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1.15 Tensor Calculus 2: Tensor Functions

1.15 Tensor Calculus 2: Tensor Functions

Section 1.15

1.15 2: Tensor Functions

1.15.1 Vector-valued functions of a vector

Consider a vector-valued of a vector

a = a(b), ai = ai (b j )

This is a function of three independent variables b1 , b2 , b3 , and there are nine partial ∂ai / ∂b j . The partial of the vector a with respect to b is defined to be a second-order tensor with these partial derivatives as its components:

∂a(b) ∂ai ≡ ei ⊗ e j (1.15.1) ∂b ∂b j

It follows from this that

−1 ∂a  ∂b  ∂a ∂b ∂ai ∂bm =   or = I, = δ ij (1.15.2) ∂b  ∂a  ∂b ∂a ∂bm ∂a j

To show this, with ai = ai (b j ), bi = bi (a j ) , note that the differential can be written as

∂a ∂a ∂b j  ∂a ∂b j   ∂a ∂b j   ∂a ∂b j  da = 1 db = 1 da = da  1  + da  1  + da  1  1 ∂ j ∂ ∂ i 1  ∂ ∂  2  ∂ ∂  3  ∂ ∂  b j b j ai  b j a1   b j a2   b j a3 

Since da1 ,da2 ,da3 are independent, one may set da2 = da3 = 0 , so that

∂a ∂b j 1 = 1 ∂b j ∂a1

Similarly, the terms inside the other brackets are zero and, in this way, one finds Eqn. 1.15.2.

1.15.2 -valued functions of a tensor

Consider a scalar valued function of a (second-order) tensor

φ = φ(T), T = T e ⊗ e ij i j .

This is a function of nine independent variables, φ = φ(Tij ) , so there are nine different partial derivatives:

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∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ ∂φ , , , , , , , , ∂T11 ∂T12 ∂T13 ∂T21 ∂T22 ∂T23 ∂T31 ∂T32 ∂T33

The of φ with respect to T is defined to be a second-order tensor with these partial derivatives as its components:

∂φ ∂φ ≡ ei ⊗ e j Partial Derivative with respect to a Tensor (1.15.3) ∂T ∂Tij

The quantity ∂φ(T) / ∂T is also called the of φ with respect to T.

Thus differentiation with respect to a second-order tensor raises the order by 2. This agrees with the idea of the gradient of a scalar where differentiation with respect to a vector raises the order by 1.

Derivatives of the Trace and Invariants

Consider now the trace: the derivative of trA , with respect to A can be evaluated as follows:

∂ ∂A ∂A ∂A trA = 11 + 22 + 33 ∂A ∂A ∂A ∂A

∂A11 ∂A22 ∂A33 = ei ⊗ e j + ei ⊗ e j + ei ⊗ e j (1.15.4) ∂Aij ∂Aij ∂Aij

= e1 ⊗ e1 + e 2 ⊗ e 2 + e3 ⊗ e3 = I

Similarly, one finds that {▲Problem 1}

2 3 ∂(trA) ∂(trA ) ∂(trA ) T = I = 2A T = 3(A 2 ) ∂A ∂A ∂A (1.15.5) ∂((trA) 2 ) ∂((trA)3 ) = 2(trA)I = 3(trA) 2 I ∂A ∂A Derivatives of Trace Functions

From these and 1.11.17, one can evaluate the derivatives of the invariants {▲Problem 2}:

∂I A = I ∂A ∂II A = I I − A T Derivatives of the Invariants (1.15.6) ∂A A ∂III 2 A = (A T ) − I A T + II I = III A −T ∂A A A A

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Derivative of the

An important relation is

∂ (det A) = (det A)A −T (1.15.7) ∂A which follows directly from 1.15.6c.

Other Relations

The total differential can be written as

∂φ ∂φ ∂φ dφ = dT + dT + dT + ∂T 11 ∂T 12 ∂T 13 11 12 13 (1.15.8) ∂φ ≡ : dT ∂T

This total differential gives an approximation to the total increment in φ when the increments of the independent variables T11 , are small.

The second partial derivative is defined similarly:

∂∂22φφ ≡eei ⊗⊗⊗ j e pq e, (1.15.9) ∂∂TT ∂TTij ∂ pq the result being in this case a fourth-order tensor.

