Tensor derivative (continuum mechanics)

The directional derivative provides a systematic way of finding these derivatives.

Derivatives with respect to vectors and second-order tensors

The definitions of directional derivatives for various situations are given below. It is assumed that the functions are sufficiently smooth that derivatives can be taken.

Derivatives of scalar valued functions of vectors

Let f(v) be a real valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the vector defined through its dot product with any vector u being

$\frac{\partial f}{\partial v} \cdot u = D f (v) [u] = {[\frac{d}{d α} f (v + α u)]}_{α = 0}$

for all vectors u. The above dot product yields a scalar, and if u is a unit vector gives the directional derivative of f at v, in the u direction.

Properties:

If $f (v) = f_{1} (v) + f_{2} (v)$ then $\frac{\partial f}{\partial v} \cdot u = (\frac{\partial f_{1}}{\partial v} + \frac{\partial f_{2}}{\partial v}) \cdot u$
If $f (v) = f_{1} (v) f_{2} (v)$ then $\frac{\partial f}{\partial v} \cdot u = (\frac{\partial f_{1}}{\partial v} \cdot u) f_{2} (v) + f_{1} (v) (\frac{\partial f_{2}}{\partial v} \cdot u)$
If $f (v) = f_{1} (f_{2} (v))$ then $\frac{\partial f}{\partial v} \cdot u = \frac{\partial f_{1}}{\partial f_{2}} \frac{\partial f_{2}}{\partial v} \cdot u$

Derivatives of vector valued functions of vectors

Let f(v) be a vector valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the second order tensor defined through its dot product with any vector u being

$\frac{\partial f}{\partial v} \cdot u = D f (v) [u] = {[\frac{d}{d α} f (v + α u)]}_{α = 0}$

for all vectors u. The above dot product yields a vector, and if u is a unit vector gives the direction derivative of f at v, in the directional u.

Properties:

If $f (v) = f_{1} (v) + f_{2} (v)$ then $\frac{\partial f}{\partial v} \cdot u = (\frac{\partial f_{1}}{\partial v} + \frac{\partial f_{2}}{\partial v}) \cdot u$
If $f (v) = f_{1} (v) \times f_{2} (v)$ then $\frac{\partial f}{\partial v} \cdot u = (\frac{\partial f_{1}}{\partial v} \cdot u) \times f_{2} (v) + f_{1} (v) \times (\frac{\partial f_{2}}{\partial v} \cdot u)$
If $f (v) = f_{1} (f_{2} (v))$ then $\frac{\partial f}{\partial v} \cdot u = \frac{\partial f_{1}}{\partial f_{2}} \cdot (\frac{\partial f_{2}}{\partial v} \cdot u)$

Derivatives of scalar valued functions of second-order tensors

Let $f (S)$ be a real valued function of the second order tensor $S$ . Then the derivative of $f (S)$ with respect to $S$ (or at $S$ ) in the direction $T$ is the second order tensor defined as $\frac{\partial f}{\partial S} : T = D f (S) [T] = {[\frac{d}{d α} f (S + α T)]}_{α = 0}$ for all second order tensors $T$ .

Properties:

If $f (S) = f_{1} (S) + f_{2} (S)$ then $\frac{\partial f}{\partial S} : T = (\frac{\partial f_{1}}{\partial S} + \frac{\partial f_{2}}{\partial S}) : T$
If $f (S) = f_{1} (S) f_{2} (S)$ then $\frac{\partial f}{\partial S} : T = (\frac{\partial f_{1}}{\partial S} : T) f_{2} (S) + f_{1} (S) (\frac{\partial f_{2}}{\partial S} : T)$
If $f (S) = f_{1} (f_{2} (S))$ then $\frac{\partial f}{\partial S} : T = \frac{\partial f_{1}}{\partial f_{2}} (\frac{\partial f_{2}}{\partial S} : T)$

Derivatives of tensor valued functions of second-order tensors

Let $F (S)$ be a second order tensor valued function of the second order tensor $S$ . Then the derivative of $F (S)$ with respect to $S$ (or at $S$ ) in the direction $T$ is the fourth order tensor defined as $\frac{\partial F}{\partial S} : T = D F (S) [T] = {[\frac{d}{d α} F (S + α T)]}_{α = 0}$ for all second order tensors $T$ .

