Compatibility Conditions for the Left Cauchy
Green Tensor Field in 3-D
Amit Acharya
OxPDE Seminar Series
Oxford, Mathematical Institute, Nov. 12, 2008
Continuum Mechanics Problem for Left/Right Cauchy
Green Compatibility (LCG/RCG)
y (i )
y(x)
x
T
Prescribed
C ( ⋅) , B ( ⋅)
as functions of
Find
y (i)
x
⎛ ∂y ⎞ ∂y
( x ) = C ( x ) ∈ Psym
⎜ ⎟
⎝ ∂x ⎠ ∂x
∂y k ∂y k
= Cαβ
α
β
∂x ∂x
T
∂y ⎛ ∂y ⎞
⎜ ⎟
∂x ⎝ ∂x ⎠
( x ) = B ( x ) ∈ Psym
∂y i ∂y j
ij
=
B
∂xα ∂xα
RCG
LCG
Are the RCG and LCG compatibility problems really
different?
LCG
RCG
∂y
= F ⇔ curl F = 0
∂x
F T F = specified ( x ) ∈ Psym
∂y
= F ⇔ curl F = 0
∂x
FF T = B
⇔ F −T F −1 = B −1
ℑ ( F ) ℑ ( F ) = specified ( x ) ∈ Psym
T
Using RCG method one has
curl ℑ ( F ( x ) ) = 0
ε ijk
∂Fan −1
∂Fak
=
ε
=0
F
0
but
need
nk
ijk
j
j
∂x
∂x
Motivation
ƒ Continuum Mechanics
• Interesting geometry question for classical kinematical measure
¾
Cauchy stress for Frame-indifferent, isotropic elastic material is a
function only of B
• Sharp contrast in uniqueness from the more studied RCG case
• Open in 3-D
ƒ Mathematics
•
•
•
•
Interesting questions involving
Geometry
Nonlinear PDE
Algebra
¾
Even in the
C∞
local existence case
Riemannian Geometry in charts
Given two coordinate patches for a Riemannian manifold, with points denoted generically by
x ⇔ xα ⇔ ( x1 , x 2 , x3 )
y ⇔ y ⇔(y ,y ,y
α
∃ 3× 3
( x ) C , ( y )C ,
satisfying
Matrix
inverse
1
2
3
y = y ( x)
)
and
⎛ ∂y ⎞
det ⎜ ⎟ ≠ 0
⎝ ∂x ⎠
sym, + def matrix fields
−1
B
:
=
C
,
( x)
( x)
∂y
= α
∂x
k
( x ) Cαβ
i
∂
y
ij
B
= α
( y)
∂x
−1
B
:
=
C
( y)
( y)
⎡ ( y) C ⎤
⎣
⎦ km
⎡ ( x) B ⎤
⎣
⎦
αβ
∂y
∂x β
m
∂y j
β
∂x
[∂y
∂x ]kα
∂y k
= α
∂x
Notational agreement:
Evaluate anything like
( x)
( y)
( ⋅)
( ⋅)
at
and
at
x
y ( x)
The question of equivalence of quadratic forms
∂p
Given two local Psym matrix fields on manifold Μ,
2
∂y
dim ( Μ ) = N , they are equivalent if one can find
N
two local charts in
related 1 − 1 with
⎛ ∂y ⎞
det ⎜ ⎟ ≠ 0 satisfying transformation rules. (WHY???)
