Nonlinear finite elements/Kinematics - time derivatives and rates

Material time derivatives
Material time derivatives are needed for many updated Lagrangian formulations of finite element analysis.

Recall that the motion can be expressed as

\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t) \quad \text{or} \qquad \mathbf{X} = \boldsymbol{\varphi}^{-1}(\mathbf{x}, t) $$

If we keep $$\mathbf{X}$$ fixed, then the velocity is given by

\mathbf{V}(\mathbf{X}, t) = \frac{\partial \boldsymbol{\varphi}}{\partial t}(\mathbf{X}, t) $$ This is the  material time derivative expressed in terms of $$\mathbf{X}$$.

The spatial version of the velocity is

\mathbf{v}(\mathbf{x}, t) = \mathbf{V}(\boldsymbol{\varphi}^{-1}(\mathbf{x}, t), t) $$

We will use the symbol $$\mathbf{v}$$ for velocity from now on by slightly abusing the notation.

We usually think of quantities such as velocity and acceleration as spatial quantities which are functions of $$\mathbf{x}$$ (rather than material quantities which are functions of $$\mathbf{X}$$).

Given the spatial velocity $$\mathbf{v}(\mathbf{x}, t)$$, if we want to find the acceleration we will have to consider the fact that $$\mathbf{x} \equiv \mathbf{x}(\mathbf{X}, t)$$, i.e., the position also changes with time. We do this by using the chain rule. Thus

\cfrac{D\mathbf{v}(\mathbf{x},t)}{Dt} = \mathbf{a}(\mathbf{x}, t) = \frac{\partial \mathbf{v}(\mathbf{x},t)}{\partial t} + \frac{\partial \mathbf{v}(\mathbf{x},t)}{\partial \mathbf{x}}\cdot \frac{\partial \boldsymbol{\varphi}(\mathbf{X},t)}{\partial t}   = \frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} \mathbf{v}\cdot\mathbf{V} = \frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} \mathbf{v}\cdot\mathbf{v} $$ Such a derivative is called the  material time derivative expressed in terms of $$\mathbf{x}$$. The second term in the expression is called the  convective derivative..

Velocity gradient
Let the velocity be expressed in  spatial form, i.e., $$\mathbf{v}(\mathbf{x}, t)$$. The spatial velocity gradient tensor is given by

\boldsymbol{l} := \frac{\partial \mathbf{v}(\mathbf{x},t)}{\partial \mathbf{x}} = \boldsymbol{\nabla} \mathbf{v} $$ The velocity gradient $$\boldsymbol{l}$$ is a second order tensor which can expressed as

\boldsymbol{l} = l_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j = \frac{\partial v_i}{\partial x_j}~\mathbf{e}_i\otimes\mathbf{e}_j $$ The velocity gradient is a measure of the relative velocity of two points in the current configuration.

Time derivative of the deformation gradient
Recall that the deformation gradient is given by

\boldsymbol{F} = \frac{\partial \boldsymbol{\varphi}}{\partial \mathbf{X}} $$ The time derivative of $$\boldsymbol{F}$$ (keeping $$\mathbf{X}$$ fixed) is

\dot{\boldsymbol{F}} = \frac{\partial }{\partial t}~\left(\frac{\partial \boldsymbol{\varphi}}{\partial \mathbf{X}}\right) = \frac{\partial }{\partial \mathbf{X}}~\left(\frac{\partial \boldsymbol{\varphi}}{\partial t}\right) = \frac{\partial \mathbf{v}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_{\circ} \mathbf{v} $$ Using the chain rule

\dot{\boldsymbol{F}} = \frac{\partial \mathbf{v}}{\partial \mathbf{x}}\cdot\frac{\partial \mathbf{x}}{\partial \mathbf{X}} = \frac{\partial \mathbf{v}}{\partial \mathbf{x}}\cdot\frac{\partial \boldsymbol{\varphi}}{\partial \mathbf{X}} = \boldsymbol{l}\cdot\boldsymbol{F} $$ Form this we get the important relation