Properties:

If $F (S) = F_{1} (S) + F_{2} (S)$ then $\frac{\partial F}{\partial S} : T = (\frac{\partial F_{1}}{\partial S} + \frac{\partial F_{2}}{\partial S}) : T$
If $F (S) = F_{1} (S) \cdot F_{2} (S)$ then $\frac{\partial F}{\partial S} : T = (\frac{\partial F_{1}}{\partial S} : T) \cdot F_{2} (S) + F_{1} (S) \cdot (\frac{\partial F_{2}}{\partial S} : T)$
If $F (S) = F_{1} (F_{2} (S))$ then $\frac{\partial F}{\partial S} : T = \frac{\partial F_{1}}{\partial F_{2}} : (\frac{\partial F_{2}}{\partial S} : T)$
If $f (S) = f_{1} (F_{2} (S))$ then $\frac{\partial f}{\partial S} : T = \frac{\partial f_{1}}{\partial F_{2}} : (\frac{\partial F_{2}}{\partial S} : T)$

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Gradient of a tensor field

The gradient, $\nabla T$ , of a tensor field $T (x)$ in the direction of an arbitrary constant vector c is defined as: $\nabla T \cdot c = lim_{α \to 0} \frac{d}{d α} T (x + α c)$ The gradient of a tensor field of order n is a tensor field of order n+1.

Cartesian coordinates

If $e_{1}, e_{2}, e_{3}$ are the basis vectors in a Cartesian coordinate system, with coordinates of points denoted by ( $x_{1}, x_{2}, x_{3}$ ), then the gradient of the tensor field $T$ is given by $\nabla T = \frac{\partial T}{\partial x_{i}} \otimes e_{i}$

Proof

The vectors x and c can be written as $x = x_{i} e_{i}$ and $c = c_{i} e_{i}$ . Let y := x + αc. In that case the gradient is given by $\begin{matrix} \nabla T \cdot c & = {\frac{d}{d α} T (x_{1} + α c_{1}, x_{2} + α c_{2}, x_{3} + α c_{3}) |}_{α = 0} \equiv {\frac{d}{d α} T (y_{1}, y_{2}, y_{3}) |}_{α = 0} \\ = {[\frac{\partial T}{\partial y_{1}} \frac{\partial y_{1}}{\partial α} + \frac{\partial T}{\partial y_{2}} \frac{\partial y_{2}}{\partial α} + \frac{\partial T}{\partial y_{3}} \frac{\partial y_{3}}{\partial α}]}_{α = 0} = {[\frac{\partial T}{\partial y_{1}} c_{1} + \frac{\partial T}{\partial y_{2}} c_{2} + \frac{\partial T}{\partial y_{3}} c_{3}]}_{α = 0} \\ = \frac{\partial T}{\partial x_{1}} c_{1} + \frac{\partial T}{\partial x_{2}} c_{2} + \frac{\partial T}{\partial x_{3}} c_{3} \equiv \frac{\partial T}{\partial x_{i}} c_{i} = \frac{\partial T}{\partial x_{i}} (e_{i} \cdot c) = [\frac{\partial T}{\partial x_{i}} \otimes e_{i}] \cdot c ◻ \end{matrix}$

Since the basis vectors do not vary in a Cartesian coordinate system we have the following relations for the gradients of a scalar field $ϕ$ , a vector field v, and a second-order tensor field $S$ . $\begin{matrix} \nabla ϕ & = \frac{\partial ϕ}{\partial x_{i}} e_{i} = ϕ_{i} e_{i} \\ \nabla v & = \frac{\partial (v_{j} e_{j})}{\partial x_{i}} \otimes e_{i} = \frac{\partial v_{j}}{\partial x_{i}} e_{j} \otimes e_{i} = v_{j, i} e_{j} \otimes e_{i} \\ \nabla S & = \frac{\partial (S_{j k} e_{j} \otimes e_{k})}{\partial x_{i}} \otimes e_{i} = \frac{\partial S_{j k}}{\partial x_{i}} e_{j} \otimes e_{k} \otimes e_{i} = S_{j k, i} e_{j} \otimes e_{k} \otimes e_{i} \end{matrix}$

Curvilinear coordinates

If $g^{1}, g^{2}, g^{3}$ are the contravariant basis vectors in a curvilinear coordinate system, with coordinates of points denoted by ( $ξ^{1}, ξ^{2}, ξ^{3}$ ), then the gradient of the tensor field $T$ is given by $\nabla T = \frac{\partial T}{\partial ξ^{i}} \otimes g^{i}$

