14
1 From Riemann manifolds to Riemann manifolds
Certainly, we agree that the various transformations have to be checked by “paper and pencil”, in
particular, by means of Examples 1.2 and 1.3. In case that we are led to “non-integrable differentials”
(namely differential forms), we have indicated this result by writing “ –
dV ” and “ –
dv” according to
the M. Planck notation. In this context, the left and right Frobenius matrices, F l and F r , have to be
seen. They are used as matrices of integrating factors which transform “imperfect differentials” –
dV
A
(namely –
dV
1 , –
dV
2 , or differential forms Ω 1 , Ω 2 ) or –
dv
α (namely –
dv
1 , –
dv
2 , or differential forms ω 1 , ω 2 )
to “perfect differentials” dU
A (namely dU
1 , dU
2 ) or du
α (namely du
1 , du
2 ). As a sample reference
of the theory of differential forms and the Frobenius Integration Theorem, we direct the interested
reader to J. A. de Azcarraga and J. M. Izquierdo (1995), M. P. do Carmo (1994), and H. Flanders (1970
p. 97). Indeed, we hope that the reader appreciates the triple notation index notation (Ricci calculus),
matrix notation (Cayley calculus), and explicit notation (Leibniz–Newton calculus). Thus, we are led
to the general eigenvalue problem as a result of simultaneous diagonalization of two positive-definite
symmetric matrices {C l , G l } or {C r , G r }, respectively. Compare with Lemma 1.5.
Lemma 1.5 (Left and right general eigenvalue problem of the Cauchy–Green deformation tensor).
For the pair of positive-definite symmetric matrices {C l , G l } or {C r , G r }, respectively, a simultaneous
diagonalization defined by
left diagonalization:
F
T
l C l F l = diag
Λ
2
1 , Λ
2
2
:= D l ,
F
T
l G l F l = I 2
versus
right diagonalization:
F
T
r C r F r = diag
λ
2
1 , λ
2
2
:= D r ,
F
T
r G r F r = I 2
(1.56)
is readily obtained from the following general eigenvalue–eigenvector problem of type left eigenvalues
and left principal stretches:
C l F l − G l F l D l = 0
⇐⇒
(C l − Λ
2
i G l )f li = 0
⇐⇒
C l − Λ
2 G l
= 0 ,
Λ
2
1,2 = Λ
2
± =
1
2
tr
C l G
−1
l
±
tr
C l G
−1
l
2 − 4det
C l G
−1
l
,
(1.57)
subject to F
T
l G l F l = I 2 , and
C r F r − G r F r D r = 0
⇐⇒
(C r − λ
2
i G r )f ri = 0
⇐⇒
C r − λ
2 G r
= 0 ,
λ
2
1,2 = λ
2
± =
1
2
tr
C r G
−1
r
±
tr
C r G
−1
r
2 − 4det
C r G
−1
r
,
(1.58)
subject to F
T
r G r F r = I 2 , and
Λ
2
1,2 = 1/λ
2
1,2 ⇐⇒ 1/Λ
2
1,2 = λ
2
1,2 .
(1.59)
End of Lemma.
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