EQUATIONS OF ELASTIC HYPERSURFACES

3. CALCULUS OF DIFFERENTIAL OPERATORS ON HYPERSURFACES 87
Since the space of tensors [ζ (q)
jk ] n×n is finite dimensional, there exists a convergent subsequence
ζ (q r)
kl
→ ζ (0)
kl
as r → ∞. Then we get an obvious contradiction
∑
c ijkl ζ (0)
ij ζ(0) kl
= 0 , |ζ (0) | = 1 .
which proves that C 0 > 0.
i,j,k,l
Theorem 3.24 The conditions (3.89) and the first equality in (3.91) imply the second equality
in (3.91) as well as the conditions (3.89) and the second equality in (3.91) imply the first
equality in (3.91).
A linear operator T in the energy functional of anisotropic elasticity (3.85) satisfies the
symmetry conditions (3.89), and (3.91). Equivalently, the corresponding elasticity tensor
T = [ c ijkl
]ijkl has the symmetries (3.90), (3.92) and, therefore, might have n+n2 (n−1) 2 /2
different entries only.
The Euler-Lagrange equation associated with the energy functional (3.83) for a linear
anisotropic elastic medium, reads
L S U = Def ∗ S TDef S U = (D S ) ∗ TD S U = 0, U ∈ TS ⋂ W 2 (S ), (3.96)
where again T = [ c ijkl is the elasticity tensor which is positive definite (cf. (3.94)) and
]ijkl
has the symmetries (3.90), (3.92).
Proof: The claimed implications in the first part of the theorem is evident in equivalent
formulation for corresponding tensors and matrices: (3.90) and (3.92).
Also the second part of the theorem is trivial.
The desired Euler-Lagrange equation (3.96) follows from (3.83), (3.85) and (3.93):
∫
E [U]:= 〈TD S U(x), D S U(x)〉 dS
S
= (TD S U(x), D S U(x)) S = ((D S ) ∗ TD S U, U) S = 0
= 4(TDef S U(x), Def S U(x)) S = 4(Def ∗ S TDef S U, U) S = 0
if and only if U ∈ TS ∩ W 2 (S ) is a solution of equation (3.96) due to the positive definiteness
of the elasticity tensor T (cf. (3.94)).
Next we will find the Euler-Lagrange equation associated with the energy functional
(3.83) for a linear isotropic elastic medium (Lamé equation) which is simpler. Such energy
functional should be invariant with respect to any rotation. For the elasticity tensor this
results into the requirement that
T(BAB −1 ) = B(TA)B −1 , ∀ A, B ∈ M n,n (R) and unitary B ⊤ = B −1 . (3.97)
Examples of linear operators (3.86) satisfying (3.91) and (3.97) include
T = TA := (Tr A)I and TA := A + A ⊤ , (3.98)