From this definition we have the following relations for the gradients of a scalar field $ϕ$ , a vector field v, and a second-order tensor field $S$ . $\begin{matrix} \nabla ϕ & = \frac{\partial ϕ}{\partial ξ^{i}} g^{i} \\ \nabla v & = \frac{\partial (v^{j} g_{j})}{\partial ξ^{i}} \otimes g^{i} \\ = (\frac{\partial v^{j}}{\partial ξ^{i}} + v^{k} Γ_{i k}^{j}) g_{j} \otimes g^{i} = (\frac{\partial v_{j}}{\partial ξ^{i}} - v_{k} Γ_{i j}^{k}) g^{j} \otimes g^{i} \\ \nabla S & = \frac{\partial (S_{j k} g^{j} \otimes g^{k})}{\partial ξ^{i}} \otimes g^{i} \\ = (\frac{\partial S_{j k}}{\partial ξ_{i}} - S_{l k} Γ_{i j}^{l} - S_{j l} Γ_{i k}^{l}) g^{j} \otimes g^{k} \otimes g^{i} \end{matrix}$

where the Christoffel symbol $Γ_{i j}^{k}$ is defined using $Γ_{i j}^{k} g_{k} = \frac{\partial g_{i}}{\partial ξ^{j}} ⟹ Γ_{i j}^{k} = \frac{\partial g_{i}}{\partial ξ^{j}} \cdot g^{k} = - g_{i} \cdot \frac{\partial g^{k}}{\partial ξ^{j}}$

Cylindrical polar coordinates

In cylindrical coordinates, the gradient is given by $\begin{matrix} \nabla ϕ = & \frac{\partial ϕ}{\partial r} e_{r} + \frac{1}{r} \frac{\partial ϕ}{\partial θ} e_{θ} + \frac{\partial ϕ}{\partial z} e_{z} \end{matrix}$

$\begin{matrix} \nabla v = & \frac{\partial v_{r}}{\partial r} e_{r} \otimes e_{r} + \frac{1}{r} (\frac{\partial v_{r}}{\partial θ} - v_{θ}) e_{r} \otimes e_{θ} + \frac{\partial v_{r}}{\partial z} e_{r} \otimes e_{z} \\ + & \frac{\partial v_{θ}}{\partial r} e_{θ} \otimes e_{r} + \frac{1}{r} (\frac{\partial v_{θ}}{\partial θ} + v_{r}) e_{θ} \otimes e_{θ} + \frac{\partial v_{θ}}{\partial z} e_{θ} \otimes e_{z} \\ + & \frac{\partial v_{z}}{\partial r} e_{z} \otimes e_{r} + \frac{1}{r} \frac{\partial v_{z}}{\partial θ} e_{z} \otimes e_{θ} + \frac{\partial v_{z}}{\partial z} e_{z} \otimes e_{z} \end{matrix}$

$\begin{matrix} \nabla S = & \frac{\partial S_{r r}}{\partial r} e_{r} \otimes e_{r} \otimes e_{r} + \frac{\partial S_{r r}}{\partial z} e_{r} \otimes e_{r} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{r r}}{\partial θ} - (S_{θ r} + S_{r θ})] e_{r} \otimes e_{r} \otimes e_{θ} \\ + & \frac{\partial S_{r θ}}{\partial r} e_{r} \otimes e_{θ} \otimes e_{r} + \frac{\partial S_{r θ}}{\partial z} e_{r} \otimes e_{θ} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{r θ}}{\partial θ} + (S_{r r} - S_{θ θ})] e_{r} \otimes e_{θ} \otimes e_{θ} \\ + & \frac{\partial S_{r z}}{\partial r} e_{r} \otimes e_{z} \otimes e_{r} + \frac{\partial S_{r z}}{\partial z} e_{r} \otimes e_{z} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{r z}}{\partial θ} - S_{θ z}] e_{r} \otimes e_{z} \otimes e_{θ} \\ + & \frac{\partial S_{θ r}}{\partial r} e_{θ} \otimes e_{r} \otimes e_{r} + \frac{\partial S_{θ r}}{\partial z} e_{θ} \otimes e_{r} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{θ r}}{\partial θ} + (S_{r r} - S_{θ θ})] e_{θ} \otimes e_{r} \otimes e_{θ} \\ + & \frac{\partial S_{θ θ}}{\partial r} e_{θ} \otimes e_{θ} \otimes e_{r} + \frac{\partial S_{θ θ}}{\partial z} e_{θ} \otimes e_{θ} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{θ θ}}{\partial θ} + (S_{r θ} + S_{θ r})] e_{θ} \otimes e_{θ} \otimes e_{θ} \\ + & \frac{\partial S_{θ z}}{\partial r} e_{θ} \otimes e_{z} \otimes e_{r} + \frac{\partial S_{θ z}}{\partial z} e_{θ} \otimes e_{z} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{θ z}}{\partial θ} + S_{r z}] e_{θ} \otimes e_{z} \otimes e_{θ} \\ + & \frac{\partial S_{z r}}{\partial r} e_{z} \otimes e_{r} \otimes e_{r} + \frac{\partial S_{z r}}{\partial z} e_{z} \otimes e_{r} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{z r}}{\partial θ} - S_{z θ}] e_{z} \otimes e_{r} \otimes e_{θ} \\ + & \frac{\partial S_{z θ}}{\partial r} e_{z} \otimes e_{θ} \otimes e_{r} + \frac{\partial S_{z θ}}{\partial z} e_{z} \otimes e_{θ} \otimes e_{z} + \frac{1}{r} [\frac{\partial S_{z θ}}{\partial θ} + S_{z r}] e_{z} \otimes e_{θ} \otimes e_{θ} \\ + & \frac{\partial S_{z z}}{\partial r} e_{z} \otimes e_{z} \otimes e_{r} + \frac{\partial S_{z z}}{\partial z} e_{z} \otimes e_{z} \otimes e_{z} + \frac{1}{r} \frac{\partial S_{z z}}{\partial θ} e_{z} \otimes e_{z} \otimes e_{θ} \end{matrix}$