Consider a scalar-valued function of a tensor, φ(A) , but now suppose that the components of A depend upon some scalar parameter t: φ = φ(A(t)) . By means of the of differentiation,

∂φ dA φ = ij (1.15.10) ∂Aij dt which in symbolic notation reads (see Eqn. 1.10.10e)

T dφ ∂φ dA  ∂φ  dA = : = tr    (1.15.11) dt ∂A dt  ∂A  dt 

Identities for Scalar-valued functions of Functions

Let C be a symmetric tensor, C = CT . Then the partial derivative of φ = φ(C(T)) with respect to T can be written as {▲Problem 3}

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∂φ ∂φ (1) = 2T for C = TT T ∂T ∂C

∂φ ∂φ (2) = 2 C for C = TTT (1.15.12) ∂T ∂T

∂φ ∂φ ∂φ ∂φ ∂φ (3) = 2T = 2 T = T + T for C = TT and symmetric T ∂T ∂C ∂C ∂C ∂C

Scalar-valued functions of a Symmetric Tensor

Consider the expression

∂φ(A) ∂φ(Aij ) B = Bij = (1.15.13) ∂A ∂Aij

If A is a symmetric tensor, there are a number of ways to consider this expression: two possibilities are that φ can be considered to be

(i) a symmetric function of the 9 variables Aij

(ii) a function of 6 independent variables: φ = φ(A11 , A12 , A13 , A22 , A23 , A33 ) where 1 A = (A + A ) = A = A 12 2 12 21 12 21 1 A = (A + A ) = A = A 13 2 13 31 13 31 1 A = (A + A ) = A = A 23 2 23 32 23 32

Looking at (i) and writing φ = φ(A11 , A12 (A12 ),, A21 (A12 ),), one has, for example,

∂φ ∂φ ∂A ∂φ ∂A ∂φ ∂φ ∂φ = 12 + 21 = + = 2 , ∂A12 ∂A12 ∂A12 ∂A21 ∂A12 ∂A12 ∂A21 ∂A12

the last equality following from the fact that φ is a symmetrical function of the Aij .

Thus, depending on how the scalar function is presented, one could write

∂φ ∂φ ∂φ (i) B11 = , B12 = , B13 = , etc. ∂A11 ∂A12 ∂A13 ∂φ 1 ∂φ 1 ∂φ (ii) B11 = , B12 = , B13 = , etc. ∂A11 2 ∂A12 2 ∂A13

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1.15.3 Tensor-valued functions of a tensor

The derivative of a (second-order) tensor A with respect to another tensor B is defined as

∂A ∂Aij ≡ ei ⊗ e j ⊗ e p ⊗ eq (1.15.14) ∂B ∂B pq and forms therefore a fourth-order tensor. The total differential dA can in this case be written as

∂A dA = : dB (1.15.15) ∂B

Consider now

∂A ∂Aij = ei ⊗ e j ⊗ e k ⊗ el ∂A ∂Akl

The components of the tensor are independent, so

∂A11 ∂A11 ∂Amn = 1, = 0,  etc. = δ mpδ nq (1.15.16) ∂A11 ∂A12 ∂Apq and so

∂A = e ⊗ e ⊗ e ⊗ e = I , (1.15.17) ∂A i j i j the fourth-order identity tensor of Eqn. 1.12.4.

Example

Consider the scalar-valued function φ of the tensor A and vector v (the “dot” can be omitted from the following and similar expressions),

φ(A, v) = v ⋅ Av

The gradient of φ with respect to v is

∂φ ∂v ∂v = ⋅ Av + v ⋅ A = Av + vA = (A + A T )v ∂v ∂v ∂v

On the other hand, the gradient of φ with respect to A is

∂φ ∂A = v ⋅ v = v ⋅ Iv = v ⊗ v ∂A ∂A

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Consider now the derivative of the inverse, ∂A −1 / ∂A . One can differentiate A −1A = 0 using the to arrive at

−1 ∂A − ∂A A = −A 1 ∂A ∂A

One needs to be careful with derivatives because of the position of the indices in 1.15.14); it looks like a post-operation of both sides with the inverse leads to −1 −1 −1 −1 −1 ∂A / ∂A = −A (∂A / ∂A)A = −Aik Ajl ei ⊗ e j ⊗ e k ⊗ el . However, this is not correct (unless A is symmetric). Using the (there is no clear symbolic notation), one has