\boldsymbol{l} = \dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1} ~. $$

Time derivative of strain
Let $$d\mathbf{X}_1$$ and $$d\mathbf{X}_2$$ be two infinitesimal material line segments in a body. Then

d\mathbf{x}_1 = \boldsymbol{F}\cdot d\mathbf{X}_1 ~; d\mathbf{x}_2 = \boldsymbol{F}\cdot d\mathbf{X}_2 $$ Hence,

d\mathbf{x}_1\cdot d\mathbf{x}_2 = (\boldsymbol{F} \cdot d\mathbf{X}_1) \cdot (\boldsymbol{F} \cdot d\mathbf{X}_2) = d\mathbf{X}_1 \cdot (\boldsymbol{F}^T \cdot \boldsymbol{F}) \cdot d\mathbf{X}_2 = d\mathbf{X}_1 \cdot \boldsymbol{C} \cdot d\mathbf{X}_2 = d\mathbf{X}_1 \cdot (2~\boldsymbol{E} + \boldsymbol{\mathit{1}}) \cdot d\mathbf{X}_2 $$ Taking the derivative with respect to $$t$$ gives us

\frac{\partial }{\partial t} (d\mathbf{x}_1 \cdot d\mathbf{x}_2) = d\mathbf{X}_1 \cdot \frac{\partial \boldsymbol{C}}{\partial t} \cdot d\mathbf{X}_2 = 2~d\mathbf{X}_1 \cdot \frac{\partial \boldsymbol{E}}{\partial t} \cdot d\mathbf{X}_2 $$

The  material strain rate tensor is defined as

\dot{\boldsymbol{E}} = \frac{\partial \boldsymbol{E}}{\partial t} = \frac{1}{2}~\frac{\partial \boldsymbol{C}}{\partial t} = \frac{1}{2}~\dot{\boldsymbol{C}} $$ Clearly,

\dot{\boldsymbol{E}} = \frac{1}{2}~\frac{\partial }{\partial t}(\boldsymbol{F}^T\cdot\boldsymbol{F}) = \frac{1}{2}~(\dot{\boldsymbol{F}}^T \cdot \boldsymbol{F} + \boldsymbol{F}^T\cdot\dot{\boldsymbol{F}}) ~. $$

Also,

\frac{1}{2}~\frac{\partial }{\partial t}(d\mathbf{x}_1\cdot d\mathbf{x}_2) = d\mathbf{X}_1 \cdot \dot{\boldsymbol{E}} \cdot d\mathbf{X}_2 = (\boldsymbol{F}^{-1}\cdot d\mathbf{x}_1) \cdot \dot{\boldsymbol{E}} \cdot(\boldsymbol{F}^{-1}\cdot d\mathbf{x}_2) = d\mathbf{x}_1 \cdot (\boldsymbol{F}^{-T}\cdot \dot{\boldsymbol{E}} \cdot \boldsymbol{F}^{-1}) \cdot d\mathbf{x}_2 $$

The  spatial rate of deformation tensor or  stretching tensor is defined as

\boldsymbol{d} = \boldsymbol{F}^{-T}\cdot \dot{\boldsymbol{E}} \cdot\boldsymbol{F}^{-1} = \frac{1}{2}~\boldsymbol{F}^{-T}\cdot \dot{\boldsymbol{C}} \cdot\boldsymbol{F}^{-1} $$ In fact, we can show that $$\boldsymbol{d}$$ is the symmetric part of the velocity gradient, i.e.,

\boldsymbol{d} = \frac{1}{2}~(\boldsymbol{l} + \boldsymbol{l}^T) $$ For rigid body motions we get $$\boldsymbol{d} = \boldsymbol{\mathit{0}}$$.

Lie derivatives
Most of the operations above can be interpreted as push-forward and pull-back operations. Also, time derivatives of these tensors can be interpreted as Lie derivatives.