Divergence of a tensor field

The divergence of a tensor field $T (x)$ is defined using the recursive relation $(\nabla \cdot T) \cdot c = \nabla \cdot (c \cdot T^{T}); \nabla \cdot v = tr (\nabla v)$

where c is an arbitrary constant vector and v is a vector field. If $T$ is a tensor field of order n > 1 then the divergence of the field is a tensor of order n− 1.

Cartesian coordinates

In a Cartesian coordinate system we have the following relations for a vector field v and a second-order tensor field $S$ . $\begin{matrix} \nabla \cdot v & = \frac{\partial v_{i}}{\partial x_{i}} = v_{i, i} \\ \nabla \cdot S & = \frac{\partial S_{i k}}{\partial x_{i}} e_{k} = S_{i k, i} e_{k} \end{matrix}$

where tensor index notation for partial derivatives is used in the rightmost expressions. Note that $\nabla \cdot S \neq \nabla \cdot S^{T} .$

For a symmetric second-order tensor, the divergence is also often written as

$\begin{matrix} \nabla \cdot S & = \frac{\partial S_{k i}}{\partial x_{i}} e_{k} = S_{k i, i} e_{k} \end{matrix}$

The above expression is sometimes used as the definition of $\nabla \cdot S$ in Cartesian component form (often also written as $div S$ ). Note that such a definition is not consistent with the rest of this article (see the section on curvilinear co-ordinates).

The difference stems from whether the differentiation is performed with respect to the rows or columns of $S$ , and is conventional. This is demonstrated by an example. In a Cartesian coordinate system the second order tensor (matrix) $S$ is the gradient of a vector function $v$ .

$\begin{matrix} \nabla \cdot (\nabla v) & = \nabla \cdot (v_{i, j} e_{i} \otimes e_{j}) = v_{i, j i} e_{i} \cdot e_{i} \otimes e_{j} = {(\nabla \cdot v)}_{, j} e_{j} = \nabla (\nabla \cdot v) \\ \nabla \cdot [{(\nabla v)}^{T}] & = \nabla \cdot (v_{j, i} e_{i} \otimes e_{j}) = v_{j, i i} e_{i} \cdot e_{i} \otimes e_{j} = \nabla^{2} v_{j} e_{j} = \nabla^{2} v \end{matrix}$

The last equation is equivalent to the alternative definition / interpretation

$\begin{matrix} {(\nabla \cdot)}_{alt} (\nabla v) = {(\nabla \cdot)}_{alt} (v_{i, j} e_{i} \otimes e_{j}) = v_{i, j j} e_{i} \otimes e_{j} \cdot e_{j} = \nabla^{2} v_{i} e_{i} = \nabla^{2} v \end{matrix}$

Curvilinear coordinates

In curvilinear coordinates, the divergences of a vector field v and a second-order tensor field $S$ are $\begin{matrix} \nabla \cdot v & = (\frac{\partial v^{i}}{\partial ξ^{i}} + v^{k} Γ_{i k}^{i}) \\ \nabla \cdot S & = (\frac{\partial S_{i k}}{\partial ξ_{i}} - S_{l k} Γ_{i i}^{l} - S_{i l} Γ_{i k}^{l}) g^{k} \end{matrix}$

