Tensor differentiation - classical tensoranalysiskintzel.net/ruhruni/pdf-files/Tensorvortrag.pdf ·...

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A short introduction into tensor algebraThe algebra of fourth-order tensors - a new tensor formalism

New rules for the tensor differentiation w.r.t. a second-order tensor

Tensor differentiation- classical tensoranalysis

Olaf Kintzel

GKSS ForschungsinstitutGeesthacht

21th May 2008

Olaf Kintzel Tensor differentiation


1 A short introduction into tensor algebra

2 The algebra of fourth-order tensors - a new tensor formalism

3 New rules for the tensor differentiation w.r.t. a second-order tensor



What is a tensor ?

Index notation Absolute notation

(scalar) a ∈ R a

(vector) ai ∈ V = R3 a

(tensor of 2. order) Aij ∈ V ×V A∈ Lin(V→ V)

(tensor of 4. order) AAAijkl ∈ V ×V ×V ×V A∈ Lin(V ×V→ V ×V)



What is a tensor ?


(scalar) a ∈ R a






What is a tensor ?


(scalar) a ∈ R a






What is a tensor ?


(scalar) a ∈ R a






What is a tensor ?


(scalar) a ∈ R a






What is a tensor ?


(scalar) a ∈ R a






What is a tensor ?


(scalar) a ∈ R a






Advantage of absolute notation

has a short and concise form.is invariant i.e. has the same expression (e.g. a, a, . . .) for anycoordinate system considered.

to ensure the invariance in index notation we have to use specialtransformation rules:

e.g. Θi → Θ̄j i.e. Θ̄i(Θj):

Āij = Akl∂Θk

∂Θ̄i∂Θl

∂Θ̄jor Āij = Akl

∂Θ̄i

∂Θk∂Θ̄j

∂Θl!







Āij = Akl∂Θk

∂Θ̄i∂Θl


∂Θ̄i

∂Θk∂Θ̄j

∂Θl!







Āij = Akl∂Θk

∂Θ̄i∂Θl


∂Θ̄i

∂Θk∂Θ̄j

∂Θl!







Āij = Akl∂Θk

∂Θ̄i∂Θl


∂Θ̄i

∂Θk∂Θ̄j

∂Θl!



In absolute notation we have to think of a tensor as a matrix with abasis attached to it:

v ∈ V means v = vi ei vi ∈ R3.The representation v ∈ V is unique irrespectively of any special basisconsidered.

If we want to compute the components for another basis (e.g. ēj) wehave to multiply (contract) with this basis:

v̄j = ēj · v = ēj · vi ei = vi (ēj · ei)Comparing this with the tensor transformation rules we immediatelysee that:

v̄j = vi∂Θi

∂Θ̄j→ ∂Θ

i

∂Θ̄j= ēj · ei







v̄j = vi∂Θi

∂Θ̄j→ ∂Θ

i








v̄j = vi∂Θi

∂Θ̄j→ ∂Θ

i








v̄j = vi∂Θi

∂Θ̄j→ ∂Θ

i




Considering a tensor of second order we have to use the dyadicproduct (⊗) to construct a tensor.E.g. A ∈ V ×V is defined by A = Aij ei ⊗ ejWe can compute the components by a left- and right-multiplicationwith the new basis considered:

E.g. Ākl = ēk ·A · ēl = Aij (ēk · ei) (ēl · ej)The operation (·) is called simple contraction.A double contraction is defined by the operation (:).For instance, we have:

A : B = Aij ei ⊗ ej : B̄klēk ⊗ ēl = AijB̄kl (ēk · ei)(ēl · ej)For the special case ēi = ei (ei · ej = δij) we get in particular:A : B = AijBij which is well-known!



