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1.15 Tensor Calculus 2: Tensor Functions

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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: Solid Mechanics Part III 124 Kelly Section 1.15 ∂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 Solid Mechanics Part III 125 Kelly Section 1.15 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 dAij φ = (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} Solid Mechanics Part III 126 Kelly Section 1.15 ∂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 Solid Mechanics Part III 127 Kelly Section 1.15 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 Solid Mechanics Part III 128 Kelly Section 1.15 ■ 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 Solid Mechanics Part III 129 Kelly Section 1.15 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).
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