More generally, $\begin{matrix} \nabla \cdot S & = [\frac{\partial S_{i j}}{\partial q^{k}} - Γ_{k i}^{l} S_{l j} - Γ_{k j}^{l} S_{i l}] g^{i k} b^{j} \\ = [\frac{\partial S^{i j}}{\partial q^{i}} + Γ_{i l}^{i} S^{l j} + Γ_{i l}^{j} S^{i l}] b_{j} \\ = [\frac{\partial S_{j}^{i}}{\partial q^{i}} + Γ_{i l}^{i} S_{j}^{l} - Γ_{i j}^{l} S_{l}^{i}] b^{j} \\ = [\frac{\partial S_{i}^{j}}{\partial q^{k}} - Γ_{i k}^{l} S_{l}^{j} + Γ_{k l}^{j} S_{i}^{l}] g^{i k} b_{j} \end{matrix}$

Cylindrical polar coordinates

In cylindrical polar coordinates $\begin{matrix} \nabla \cdot v = & \frac{\partial v_{r}}{\partial r} + \frac{1}{r} (\frac{\partial v_{θ}}{\partial θ} + v_{r}) + \frac{\partial v_{z}}{\partial z} \\ \nabla \cdot S = & \frac{\partial S_{r r}}{\partial r} e_{r} + \frac{\partial S_{r θ}}{\partial r} e_{θ} + \frac{\partial S_{r z}}{\partial r} e_{z} \\ + & \frac{1}{r} [\frac{\partial S_{θ r}}{\partial θ} + (S_{r r} - S_{θ θ})] e_{r} + \frac{1}{r} [\frac{\partial S_{θ θ}}{\partial θ} + (S_{r θ} + S_{θ r})] e_{θ} + \frac{1}{r} [\frac{\partial S_{θ z}}{\partial θ} + S_{r z}] e_{z} \\ + & \frac{\partial S_{z r}}{\partial z} e_{r} + \frac{\partial S_{z θ}}{\partial z} e_{θ} + \frac{\partial S_{z z}}{\partial z} e_{z} \end{matrix}$

Curl of a tensor field

The curl of an order-n > 1 tensor field $T (x)$ is also defined using the recursive relation $(\nabla \times T) \cdot c = \nabla \times (c \cdot T); (\nabla \times v) \cdot c = \nabla \cdot (v \times c)$ where c is an arbitrary constant vector and v is a vector field.

Curl of a first-order tensor (vector) field

Consider a vector field v and an arbitrary constant vector c. In index notation, the cross product is given by $v \times c = ε_{i j k} v_{j} c_{k} e_{i}$ where $ε_{i j k}$ is the permutation symbol, otherwise known as the Levi-Civita symbol. Then, $\nabla \cdot (v \times c) = ε_{i j k} v_{j, i} c_{k} = (ε_{i j k} v_{j, i} e_{k}) \cdot c = (\nabla \times v) \cdot c$ Therefore, $\nabla \times v = ε_{i j k} v_{j, i} e_{k}$

Curl of a second-order tensor field

For a second-order tensor $S$ $c \cdot S = c_{m} S_{m j} e_{j}$ Hence, using the definition of the curl of a first-order tensor field, $\nabla \times (c \cdot S) = ε_{i j k} c_{m} S_{m j, i} e_{k} = (ε_{i j k} S_{m j, i} e_{k} \otimes e_{m}) \cdot c = (\nabla \times S) \cdot c$ Therefore, we have $\nabla \times S = ε_{i j k} S_{m j, i} e_{k} \otimes e_{m}$

Identities involving the curl of a tensor field

The most commonly used identity involving the curl of a tensor field, $T$ , is $\nabla \times (\nabla T) = 0$ This identity holds for tensor fields of all orders. For the important case of a second-order tensor, $S$ , this identity implies that $\nabla \times (\nabla S) = 0 ⟹ S_{m i, j} - S_{m j, i} = 0$

Derivative of the determinant of a second-order tensor

The derivative of the determinant of a second order tensor $A$ is given by $\frac{\partial}{\partial A} det (A) = det (A) {[A^{- 1}]}^{T} .$

In an orthonormal basis, the components of $A$ can be written as a matrix A. In that case, the right hand side corresponds the cofactors of the matrix.

