1.15 Tensor Calculus 2: Tensor Functions
Section 1.15
1.15 Tensor Calculus 2: Tensor Functions
1.15.1 Vector-valued functions of a vector
Consider a vector-valued function 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 derivatives ∂ai / ∂b j . The partial derivative 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 Scalar-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 partial derivative 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 gradient 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 field 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 Determinant
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 chain rule 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 Symmetric Tensor 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 product rule 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 index notation (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 Directional Derivative
The directional derivative was introduced in §1.6.11. The ideas introduced there can be extended to tensors. 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 matrix 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 tangent 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 operator 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 Tensor Calculus
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 invariant 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 norm 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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