The algebra of fourth-order tensors

For a tensor of fourth order we could have:

A = a⊗ b⊗ c⊗ d = ai bj ck dl ei ⊗ ej ⊗ ek ⊗ el

The double contraction is defined by:

From the right: A : B = a⊗ b⊗ c⊗ d : B = a⊗ b (c ·B · d)From the left: B : A = B : a⊗ b⊗ c⊗ d = (a ·B · b) c⊗ d

The double contraction of two fourth-order tensors is defined by:

A1 : A2 = a1 ⊗ b1 ⊗ c1 ⊗ d1 : a2 ⊗ b2 ⊗ c2 ⊗ d2= (c1 · a2)(d1 · b2) a1 ⊗ b1 ⊗ c2 ⊗ d2

Obviously, the double contraction commutes:(A : B) : C = A : (B : C) !



Now we make the following definitions:

A q aB = a⊗ b⊗ c⊗ d q aB = (b ·B · c) a⊗ dA a qB = a⊗ b⊗ c⊗ d a qB = (a ·B · d) b⊗ cThe double contraction of two fourth-order tensor is defined by:

A1 q aA2 = a1 ⊗ b1 ⊗ c1 ⊗ d1 q aa2 ⊗ b2 ⊗ c2 ⊗ d2= (b1 · a2)(c1 · d2) a1 ⊗ b2 ⊗ c2 ⊗ d1

A1 a qA2 = a1 ⊗ b1 ⊗ c1 ⊗ d1 a qa2 ⊗ b2 ⊗ c2 ⊗ d2= (a1 · b2)(d1 · c2) a2 ⊗ b1 ⊗ c1 ⊗ d2

As can be proved, these rules are also associative i.e.:(A q aB) q aC = A q a(B q aC) and (A a qB) a qC = A a q(B a qC)To put it short:( q) stands for the inner two basis vectors and ( a) for the outer twobasis vectors of a fourth-order tensor.Obviously, this distinction is meaningless for two-order tensors!



To construct a fourth-order tensor from two second-order tensors, wedefine three different tensor products:

A = A⊗B = a1 ⊗ a2 ⊗ b1 ⊗ b2 (Standard!)A = A×B = a1 ⊗ b1 ⊗ b2 ⊗ a2 (New!)A = A �× B = a1 ⊗ b2 ⊗ a2 ⊗ b2 (New!)

Sometimes, other notations are used, e.g.:

A×B = A ⊗B and A �× B = A ⊗Be.g. used in the STEINMANN-group.

But we think that the selected notation is better to distinguish!



Summary

Hence, we have 2 different systems of rules:

The classical rule: A : B, Here: A : A−1 = I �× Ii.e. the identity element is I �× I.

The new rules: A q aB, A a qB, Here: A q aA−1 = I⊗ IA a qA−1 = I⊗ I

i.e. the identity element is I⊗ I.



Transposition rules

In close connection to the three rules of double contraction, we havein fact two different systems:

The classical system: Consider for convenience A = A⊗B.Here, we can define four different transposition rules:Adl = AT ⊗B, Adr = A⊗BT , Ad = AT ⊗BT , AD = B⊗A

The new system: Consider for convenience A = A×B.Here, we can define four different transposition rules:Ato = AT ×B, Ati = A×BT , At = AT ×BT , AT = B×A



Transposition rules






Transposition rules






(A⊗B)T = AT ⊗BT (A⊗B)D = B⊗A(A×B)T = B×A (A×B)D = BT ×AT(A 2× B)T = B 2× A (A 2× B)D = AT 2× BT

(A⊗B)ti = A 2× B (A⊗B)dl = AT ⊗B(A×B)ti = A×BT (A×B)dl = B 2× A(A 2× B)ti = A⊗B (A 2× B)dl = B×A

(A⊗B)to = BT 2× AT (A⊗B)dr = A⊗BT(A×B)to = AT ×B (A×B)dr = A 2× B(A 2× B)to = BT ⊗AT (A 2× B)dr = A×B

(A⊗B)t = BT ⊗AT (A⊗B)d = AT ⊗BT(A×B)t = AT ×BT (A×B)d = B×A(A 2× B)t = BT 2× AT (A 2× B)d = B 2× A

Tabelle 1: Transpositionsoperationen angewendet auf Tensoren vierter Stufe.

1

Transposition operations



Symmetry properties

A fourth-order tensor A has minor symmetry if:

A = Adl and A = Adr or A = Ati and A = Ato

A fourth-order tensor A has major symmetry if:

A = AD or A = AT

A fourth-order tensor A is supersymmetric if:

it has minor and major symmetry!