⎝ ∂x ⎠
On tangent space Tp of p0 ∈ Μ spanned by
y1 = const
∂p
∂y1
y 2 = const
p0
Μ
0
⎛ ∂p ∂p ⎞
⎛ ∂p ∂p ⎞
⎜ 1 , 2 ⎟ or ⎜ 1 , 2 ⎟ any vector a ∈ Tp0
⎝ ∂x ∂x ⎠
⎝ ∂y ∂y ⎠
i
i
∂
∂
∂
∂
∂
y
y
p
p
p
a = ( x ) aα α = ( x ) aα i α = ( y ) a i i ⇒ ( y ) a i = ( x ) aα α
∂x
∂y ∂x
∂y
∂x
Now, let there be a quadratic form for each chart s.t. for a,b
α
a
( x)
( x )Cαβ
β
b
=: physical scalar indep. of chart
( x)
But, no chart is special; ∴ ( x ) aα
( x )Cαβ
β
i
b
=
a
( x)
( y)
( y )Cij
j
b
( y)
α
( x ) a ( x )b
β
⎡ ∂y i
∂y j
⎢ ∂xα ( y )Cij ∂x β −
⎣
∀ ( x ) a α ,( x ) b β
⎤
C
( x ) αβ ⎥ = 0
⎦
Mapping compatibility question to Riemannian
Geometry
( x ) Cαβ
∂y k
= α
∂x
i
∂
y
ij
B
= α
( y)
∂x
⎡ ( y) C ⎤
⎣
⎦ km
⎡ ( x) B ⎤
⎣
⎦
αβ
∂y m
∂x β
∂y j
∂x β
choose
( y) C
≡I
RCG compatibility
given C , B
find y x
( )
( x) B ≡ I
as fns. of
x
LCG compatibility
Overdetermined Problems
∂y k ∂y k
= Cαβ
α
β
∂x ∂x
∂y i ∂y j
ij
B
=
∂xα ∂xα
Machinery of Riemannian Geometry
thanks to Christoffel
( z) Γ
i
rs
ip
B
⎡ ∂ ( z )Crp
( z)
⎢
2 ⎢⎣ ∂z s
:=
Necessary condition for existence of
∂y i
∂xα ∂x β
=
ρ
Γ
αβ
( x)
∂y i
∂x ρ
−1
B
=
C
(z)
(z)
z
Recall notation, for any patch
+
y ( x)
−
∂
( z )Csp
∂z
r
−
∂
( z )Crs
∂z p
⎤
⎥
⎥⎦
satisfying metric transformation rules is
r
s
∂
∂
y
y
i
Γ
rs
( y)
∂xα ∂x β
Roughly:
For RCG
( y) C ≡ I
⇒
i
Γ
( y ) rs ≡ 0
For LCG
( x) B ≡ I
⇒
ρ
Γ
( x ) αβ ≡ 0
Linear problem for
∂y ∂x
Quasilinear problem for
∂y ∂x
!!!!
RCG compatibility
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
ƒ
Riemann
Christoffel
Brothers Cosserat
….....
……
Shield
Deturck and Yang
Interesting associated facts, especially uniqueness question
• if two deformations have same RCG field, then they differ by rigid
deformation
¾
Reshetnyak (according to Ball and James)
– Inadequacy of single-well energy for prediction of microstructure with compatible elastic
deformation (Ball & James)
¾
Friesecke, Muller, James
• An invertible tensor field F may have nonvanishing curl even if its RCG
field (FTF) is compatible
Complete Integrability of Pfaff PDE (T.Y. Thomas, 1934)
Theorem: Consider PDE
∂wi
i
ψ
x
=
i = 1 to R ; α = 1 to n
(
)
α ( w( x), x)
α
∂x
ψ αi ∈ C1 ( Ω ) , Ω open connected subset of R × n
Suppose the integrability condition
i
i
∂ψ αi j ∂ψ αi ∂ψ β j ∂ψ β
ψβ + β =
ψα + α
j
j
∂w
∂x
∂w
∂x
holds in Ω . (motivated by equality of second partial derivs.)
Then for arbitrary ( w0 , x0 ) ∈ Ω , ∃ a unique local solution around x0 satisfying
w ( x0 ) = w0 . Therefore, solution allows R arbitrary constants to be specified.
RCG compatibility
Motivated by necessary condition, consider
∂y i
i
=
u
α
∂xα
∂uαi
=
β
∂x
( x) B
C γµ ⎡ ∂Cαµ ∂Cβµ ∂Cαβ ⎤
uγ = uγ
+ α − µ ⎥
⎢
β
( x)Γ
∂x
∂x ⎦
2 ⎣ ∂x
Integrability Condition (nice and 'separably' factored in x-dependent terms and
γ
αβ
i
i
u -dependent terms)
µ
µ
⎡
⎤
∂
∂
Γ
Γ
αβ
αρ
( x)
( x)
i
µ
γ
µ
γ
uµ ⎢
−
+ ( x ) Γ γρ ( x ) Γ αβ − ( x ) Γ γβ ( x ) Γ αρ ⎥ = 0
ρ
β
∂x
⎢⎣ ∂x
⎥⎦
∴ require Riemann-Christoffel curvature tensor to vanish. Guarantees existence of
u with arbitrarily specifiable value u0 at one point, and because of symmetry of Γ
in lower indices, of y.