−1 ∂ ∂Aim −1 Amj Amj = −Aim (ei ⊗ e j ⊗ e k ⊗ el ) ∂Akl ∂Akl −1 ∂ ∂Aim −1 −1 Amj −1 → Amj Ajn = −Aim Ajn ∂Akl ∂Akl −1 (1.15.18) ∂Aim −1 −1 → δ mn = −Aim δ mkδ jl Ajn ∂Akl ∂ −1 Aij −1 −1 → = −Aik Al j (ei ⊗ e j ⊗ e k ⊗ el ) ∂Akl ■

1.15.4 The

The directional derivative was introduced in §1.6.11. The ideas introduced there can be extended to . For example, the directional derivative of the trace of a tensor A, in the direction of a tensor T, is

d d ∂ A (trA)[T] = tr(A + εT) = (trA + εtrT) = trT (1.15.19) dε ε =0 dε ε =0

As a further example, consider the scalar function φ(A) = u ⋅ Av , where u and v are constant vectors. Then

d ∂ Aφ(A,u, v)[T] = [u ⋅ (A + εT)v] = u ⋅ Tv (1.15.20) dε ε =0

Also, the gradient of φ with respect to A is

∂φ ∂ = (u ⋅ Av) = u ⊗ v (1.15.21) ∂A ∂A

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and it can be seen that this is an example of the more general relation

∂φ ∂ φ[T] = : T (1.15.22) A ∂A which is analogous to 1.6.41. Indeed,

∂φ ∂ φ[w] = ⋅ w x ∂x ∂φ ∂ φ[T] = : T (1.15.23) A ∂A ∂v ∂ v[w] = w u ∂u

Example (the Directional Derivative of the Determinant)

It was shown in §1.6.11 that the directional derivative of the determinant of the 2× 2 A, in the direction of a second matrix T, is

∂ A (det A)[T] = A11T22 + A22T11 − A12T21 − A21T12

This can be seen to be equal to det A (A −T : T), which will now be proved more generally for tensors A and T:

d ∂ A (det A)[T] = det(A + εT) dε ε =0 d = det[A(I + εA −1T)] dε ε =0 d = det A det(I + εA −1T) dε ε =0

The last line here follows from (1.10.16a). Now the characteristic equation for a tensor B is given by (1.11.4, 1.11.5),

(λ1 − λ)(λ2 − λ)(λ3 − λ) = 0 = det(B − λI)

−1 where λi are the three eigenvalues of B. Thus, setting λ = −1 and B = εA T ,

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d ∂ ( ) = + λ −1 + λ −1 + λ −1 A det A [T] det A (1 1 εA T )(1 2 εA T )(1 3 εA T ) dε ε =0 d = + ε λ + ε λ + ε λ det A (1 1 A−1T )(1 2 A−1T )(1 3 A−1T ) dε ε =0 = λ + λ + λ det A ( 1 A−1T 2 A−1T 3 A−1T ) = det A tr(A −1T) and, from (1.10.10e),

−T ∂ A (det A)[T] = det A(A : T) (1.15.24) ■

Example (the Directional Derivative of a vector function)

Consider the n homogeneous algebraic equations f (x) = o :

f1 (x1 , x2 ,, xn ) = 0 f (x , x ,, x ) = 0 2 1 2 n 

f n (x1 , x2 ,, xn ) = 0

The directional derivative of f in the direction of some vector u is

d ∂ xf (x)[u] = f (z(ε )) (z = x + εu) dε ε =0  ∂f (z) dz  =   (1.15.25)  ∂z dε ε =0 = Ku where K, called the matrix of the system, is

∂f1 / ∂x1 ∂f1 / ∂x2  ∂f1 / ∂xn    ∂f ∂f 2 / ∂x1 ∂f 2 / ∂x2 ∂f 2 / ∂xn K = =   , ∂ f [u] = (gradf )u ∂x     x   ∂f n / ∂x1  ∂f n / ∂xn  which can be compared to (1.15.23c). ■

Properties of the Directional Derivative

The directional derivative is a linear and so one can apply the usual product rule. For example, consider the directional derivative of A −1 in the direction of T:

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−1 d −1 ∂ A (A )[T] = (A + εT) dε ε =0

−1 To evaluate this, note that ∂ A (A A)[T] = ∂ A (I)[T] = 0 , since I is independent of A. The −1 −1 product rule then gives ∂ A (A )[T]A = −A ∂ A (A)[T], so that

−1 −1 −1 −1 −1 ∂ A (A )[T] = −A ∂ A A[T]A = −A TA (1.15.26)