(A⊗B)C = A⊗ (BC) C (A⊗B) = (CA)⊗B(A×B)C = (AC)×B C (A×B) = (CA)×B(A 2× B)C = A 2× (BC) C (A 2× B) = (CA) 2× B

Simple contraction of fourth- and second-order tensors



(A⊗B) : C = A (B : C) C : (A⊗B) = (C : A)B(A⊗B) q aC = ACB C q a(A⊗B) = AT CBT(A⊗B) a qC = AT CBT C a q(A⊗B) = ACB(A×B) : C = ACT BT C : (A×B) = BT CT A(A×B) q aC = (B : C)A C q a(A×B) = (A : C)B(A×B) a qC = (A : C)B C a q(A×B) = (C : B)A(A 2× B) : C = ACBT C : (A 2× B) = AT CB(A 2× B) q aC = ACT B C q a(A 2× B) = BCT A(A 2× B) a qC = BCT A C a q(A 2× B) = ACT B

Double contraction of fourth- and second-order tensors



(A⊗B) : (C⊗D) = (B : C)A⊗D(A⊗B) r b(C⊗D) = (AC)⊗ (DB)(A⊗B) b r(C⊗D) = (CA)⊗ (BD)(A⊗B) : (C×D) = A⊗ (DT BT C) (A×B) : (C⊗D) = (ACT BT )⊗D(A⊗B) r b(C×D) = (ACB)×D (A×B) r b(C⊗D) = A× (CT BDT )(A⊗B) b r(C×D) = C× (AT DBT ) (A×B) b r(C⊗D) = (CAD)×B(A×B) : (C×D) = (AD) 2× (BC)(A×B) r b(C×D) = (B : C)A×D(A×B) b r(C×D) = (A : D)C×B(A⊗B) : (C 2× D) = A⊗ (CT BD) (A 2× B) : (C⊗D) = (ACBT )⊗D(A⊗B) r b(C 2× D) = (AC) 2× (DB) (A 2× B) r b(C⊗D) = (ADT ) 2× (CT B)(A⊗B) b r(C 2× D) = (CBT ) 2× (AT D) (A 2× B) b r(C⊗D) = (CA) 2× (BD)(A 2× B) : (C 2× D) = (AC) 2× (BD)(A 2× B) r b(C 2× D) = (ADT )⊗ (CT B)(A 2× B) b r(C 2× D) = (CBT )⊗ (AT D)(A×B) : (C 2× D) = (AD)× (BC) (A 2× B) : (C×D) = (AC)× (BD)(A×B) r b(C 2× D) = A× (DBT C) (A 2× B) r b(C×D) = (ACT B)×D(A×B) b r(C 2× D) = (CAT D)×B (A 2× B) b r(C×D) = C× (BDT A)

Tabelle 1: Doppelte Überschiebungen von Tensoren vierter Stufe.

1

Double contraction of fourth-order tensorsOlaf Kintzel Tensor differentiation


I d e n t i t y t e n s o r s o f 4 . o r d e r

( I I ) ( I I ) ( I I )




( I I ) : A = A

( I I ) A = A

( I I ) A = A

( I I ) ( I I ) ( I I )




( I I ) : A = A

( I I ) A = A

( I I ) A = A

( I I ) : A = A T

( I I ) A = A T

( I I ) A = A T

( I I ) ( I I ) ( I I )




( I I ) : A = A

( I I ) A = A

( I I ) A = A

( I I ) : A = A T

( I I ) A = A T

( I I ) A = A T

( I I ) : A = t r ( A )

( I I ) A = t r ( A )

( I I ) A = t r ( A )

( I I ) ( I I ) ( I I )



Tensor differentiation

The differentiation of a tensor function with respect to a second-ordertensor plays an important role for the modeling of materialse.g. for the construction of tangent operators.

As example we will consider the exponential function:

Example: F(A) = exp(A) = I + A + 12! A2 + 13! A

3 + . . .

Normally, we make a small pertubation � of A in the direction of X.