Remains to be shown that uαi uβi = Cαβ
RCG Compatibility
Assign u ( x0 ) such that u T u ( x0 ) = C ( x0 ) .
Continuity ⇒ u invertible locally around x0 .
Define v = u −1 ; noting δ ij = viα Cαβ v βj ( x0 )
∂
α
β
β ρ
v
C
v
v
=
j vi
αβ j )
µ ( i
∂x
⎡ ∂Cρβ
α
α ⎤
⎢ ∂x µ − Cαβ ( x ) Γ ρµ − Cρα ( x ) Γ βµ ⎥ = 0 !!!!
⎣
⎦
covariant derivative of covariant metric tensor
Ricci: metric tensors covariantly constant
(merely smoothness, and defn. of Γ !!!)
T
α
β
∴δ ij = vi Cαβ v j
⎛ ∂y ⎞ ∂y
⇒ u u=C =⎜ ⎟
⎝ ∂x ⎠ ∂x
T
locally
Left Cauchy Green Compatibility
ƒ
2-D
• Blume (1989)
¾ Formulation based on Polar Decomposition (for both 2/3-D)
– Find a rotation tensor field
¾ Compatibility condition for 2-d problem is derived
¾ ‘Explicit’ characterization of the condition
¾ Uniqueness is analyzed
• Duda & Martins (1995)
¾ Plane case
¾ Polar Decomposition
¾ Analysis of possible cases; construction of the rotation field
¾ Insightful and detailed analysis of the uniqueness question
¾ Demonstration of nonuniqueness through constructive examples
systematically using Thomas
ƒ
3-D - open
• Acharya (1999)
¾ geometric formulation
¾ provides condition for local existence in 3-d
¾ Much can be done in ‘explicit’ characterization of the existence condition
LCG Compatibility
Consider necessary condition for existence of y i such that
∂y i ∂y j
ij
αβ
αβ
δ
x
B
x
holds.
(
and
B
x
constant)
=
=
(
)
(
)
(
)
α
α
( y)
( x)
∂x ∂x
Around arbitrary x0 , the map y is then locally invertible and so
∂Brp
∂Brp ∂xα
= α
s
∂y
∂x ∂y s
where ⎡⎣ ( y ) B −1 ⎤⎦ =: Brp = ( y )Crp
rp
ip
∂ ( z )Csp
∂ ( z )Crs ⎤
B
⎡ ∂ ( z )Crp
( z)
−1
i
Recall ( z ) B = ( z )C
+
−
⎢
⎥
s
r
( z ) Γ rs :=
2 ⎢⎣ ∂z
∂z
∂z p ⎥⎦
i
r
s
∂y i
ρ ∂y
i ∂y ∂y
= ( x ) Γ αβ ρ − ( y ) Γ rs α β
α
β
∂x ∂x
∂x
∂x ∂x
∂y i
B im ⎡ ∂Brm ∂y r ∂Bsm ∂y s ∂Brs ∂x ρ ∂y r ∂y s ⎤
ρ
∴ ( x ) Γ αβ ≡ 0, and α β = −
+ α
− ρ
⎢
β
α
β
2 ⎣ ∂x ∂x
∂x ∂x
∂x ∂x
∂x ∂y m ∂xα ∂x β ⎥⎦
Left Cauchy Green Compatibility
Governing PDE System
Original – nonlinear, first order system
Formulated as – Quasilinear, Pfaffian system
i
j
uρ uρ
= B
⎡ ( y) B⎤
⎣
⎦
ij
ij
( x ) =: [ B ] ( x )
ij
⎡ ( y ) B −1 ⎤ =: Brp = ( y )Crp
⎣
⎦ rp
∂y i
∂xα
= uαi
∂uαi
∂x β
B im
= −
2
Matrix inverse function
⎡ ∂Brm r
⎢ ∂x β uα
⎣
+
∂Bsm s
uβ
α
∂x
−
∂Brs ρ
r s ⎤
ℑm ( u ) uα uβ ⎥
ρ
∂x
⎦
Setup
Seek functions
wi ( x ) , i = 1, 2,… , R
that satisfy
∂wi
i
x
=
ψ
α = 1, 2,… , n
(
)
α ( w( x) , x)
α
∂x
(for definiteness think of domain of ψ αi to be open connected set of
Refer to domain of ψ αi as ( z , x )
Associate wi → uαi
⎛
∂uαi
i
= Aαβ ⎜ u,
β
⎜
∂x
⎝
( y) B
; so R = 3 × 3 = 9 ;
( x),
∂ ( y )C
⎞
( x ) ⎟⎟
∂x
⎠
;
n = 3.
