosh x − cosh x sinh x sin θ +cos x sin x cos θ − cos x sin x cosh x − cosh x sin x sin θ − sinh x cos x cos θ +sinh x cos x cosh x , (A.21a) Q θ e = cosh x sinh x sin θ − cos x sin x cos θ +cos x sin x cosh x − ( sin x − cosh x ) cos θ − sin x cosh x cosh x cos x sin θ +sinh x sin x cos θ − sinh x sin x cosh x , (A.21b) Q θ e = cosh x sin x sin θ +sinh x cos x cos θ − sinh x cos x cosh x − cosh x cos x sin θ − sinh x sin x cos θ +sinh x sin x cosh x cos θ +cosh x − x . (A.21c)22ecall, that if a surface X is parametrized conformally, i.e., it holds (cid:107) ∂ X (cid:107) = (cid:107) ∂ X (cid:107) and (cid:104) ∂ X, ∂ X (cid:105) = 0,then it is a minimal surface (i.e. has vanishing mean curvature everywhere) if and only if ∆ X ≡ X θ are, indeed,minimal surfaces, we compute ∆ X θ (A.3) = cos θ · ∆ X cat (A.9) ≡ . (A.22)Furthermore, if a minimal surface is parametrized conformally, then the same holds for its correspondingGauss map, cf. [14, p.74]. Indeed, all members of the associate family X θ are minimal surfaces and for theirGauss maps (which all coincide) n : ω → S it holdsI n = [D n ] T D n = 1cosh x · , (A.23)which shows that n is also parametrized conformally.Let us mention, that the constancy of the rotation angle can also be achieved without applying Lemma4.1. For that purpose, let us here call the in-plane drill rotation (cid:98) Q ( x ) ∈ SO(3) which fulfillsD X θ ( x ) = (cid:98) Q ( x ) D X cat ( x ) and (cid:98) Q ( x ) n ( x ) = n ( x ) , where n ( x ) := n X θ ( x ) = n X cat ( x ) . (A.24)Thus, as in (4.13), it follows(D X θ (cid:12)(cid:12) n ) = (cid:98) Q (D X cat (cid:12)(cid:12) n ) ⇒ (cid:98) Q = (D X θ (cid:12)(cid:12) n )(D X cat (cid:12)(cid:12) n ) − , (A.25)and a direct computation gives the entries of (cid:98) Q ( x ), which, indeed, coincide with (A.21). From the uniquenessof the Euler-Rodrigues representation it then follows that the corresponding rotation angle (cid:98) α ( x ) is constantand is given by − θ (or differs from it by an integer multiple of 2 π ). Indeed, with (2.17) we havesin( (cid:98) α ( x )) (2.17) = − tr (cid:16) Anti( n ( x )) (cid:98) Q ( x ) (cid:17) (cid:98) Q = Q θ = (A.21) − sin θ and cos( (cid:98) α ( x )) (2.17) = − tr (cid:98) Q ( x ) − (cid:98) Q = Q θ = (A.21) cos θ.θ.