Proof

Let $A$ be a second order tensor and let $f (A) = det (A)$ . Then, from the definition of the derivative of a scalar valued function of a tensor, we have $\begin{matrix} \frac{\partial f}{\partial A} : T & = {\frac{d}{d α} det (A + α T) |}_{α = 0} \\ = {\frac{d}{d α} det [α A (\frac{1}{α} I + A^{- 1} \cdot T)] |}_{α = 0} \\ = {\frac{d}{d α} [α^{3} det (A) det (\frac{1}{α} I + A^{- 1} \cdot T)] |}_{α = 0} . \end{matrix}$

The determinant of a tensor can be expressed in the form of a characteristic equation in terms of the invariants $I_{1}, I_{2}, I_{3}$ using $det (λ I + A) = λ^{3} + I_{1} (A) λ^{2} + I_{2} (A) λ + I_{3} (A) .$

Using this expansion we can write $\begin{matrix} \frac{\partial f}{\partial A} : T & = {\frac{d}{d α} [α^{3} det (A) (\frac{1}{α^{3}} + I_{1} (A^{- 1} \cdot T) \frac{1}{α^{2}} + I_{2} (A^{- 1} \cdot T) \frac{1}{α} + I_{3} (A^{- 1} \cdot T))] |}_{α = 0} \\ = {det (A) \frac{d}{d α} [1 + I_{1} (A^{- 1} \cdot T) α + I_{2} (A^{- 1} \cdot T) α^{2} + I_{3} (A^{- 1} \cdot T) α^{3}] |}_{α = 0} \\ = {det (A) [I_{1} (A^{- 1} \cdot T) + 2 I_{2} (A^{- 1} \cdot T) α + 3 I_{3} (A^{- 1} \cdot T) α^{2}] |}_{α = 0} \\ = det (A) I_{1} (A^{- 1} \cdot T) . \end{matrix}$

Recall that the invariant $I_{1}$ is given by $I_{1} (A) = tr A .$

Hence, $\frac{\partial f}{\partial A} : T = det (A) tr (A^{- 1} \cdot T) = det (A) {[A^{- 1}]}^{T} : T .$

Invoking the arbitrariness of $T$ we then have $\frac{\partial f}{\partial A} = det (A) {[A^{- 1}]}^{T} .$

Derivatives of the invariants of a second-order tensor

The principal invariants of a second order tensor are $\begin{matrix} I_{1} (A) & = tr A \\ I_{2} (A) & = \frac{1}{2} [(tr A)^{2} - tr A^{2}] \\ I_{3} (A) & = det (A) \end{matrix}$

The derivatives of these three invariants with respect to $A$ are $\begin{matrix} \frac{\partial I_{1}}{\partial A} & = 1 \\ \frac{\partial I_{2}}{\partial A} & = I_{1} 1 - A^{T} \\ \frac{\partial I_{3}}{\partial A} & = det (A) {[A^{- 1}]}^{T} \\ = I_{2} 1 - A^{T} (I_{1} 1 - A^{T}) = {(A^{2} - I_{1} A + I_{2} 1)}^{T} \end{matrix}$

Proof

From the derivative of the determinant we know that $\frac{\partial I_{3}}{\partial A} = det (A) {[A^{- 1}]}^{T} .$

For the derivatives of the other two invariants, let us go back to the characteristic equation $det (λ 1 + A) = λ^{3} + I_{1} (A) λ^{2} + I_{2} (A) λ + I_{3} (A) .$

Using the same approach as for the determinant of a tensor, we can show that $\frac{\partial}{\partial A} det (λ 1 + A) = det (λ 1 + A) {[(λ 1 + A)^{- 1}]}^{T} .$

Now the left hand side can be expanded as $\begin{matrix} \frac{\partial}{\partial A} det (λ 1 + A) & = \frac{\partial}{\partial A} [λ^{3} + I_{1} (A) λ^{2} + I_{2} (A) λ + I_{3} (A)] \\ = \frac{\partial I_{1}}{\partial A} λ^{2} + \frac{\partial I_{2}}{\partial A} λ + \frac{\partial I_{3}}{\partial A} . \end{matrix}$

Hence $\frac{\partial I_{1}}{\partial A} λ^{2} + \frac{\partial I_{2}}{\partial A} λ + \frac{\partial I_{3}}{\partial A} = det (λ 1 + A) {[(λ 1 + A)^{- 1}]}^{T}$ or, $(λ 1 + A)^{T} \cdot [\frac{\partial I_{1}}{\partial A} λ^{2} + \frac{\partial I_{2}}{\partial A} λ + \frac{\partial I_{3}}{\partial A}] = det (λ 1 + A) 1 .$