∂y i
i
=
u
α
∂xα
R
×
n
)
Sufficient condition for local existence:
the completely-integrable situation
Hypothesis: Suppose
i
i
⎛ ∂Aαβ
∂Aαβ
∂Aαγi k ∂Aαγi ⎞
k
F ( u, x ) := ⎜ k Aµγ + γ − k Aµβ − µ ⎟ ( u , x ) ≡ 0 locally in ( u, x )
⎜ ∂u
⎟
x
u
x
∂
∂
∂
µ
µ
⎝
⎠
(seek symmetry in β ,γ for each i, α ⇒ 27 nonlinear algebraic equations)
(1)
If so, Thomas guarantees solution to u and therefore y with arbitrary data
at one x0 .
So - specify conditions on B field for when identity can be satisfied:
Unlike RCG case, F (1) is
cumbersome (downright scary!) nonlinear in u does not readily
SEPARABLY factorize into at least 1 solely x − dependent term
- need separability (seems to me) for identity with control only on x − dependent terms
- need (algebraic-geometric?) theorem for when this can happen
given field ( y ) B ( x ) and algebraic structure of array A.
Sufficient condition for local existence:
the completely-integrable situation
Initial data to match uu T ( x0 ) = ( y ) B ( x0 ) ∈ Pym can be constructed. Then local
diffeomorphism y around x0 satisfying ∂y ∂x = u exists (and u is invertible)
ip
⎛
B
x ⎡ ∂ ( y )Crp x ∂ ( y )Csp x ∂ ( y )Crs x ⎤ ⎞
( y)
i
⎟ ( y′ )
Now define ( y ) Γ rs ( y′ ) := ⎜
+
−
⎢
⎥
s
r
p
⎜
⎟
2
∂
y
∂
y
∂
y
⎢
⎥
⎣
⎦
⎝
⎠
(Notation: y −1 =: x) Then,
Since u invertible,
i
∂uα
i
r s
T
′
′
′
⎡
⎤
=
−
Γ
x
y
y
u
u
x
y
(
)
(
)
(
)
(
)
(
)
∂
[
]
∂
∂
y
y
α
β
⎛
⎞
β
( y ) rs
⎣
⎦
∂x
u −1 = 0 ⇒ ⎜ ⎟ = B
∂x
∂x ⎝ ∂x ⎠
∂
ij α β
−1
Define v = u , consider m ⎡⎣( B vi v j ) x ⎤⎦
∂y
ij
⎡
⎤ α β
B
∂
∂
( y)
ij α β
kj
i
ik
j
⎡ B vi v j ⎤⎦ = ⎢
+ ( y ) B ( y ) Γ km + ( y ) B ( y ) Γ km ⎥ vi v j = 0
m ⎣
m
∂y
⎢⎣ ∂y
⎥⎦
Ricci: covariant deriv. of
contravariant Metric tensor = 0
Sufficient condition for NOT completely integrable case
Let
F ( ) ( z , x ) ≠ 0 identically.
1
( F ( ) defines complete integrability condition)
1
⎛ R ∂F ( j ) i ∂F ( j ) ⎞
Define Fα
( z, x ) := ⎜⎜ ∑ i ψ α + α ⎟⎟ ( z, x )
∂x ⎠
⎝ i =1 ∂z
and consider two integers N , R with
1 ≤ N ≤ R ; 1 ≤ M ≤ R.
( j +1)
Assume that
there exist M equations in the sets F ( ) = 0
1
through
F(
N)
= 0 denoted by
Gλ = 0, λ = 1 to M ,
and M of the variables z i (from the list z i , i = 1 to R ) denoted by
z i , i = 1 to M identified through a known one-to-one map κ : {1, 2,… , M } → {1, 2,… R}
by z i := z
which satisfy
κ (i )
⎡ ∂Gλ
⎤
det ⎢ i ( z, x ) ⎥ ≠ 0 locally in ( z, x ) space.