Thus, we speak of a directional derivative (GATEAUX-derivative).

The linear term with respect to � represents the derivative of F(A)with respect to A:

∂ F(A + �X)∂ �

∣∣∣�=0

= F(A),A : X







3 + . . .




∂ F(A + �X)∂ �

∣∣∣�=0

= F(A),A : X







3 + . . .




∂ F(A + �X)∂ �

∣∣∣�=0

= F(A),A : X







3 + . . .




∂ F(A + �X)∂ �

∣∣∣�=0

= F(A),A : X







3 + . . .




∂ F(A + �X)∂ �

∣∣∣�=0

= F(A),A : X



If we use this method for the exponential function, we get:

F(A),A : X =∂(I+A+�X+ 12! (A

2+� (AX+XA)+�2 X2)+...)

∂�

∣∣∣�=0

= (X + 12! (AX + XA) +13! (A

2X + AXA + XA2) + . . .

= F(A),A : X

By an extraction of X from this relation we can finally derive thederivative F(A),A.

Now we want to present an alternative procedure which is mucheasier to use.




F(A),A : X =∂(I+A+�X+ 12! (A

2+� (AX+XA)+�2 X2)+...)

∂�

∣∣∣�=0

= (X + 12! (AX + XA) +13! (A

2X + AXA + XA2) + . . .

= F(A),A : X






F(A),A : X =∂(I+A+�X+ 12! (A

2+� (AX+XA)+�2 X2)+...)

∂�

∣∣∣�=0

= (X + 12! (AX + XA) +13! (A

2X + AXA + XA2) + . . .

= F(A),A : X





In index notation the derivative of the components Aik with respect tothe components Ajl is given by:

∂ Aik∂Ajl

= δij δlk

The crucial step is now that we transform this relation into absolutenotation by placing the basis of Ajl at the center:

∂A∂A

= δij δlk ei ⊗ ej ⊗ el ⊗ ek = I⊗ I.

By placing the basis of the denominator at the center of thefourth-order tensor the product rule of differentiation is fullfilled:

(C(A)D(A)),A q a∆A = (C,A D + CD,A ) q a∆A.Furthermore, the chain rule of differentiation is also fullfilled.

A main presumption is that we have to use the new rules ( q a, a q).Olaf Kintzel Tensor differentiation



∂ Aik∂Ajl

= δij δlk


∂A∂A







∂ Aik∂Ajl

= δij δlk


∂A∂A







∂ Aik∂Ajl

= δij δlk


∂A∂A







∂ Aik∂Ajl

= δij δlk


∂A∂A






From the relation A,A = I⊗ I we are able to obtain the derivative of apower of A in an straightforward manner e.g. considering A2:

A2,A = A,A A + AA,A = (I⊗ I)A + A(I⊗ I) = I⊗A + A⊗ Iby means of the product rule and the simple contraction rules!

Or the derivative of the inverse of A:

A−1A = I⇒ A−1,AA + A−1A,A = 0⇔ A−1,AA = −A−1(I⊗ I) = −A−1 ⊗ I⇔ A−1,A = −A−1 ⊗A−1

In full analogy we can compute A,A−1 :

(A−1A),A−1 = 0⇒ A,A−1 = −A⊗A








(A−1A),A−1 = 0⇒ A,A−1 = −A⊗A








(A−1A),A−1 = 0⇒ A,A−1 = −A⊗A



The derivative of the transpose of A is computed simply by using thetransposition rules:

AT,A = (A,A )to = (I⊗ I)to = I �× I

A,AT = (A,A )ti = (I⊗ I)ti = I �× I

We can obtain the derivative of A−T with respect to A as follows:

A−T,A−1 q aA−1,A = (I �× I) q a(−A−1 ⊗A−1) = −A−1 �× A−1by means of the chain rule of differentiation.