Expanding the right hand side and separating terms on the left hand side gives $(λ 1 + A^{T}) \cdot [\frac{\partial I_{1}}{\partial A} λ^{2} + \frac{\partial I_{2}}{\partial A} λ + \frac{\partial I_{3}}{\partial A}] = [λ^{3} + I_{1} λ^{2} + I_{2} λ + I_{3}] 1$

or, $\begin{matrix} [\frac{\partial I_{1}}{\partial A} λ^{3} & + \frac{\partial I_{2}}{\partial A} λ^{2} + \frac{\partial I_{3}}{\partial A} λ] 1 + A^{T} \cdot \frac{\partial I_{1}}{\partial A} λ^{2} + A^{T} \cdot \frac{\partial I_{2}}{\partial A} λ + A^{T} \cdot \frac{\partial I_{3}}{\partial A} \\ = [λ^{3} + I_{1} λ^{2} + I_{2} λ + I_{3}] 1 . \end{matrix}$

If we define $I_{0} := 1$ and $I_{4} := 0$ , we can write the above as $\begin{matrix} [\frac{\partial I_{1}}{\partial A} λ^{3} & + \frac{\partial I_{2}}{\partial A} λ^{2} + \frac{\partial I_{3}}{\partial A} λ + \frac{\partial I_{4}}{\partial A}] 1 + A^{T} \cdot \frac{\partial I_{0}}{\partial A} λ^{3} + A^{T} \cdot \frac{\partial I_{1}}{\partial A} λ^{2} + A^{T} \cdot \frac{\partial I_{2}}{\partial A} λ + A^{T} \cdot \frac{\partial I_{3}}{\partial A} \\ = [I_{0} λ^{3} + I_{1} λ^{2} + I_{2} λ + I_{3}] 1 . \end{matrix}$

Collecting terms containing various powers of λ, we get $\begin{matrix} λ^{3} & (I_{0} 1 - \frac{\partial I_{1}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{0}}{\partial A}) + λ^{2} (I_{1} 1 - \frac{\partial I_{2}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{1}}{\partial A}) + \\ λ (I_{2} 1 - \frac{\partial I_{3}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{2}}{\partial A}) + (I_{3} 1 - \frac{\partial I_{4}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{3}}{\partial A}) = 0 . \end{matrix}$

Then, invoking the arbitrariness of λ, we have $\begin{matrix} I_{0} 1 - \frac{\partial I_{1}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{0}}{\partial A} & = 0 \\ I_{1} 1 - \frac{\partial I_{2}}{\partial A} 1 - I_{2} 1 - \frac{\partial I_{3}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{2}}{\partial A} & = 0 \\ I_{3} 1 - \frac{\partial I_{4}}{\partial A} 1 - A^{T} \cdot \frac{\partial I_{3}}{\partial A} & = 0 . \end{matrix}$

This implies that $\begin{matrix} \frac{\partial I_{1}}{\partial A} & = 1 \\ \frac{\partial I_{2}}{\partial A} & = I_{1} 1 - A^{T} \\ \frac{\partial I_{3}}{\partial A} & = I_{2} 1 - A^{T} (I_{1} 1 - A^{T}) = {(A^{2} - I_{1} A + I_{2} 1)}^{T} \end{matrix}$

Derivative of the second-order identity tensor

Let $1$ be the second order identity tensor. Then the derivative of this tensor with respect to a second order tensor $A$ is given by $\frac{\partial 1}{\partial A} : T = 0 : T = 0$ This is because $1$ is independent of $A$ .

Derivative of a second-order tensor with respect to itself

Let $A$ be a second order tensor. Then $\frac{\partial A}{\partial A} : T = {[\frac{\partial}{\partial α} (A + α T)]}_{α = 0} = T = I : T$

Therefore, $\frac{\partial A}{\partial A} = I$

Here $I$ is the fourth order identity tensor. In index notation with respect to an orthonormal basis $I = δ_{i k} δ_{j l} e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$

This result implies that $\frac{\partial A^{T}}{\partial A} : T = I^{T} : T = T^{T}$ where $I^{T} = δ_{j k} δ_{i l} e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$

Therefore, if the tensor $A$ is symmetric, then the derivative is also symmetric and we get $\frac{\partial A}{\partial A} = I^{(s)} = \frac{1}{2} (I + I^{T})$ where the symmetric fourth order identity tensor is $I^{(s)} = \frac{1}{2} (δ_{i k} δ_{j l} + δ_{i l} δ_{j k}) e_{i} \otimes e_{j} \otimes e_{k} \otimes e_{l}$