⎣ ∂z
⎦
Not Completely Integrable Case, contd.
Denote remaining R − M =: P variables z as zˆ i , i = 1 to P, defined by
µ : {1, 2,… , P} → {1, 2,… , R}
µ i
zˆ i := z ( ) .
Assume that around a point ( zˆ0 , x0 ) := ( zˆ10 ,… , zˆ0P , x01 ,… , x0n ) , the solution
z i = ϕ i ( zˆ10 ,… , zˆ0P , x01 ,… , x0n ) , i = 1, 2,… , M , of Gλ = 0, λ = 1, 2,… , M
satisfies all the equations of the sets
F (1) = 0
through
F ( N +1) = 0
identically in a neighborhood of ( zˆ0 , x0 ) in ( zˆ, x ) space.
Then ∃ a function w ( x ) satisfying
∂wi
i
=
α = 1, 2,… , n, i = 1, 2,… , R
x
ψ
(
)
α (w( x), x)
α
∂x
determined by P constants.
Let Π ( z , zˆ, x )
Proof: Not completely Integrable case
( z, x ) . Define,
Gλ ( z , zˆ, x ) = Gλ ( Π ( z , zˆ, x ) ) , λ = 1, 2,… , M
ψ αi ( z , zˆ, x ) = ψ ακ (i ) ( Π ( z , zˆ, x ) ) , i = 1, 2,… , M , α = 1, 2,… , n
ψˆαj ( z , zˆ, x ) = ψ αµ ( j ) ( Π ( z , zˆ, x ) ) , j = 1, 2,… , P, α = 1, 2,… , n
N +1
Since F ( ) (ϕ ( zˆ, x ) , zˆ, x ) ≡ 0 in ( zˆ, x )
⎡ ∂Gλ i ∂Gλ j ∂Gλ ⎤
(I )
⎢ ∂z i ψ α + ∂zˆ j ψˆα + ∂xα ⎥ (ϕ ( zˆ, x ) , zˆ, x ) = 0
⎣
⎦
For arbitrarily fixed path f ( t ) with f ( 0 ) = x∗ ∃ solution to ODE
⎡ j
dg j
t
=
( ) ⎢ψˆα
dt
⎣
df α ⎤
(ϕ ( g , f ) , g , f ) dt ⎥ ( t ) ; g j ( 0 ) = zˆ∗ (arbitrary),
⎦
and g satisfies Gλ
(ϕ ( g , f ) , g , f ) ( t ) = 0 for t in some local interval around 0.
Proof: Not Completely Integrable Case
Differentiating w.r.t. t , combining with ( I ) , and using arbitrariness of path f ,
zˆ∗ , and x∗ yields,
ψ α (ϕ ( zˆ, x ) , zˆ, x ) −
i
locally in ( zˆ, x ) space.
∂ϕ i
j
ˆ
ˆ
ˆ, x ) , zˆ, x ) −
ψ
z
x
,
(
)
α (ϕ ( z
j
∂zˆ
∂ϕ i
zˆ, x ) ≡ 0
α (
∂x
( II )
F ( ) (ϕ ( zˆ, x ) , zˆ, x ) ≡ 0 combined with ( II ) implies that the conditions of
1
complete integrability for
∂g j
j
ˆ
=
ψ
x
(
)
( III )
α ϕ ( g ( x) , x), g ( x), x
∂xα
are satisfied. Thus ∃ solution to ( III ) with P arbitrary constants.
(
)
Then it can be shown that
( h, g , Id )( x )
uu T ( x ) = B ( x )
1
j = 1, 2,… , P
∂hi
κ i
satisfy α ( x ) = ψ ( )
∂x
accommodated by P
constants, stick in
in F ( ) . Then use
standard idea with Ricci
to show B compatibility.
hi ( x ) := ϕ i ( g ( x ) , x ) i = 1, 2,… , M
g j ( x)
If u0u0T = B ( x0 ) cannot be
∂g j
µ( j)
x
=
ψ
(
)
∂xα
( h, g , Id )( x ) .
Comments
ƒ Proof adapted from Eisenhart, 1927, Veblen and
Thomas, 1926
• They claim necessity as well; I couldn’t
ƒ Guess:
• Related to Cartan’s Method of Equivalence
• Modern treatment – R B Gardner CBMS-NSF-SIAM