For symmetric (skew-symmetric) tensors we can use the chain rule:

F(A),A∣∣∣A=AT

= F(A),A q a 12(A,A +A,AT )︸︷︷︸S

= F(A),A q aSF(A),A

∣∣∣A=−AT

= F(A),A q a 12(A,A−A,AT )︸︷︷︸A

= F(A),A q aAOlaf Kintzel Tensor differentiation



AT,A = (A,A )to = (I⊗ I)to = I �× I

A,AT = (A,A )ti = (I⊗ I)ti = I �× I




F(A),A∣∣∣A=AT

= F(A),A q a 12(A,A +A,AT )︸︷︷︸S

= F(A),A q aSF(A),A

∣∣∣A=−AT

= F(A),A q a 12(A,A−A,AT )︸︷︷︸A




AT,A = (A,A )to = (I⊗ I)to = I �× I

A,AT = (A,A )ti = (I⊗ I)ti = I �× I




F(A),A∣∣∣A=AT

= F(A),A q a 12(A,A +A,AT )︸︷︷︸S

= F(A),A q aSF(A),A

∣∣∣A=−AT

= F(A),A q a 12(A,A−A,AT )︸︷︷︸A




AT,A = (A,A )to = (I⊗ I)to = I �× I

A,AT = (A,A )ti = (I⊗ I)ti = I �× I




F(A),A∣∣∣A=AT

= F(A),A q a 12(A,A +A,AT )︸︷︷︸S

= F(A),A q aSF(A),A

∣∣∣A=−AT

= F(A),A q a 12(A,A−A,AT )︸︷︷︸A




AT,A = (A,A )to = (I⊗ I)to = I �× I

A,AT = (A,A )ti = (I⊗ I)ti = I �× I




F(A),A∣∣∣A=AT

= F(A),A q a 12(A,A +A,AT )︸︷︷︸S

= F(A),A q aSF(A),A

∣∣∣A=−AT

= F(A),A q a 12(A,A−A,AT )︸︷︷︸A



Let us now take a look at the derivative of the exponential function:

exp(A),A =∂(I + A + 12! A

2 + 13! A3 + . . .)

∂A

= I⊗I+ 12! (A⊗I+I⊗A)+ 13! (A2⊗I+A⊗A+I⊗A2)+ . . .

If A is symmetric, we actually have:

exp(A),A = 12 (I⊗ I+ I �× I+ 12! (A⊗ I+A �× I+ I⊗A+ I �× A) + . . .)



The differentiation of a trace of A

At first, we must consider that a trace of A is already contracted.

Therefore, we must use the new contraction rule ( a q).Example: F (A) = tr(A2) = A2 : I = A2 a qIThe last relation is very useful, because now we are able to computethe derivative (tr(A2)),A very easily:

F (A),A = (A2,A) a qI = (A⊗ I + I⊗A) a qI = 2 ATby means of the double contraction rules.

For a combination of scalar-valued and second-order tensor functionswe simply have:

(F (A) F(A)),A = F (A) (F(A),A ) + F(A)× (F (A),A )by means of the product rule of differentiation.



Let us conclude with a more complex example:

Given is the tensor function:

F(A) = A−1 tr(A2) + AT tr(A3)

The derivative of F(A) with respect to A reads as:

F(A),A =(−A−1 ⊗A−1) tr(A2) + A−1 × 2 AT + (I �× I) tr(A3) + AT × 3 A2T

Conclusion

The new tensor differentiation rules are easy to use and very efficient!

For further informations please see:Kintzel & Başar 2006, ZAMM 86(4), pp. 291-311

Thank you for your kind attention!!Olaf Kintzel Tensor differentiation




F(A) = A−1 tr(A2) + AT tr(A3)



Conclusion







F(A) = A−1 tr(A2) + AT tr(A3)



Conclusion







F(A) = A−1 tr(A2) + AT tr(A3)



Conclusion







F(A) = A−1 tr(A2) + AT tr(A3)



Conclusion




A short introduction into tensor algebraThe algebra of fourth-order tensors - a new tensor formalismNew rules for the tensor differentiation w.r.t. a second-order tensor

Tensor differentiation - classical tensoranalysiskintzel.net/ruhruni/pdf-files/Tensorvortrag.pdf ·...

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Transcript of Tensor differentiation - classical tensoranalysiskintzel.net/ruhruni/pdf-files/Tensorvortrag.pdf ·...