1-4 Euler–Lagrange deformation tensor 29
1-4 Euler–Lagrange deformation tensor
“Approach your problems from the right end and begin with the answers.
Then, one day, perhaps you will find the final question.”
(The Hermit Clad in Crane Feathers, in R. van Gulik’s The Chinese Maze Murders.)
A first additive measure of deformation: the Euler–Lagrange deformation tensor, relations between the
Cauchy–Green and the Euler-Lagrange deformation tensor.
The first additive measure of deformation is based upon the scale differences ds
2
−dS
2 versus dS
2
−ds
2 ,
which are represented by pullback U
M
→ u
µ = f
µ (U
M ) or pushforward u
µ
→ U
M = F
M (u
µ ), in
particular, by
ds
2
− dS
2 = dU
T
J
T
l G r J l − G l
dU versus dS
2
− ds
2 = du
T
J
T
r G l J r − G r
du . (1.111)
Accordingly, we are led to the deformation measures of Box 1.17, which have been introduced by
L. Euler and J. L. Lagrange, called strain.
Box 1.17 (Left versus right Euler–Lagrange deformation tensor).
Left EL deformation tensor :
Right EL deformation tensor :
ds
2 − dS
2 = dU
T
“
J
T
l G r J l − G l
”
dU ,
dS
2 − ds
2 = du
T
“
J
T
r G l J r − G r
”
du ,
1
2
`
ds
2 − dS
2 ´
= +dU
T E l dU ,
1
2
`
dS
2 − ds
2 ´
= −du
T E r du ,
E l :=
1
2
“
J
T
l G r J l − G l
”
.
E r :=
1
2
“
G r − J
T
r G l J r
”
.
(1.112)
Question.
Question: “What is the role of strain in the context of the pair of matrices {E l , G l } and
{E r , G r }, respectively?” Answer: “{E l , E r } are symmetric matrices and {G l , G r } are symetric,
positive-definite matrices. Thus, according to a standard lemma of matrix algebra, both
matrices can be simultaneously diagonalized, one matrix being the unit matrix. With the
reference to the general eigenvalue we experienced for the Cauchy–Green deformation tensor,
we arrive at Lemma 1.7.”
Lemma 1.7 (Left and right general eigenvalue problem of the Euler–Lagrange deformation tensor).
For the pair of symmetric matrices {E l , G l } or {E r , G r }, where {G l , G r } are positive-definite matrices,
a simultaneous diagonalization, namely
F
T
l E l F l = diag
K 1 , K 2
, F
T
l G l F l = I 2 versus F
T
r E r F r = diag
κ 1 , κ 2
, F
T
r G r F r = I 2 , (1.113)
is immediately obtained from the left and right general eigenvalue–eigenvector problems
E l F l − G l F l diag
K 1 , K 2
= 0
E r F r − G r F r diag
κ 1 , κ 2
= 0
⇔
⇔
E l − K i G l
f li = 0 (∀i ∈ {1, 2})
E r − κ i G r
f ri = 0 (∀i ∈ {1, 2})
⇔
and
⇔
E l − K i G l
= 0
F
T
l G l F l = I 2
,
E r − κ i G r
= 0
F
T
r G r F r = I 2
,
K 1,2 = K +,− =
1
2
tr
E l G
−1
l
±
κ 1,2 = κ +,− =
1
2
tr
E r G
−1
r
±
±
tr
E l G
−1
l
2 − 4det
E l G
−1
l
,
±
tr
E r G
−1
r
2 − 4det
E r G
−1
r
.
(1.114)
End of Lemma.
1-4 Euler–Lagrange deformation tensor
“Approach your problems from the right end and begin with the answers.
Then, one day, perhaps you will find the final question.”
(The Hermit Clad in Crane Feathers, in R. van Gulik’s The Chinese Maze Murders.)
A first additive measure of deformation: the Euler–Lagrange deformation tensor, relations between the
Cauchy–Green and the Euler-Lagrange deformation tensor.
The first additive measure of deformation is based upon the scale differences ds
2
−dS
2 versus dS
2
−ds
2 ,
which are represented by pullback U
M
→ u
µ = f
µ (U
M ) or pushforward u
µ
→ U
M = F
M (u
µ ), in
particular, by
ds
2
− dS
2 = dU
T
J
T
l G r J l − G l
dU versus dS
2
− ds
2 = du
T
J
T
r G l J r − G r
du . (1.111)
Accordingly, we are led to the deformation measures of Box 1.17, which have been introduced by
L. Euler and J. L. Lagrange, called strain.
Box 1.17 (Left versus right Euler–Lagrange deformation tensor).
Left EL deformation tensor :
Right EL deformation tensor :
ds
2 − dS
2 = dU
T
“
J
T
l G r J l − G l
”
dU ,
dS
2 − ds
2 = du
T
“
J
T
r G l J r − G r
”
du ,
1
2
`
ds
2 − dS
2 ´
= +dU
T E l dU ,
1
2
`
dS
2 − ds
2 ´
= −du
T E r du ,
E l :=
1
2
“
J
T
l G r J l − G l
”
.
E r :=
1
2
“
G r − J
T
r G l J r
”
.
(1.112)
Question.
Question: “What is the role of strain in the context of the pair of matrices {E l , G l } and
{E r , G r }, respectively?” Answer: “{E l , E r } are symmetric matrices and {G l , G r } are symetric,
positive-definite matrices. Thus, according to a standard lemma of matrix algebra, both
matrices can be simultaneously diagonalized, one matrix being the unit matrix. With the
reference to the general eigenvalue we experienced for the Cauchy–Green deformation tensor,
we arrive at Lemma 1.7.”
Lemma 1.7 (Left and right general eigenvalue problem of the Euler–Lagrange deformation tensor).
For the pair of symmetric matrices {E l , G l } or {E r , G r }, where {G l , G r } are positive-definite matrices,
a simultaneous diagonalization, namely
F
T
l E l F l = diag
K 1 , K 2
, F
T
l G l F l = I 2 versus F
T
r E r F r = diag
κ 1 , κ 2
, F
T
r G r F r = I 2 , (1.113)
is immediately obtained from the left and right general eigenvalue–eigenvector problems
E l F l − G l F l diag
K 1 , K 2
= 0
E r F r − G r F r diag
κ 1 , κ 2
= 0
⇔
⇔
E l − K i G l
f li = 0 (∀i ∈ {1, 2})
E r − κ i G r
f ri = 0 (∀i ∈ {1, 2})
⇔
and
⇔
E l − K i G l
= 0
F
T
l G l F l = I 2
,
E r − κ i G r
= 0
F
T
r G r F r = I 2
,
K 1,2 = K +,− =
1
2
tr
E l G
−1
l
±
κ 1,2 = κ +,− =
1
2
tr
E r G
−1
r
±
±
tr
E l G
−1
l
2 − 4det
E l G
−1
l
,
±
tr
E r G
−1
r
2 − 4det
E r G
−1
r
.
(1.114)
End of Lemma.