Derivative of the inverse of a second-order tensor

Let $A$ and $T$ be two second order tensors, then $\frac{\partial}{\partial A} (A^{- 1}) : T = - A^{- 1} \cdot T \cdot A^{- 1}$ In index notation with respect to an orthonormal basis $\frac{\partial A_{i j}^{- 1}}{\partial A_{k l}} T_{k l} = - A_{i k}^{- 1} T_{k l} A_{l j}^{- 1} ⟹ \frac{\partial A_{i j}^{- 1}}{\partial A_{k l}} = - A_{i k}^{- 1} A_{l j}^{- 1}$ We also have $\frac{\partial}{\partial A} (A^{- T}) : T = - A^{- T} \cdot T^{T} \cdot A^{- T}$ In index notation $\frac{\partial A_{j i}^{- 1}}{\partial A_{k l}} T_{k l} = - A_{j k}^{- 1} T_{l k} A_{l i}^{- 1} ⟹ \frac{\partial A_{j i}^{- 1}}{\partial A_{k l}} = - A_{l i}^{- 1} A_{j k}^{- 1}$ If the tensor $A$ is symmetric then $\frac{\partial A_{i j}^{- 1}}{\partial A_{k l}} = - \frac{1}{2} (A_{i k}^{- 1} A_{j l}^{- 1} + A_{i l}^{- 1} A_{j k}^{- 1})$

Proof

Recall that $\frac{\partial 1}{\partial A} : T = 0$

Since $A^{- 1} \cdot A = 1$ , we can write $\frac{\partial}{\partial A} (A^{- 1} \cdot A) : T = 0$

Using the product rule for second order tensors $\frac{\partial}{\partial S} [F_{1} (S) \cdot F_{2} (S)] : T = (\frac{\partial F_{1}}{\partial S} : T) \cdot F_{2} + F_{1} \cdot (\frac{\partial F_{2}}{\partial S} : T)$

we get $\frac{\partial}{\partial A} (A^{- 1} \cdot A) : T = (\frac{\partial A^{- 1}}{\partial A} : T) \cdot A + A^{- 1} \cdot (\frac{\partial A}{\partial A} : T) = 0$ or, $(\frac{\partial A^{- 1}}{\partial A} : T) \cdot A = - A^{- 1} \cdot T$

Therefore, $\frac{\partial}{\partial A} (A^{- 1}) : T = - A^{- 1} \cdot T \cdot A^{- 1}$

Integration by parts

Another important operation related to tensor derivatives in continuum mechanics is integration by parts. The formula for integration by parts can be written as $\int_{Ω} F \otimes \nabla G d Ω = \int_{Γ} n \otimes (F \otimes G) d Γ - \int_{Ω} G \otimes \nabla F d Ω$

where $F$ and $G$ are differentiable tensor fields of arbitrary order, $n$ is the unit outward normal to the domain over which the tensor fields are defined, $\otimes$ represents a generalized tensor product operator, and $\nabla$ is a generalized gradient operator. When $F$ is equal to the identity tensor, we get the divergence theorem $\int_{Ω} \nabla G d Ω = \int_{Γ} n \otimes G d Γ .$

We can express the formula for integration by parts in Cartesian index notation as $\int_{Ω} F_{i j k . . . .} G_{l m n . . ., p} d Ω = \int_{Γ} n_{p} F_{i j k . . .} G_{l m n . . .} d Γ - \int_{Ω} G_{l m n . . .} F_{i j k . . ., p} d Ω .$

For the special case where the tensor product operation is a contraction of one index and the gradient operation is a divergence, and both $F$ and $G$ are second order tensors, we have $\int_{Ω} F \cdot (\nabla \cdot G) d Ω = \int_{Γ} n \cdot (G \cdot F^{T}) d Γ - \int_{Ω} (\nabla F) : G^{T} d Ω .$

In index notation, $\int_{Ω} F_{i j} G_{p j, p} d Ω = \int_{Γ} n_{p} F_{i j} G_{p j} d Γ - \int_{Ω} G_{p j} F_{i j, p} d Ω .$

Derivatives with respect to vectors and second-order tensors

Derivatives of scalar valued functions of vectors

Derivatives of vector valued functions of vectors

Derivatives of scalar valued functions of second-order tensors

Derivatives of tensor valued functions of second-order tensors

Gradient of a tensor field

Cartesian coordinates

Curvilinear coordinates

Cylindrical polar coordinates

Divergence of a tensor field

Cartesian coordinates

Curvilinear coordinates

Cylindrical polar coordinates

Curl of a tensor field

Curl of a first-order tensor (vector) field

Curl of a second-order tensor field

Identities involving the curl of a tensor field

Derivative of the determinant of a second-order tensor

Derivatives of the invariants of a second-order tensor

Derivative of the second-order identity tensor

Derivative of a second-order tensor with respect to itself

Derivative of the inverse of a second-order tensor

Integration by parts

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