Another important property of the directional derivative is the chain rule, which can be applied when the function is of the form f (x) = fˆ(B(x)). To derive this rule, consider (see §1.6.11)

f (x + u) ≈ f (x) + ∂ xf[u], (1.15.27) where terms of order o(u) have been neglected, i.e.

o(u) lim = 0 . u →0 u

The left-hand side of the previous expression can also be written as

fˆ(B(x + u)) ≈ fˆ(B(x) + ∂ B[u]) x ˆ ˆ ≈ f (B(x))+ ∂ Bf (B)[∂ xB[u]]

Comparing these expressions, one arrives at the chain rule,

ˆ ∂ xf[u] = ∂ Bf (B)[∂ xB[u]] Chain Rule (1.15.28)

As an application of this rule, consider the directional derivative of det A −1 in the direction T; here, f is det A −1 and fˆ = fˆ(B(A)). Let B = A −1 and fˆ = det B . Then, from Eqns. 1.15.24, 1.15.25, 1.10.3h, f,

−1 −1 ∂ A (det A )[T] = ∂ B (det B)[∂ A A [T]] = (det B)(B −T : (− A −1TA −1 )) (1.15.29) = −det A −1 (A T : (A −1TA −1 )) = −det A −1 (A −T : T)

1.15.5 Formal Treatment of

Following on from §1.6.12 and §1.14.6, a scalar function f :V 2 → R is differentiable at A ∈V 2 if there exists a second order tensor Df (A)∈V 2 such that

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f (A + H) = f (A)+ Df (A): H + o( H ) for all H ∈V 2 (1.15.30)

In that case, the tensor Df (A) is called the derivative of f at A. It follows from this that Df (A) is that tensor for which

d 2 ∂ A f [B] = Df (A): B = f (A + εB) for all B ∈V (1.15.31) dε ε =0

For example, from 1.15.24,

−T −T ∂ A (det A)[T] = det A(A : T) = (det A A ): T (1.15.32) from which it follows, from 1.15.31, that

∂ det A = det A A −T (1.15.33) ∂A which is 1.15.7.

Similarly, a tensor-valued function T :V 2 → V 2 is differentiable at A ∈V 2 if there exists a fourth order tensor DT(A)∈V 4 such that

T(A + H) = T(A)+ DT(A)H + o( H ) for all H ∈V 2 (1.15.34)

In that case, the tensor DT(A) is called the derivative of T at A. It follows from this that DT(A) is that tensor for which

d 2 ∂ A T[B] = DT(A): B = T(A + εB) for all B ∈V (1.15.35) dε ε =0

1.15.6 Problems

1. Evaluate the derivatives (use the chain rule for the last two of these) ∂(trA 2 ) ∂(trA 3 ) ∂((trA) 2 ) ∂((trA) 2 ) , , , ∂A ∂A ∂A ∂A 2. Derive the derivatives of the invariants, Eqn. 1.15.5. [Hint: use the Cayley-Hamilton theorem, Eqn. 1.11.15, to express the derivative of the third in terms of the third invariant.] 3. (a) Consider the scalar valued function φ = φ(C(F)), where C = F T F . Use the chain rule

∂φ ∂φ ∂Cmn = ei ⊗ e j ∂F ∂Cmn ∂Fij to show that

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∂φ ∂φ ∂φ ∂φ = 2F , = 2Fik ∂F ∂C ∂Fij ∂Ckj (b) Show also that ∂φ ∂φ ∂φ = 2U = 2 U ∂U ∂C ∂C for C = UU with U symmetric.

[Hint: for (a), use the index notation: first evaluate ∂Cmn / ∂Fij using the product rule,

then evaluate ∂φ / ∂Fij using the fact that C is symmetric.] 4. Show that −1 −1 ∂A − − ∂A − − − (a) : B = −A 1BA 1 , (b) : A ⊗ A 1 = −A 1 ⊗ A 1 ∂A ∂A 5. Show that ∂A T : B = B T ∂A 6. By writing the of a tensor A , 1.10.14, where A is symmetric, in terms of the trace (see 1.10.10), show that ∂ A A = ∂A A 7. Evaluate 2 (i) ∂ A (A )[T] 2 (ii) ∂ A (trA )[T] (see 1.10.10e) 8. Derive 1.15.29 by using the definition of the directional derivative and the relation 1.15.7, ∂(det A)/ ∂A = (det A)A −T .

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