Recall that the push-forward of the strain tensor from the material configuration to the spatial configuration is given by

\boldsymbol{e} = \phi_{*} [\boldsymbol{E}] = \boldsymbol{F}^{-T}\cdot\boldsymbol{E}\cdot\boldsymbol{F}^{-1} $$ The pull-back of the spatial strain tensor to the material configuration is given by

\boldsymbol{E} = \phi^{*} [\boldsymbol{e}] = \boldsymbol{F}^T \cdot \boldsymbol{e} \cdot \boldsymbol{F} $$

Therefore, the rate of deformation tensor is a push-forward of the material strain rate tensor, i.e.,

\boldsymbol{d} = \boldsymbol{F}^{-T}\cdot\dot{\boldsymbol{E}}\cdot\boldsymbol{F}^{-1} = \phi_{*}[\dot{\boldsymbol{E}}] $$ Similarly, the material strain rate tensor is a pull-back of the rate of deformation tensor to the material configuration, i.e.,

\dot{\boldsymbol{E}} = \boldsymbol{F}^T \cdot \boldsymbol{d} \cdot \boldsymbol{F} = \phi^{*} [\boldsymbol{d}] $$

Now,

\boldsymbol{E} = \phi^{*}[\boldsymbol{e}] \quad \implies \quad \dot{\boldsymbol{E}} = \frac{\partial }{\partial t} \left(\phi^{*}[\boldsymbol{e}]\right) $$ Also,

\boldsymbol{d} = \phi_{*}[\dot{\boldsymbol{E}}] = \phi_{*} \left[\frac{\partial }{\partial t} \left(\phi^{*}[\boldsymbol{e}]\right)\right] $$ Therefore the rate of deformation tensor can be obtained by first pulling back $$\boldsymbol{e}$$ to the reference configuration, taking a material time derivative in that configuration, and then pushing forward the result to the current configuration.

Such an operation is called a  Lie derivative. In general, the Lie derivative of a spatial tensor $$\mathbf{g}$$ is defined as

\mathcal{L}_{\phi}[\boldsymbol{g}] := \phi_{*} \left[\frac{\partial }{\partial t} \left(\phi^{*}[\boldsymbol{g}]\right)\right] ~. $$

Spin tensor
The velocity gradient tensor can be additively decomposed into a symmetric part and a skew part:

\boldsymbol{l} = \frac{1}{2}~(\boldsymbol{l} + \boldsymbol{l}^T) + \frac{1}{2}(\boldsymbol{l} - \boldsymbol{l}^T) = \boldsymbol{d} + \boldsymbol{w} $$ We have seen that $$\boldsymbol{d}$$ is the rate of deformation tensor. The quantity $$\boldsymbol{w}$$ is called the  spin tensor.

Note that $$\boldsymbol{d}$$ is symmetric while $$\boldsymbol{w}$$ is skew symmetric, i.e.,

\boldsymbol{d} = \boldsymbol{d}^T ~; \boldsymbol{w} = -\boldsymbol{w}^T ~. $$

So see why $$\boldsymbol{w}$$ is called a "spin", recall that

\boldsymbol{l} = \dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1} $$ Therefore,

\boldsymbol{w} = \frac{1}{2}(\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1} - \boldsymbol{F}^{-T}\cdot\dot{\boldsymbol{F}}^T) $$ Also,

\boldsymbol{F} = \boldsymbol{R}\cdot\boldsymbol{U} \quad \implies \quad \dot{\boldsymbol{F}} = \dot{\boldsymbol{R}}\cdot\boldsymbol{U} + \boldsymbol{R}\cdot\dot{\boldsymbol{U}} $$ Therefore,

\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1} = (\dot{\boldsymbol{R}}\cdot\boldsymbol{U} + \boldsymbol{R}\cdot\dot{\boldsymbol{U}})\cdot (\boldsymbol{U}^{-1}\cdot\boldsymbol{R}^T) = \dot{\boldsymbol{R}}\cdot\boldsymbol{R}^T + \boldsymbol{R}\cdot\dot{\boldsymbol{U}}\cdot\boldsymbol{U}^{-1}\cdot\boldsymbol{R}^T $$ and

\boldsymbol{F}^{-T}\cdot\dot{\boldsymbol{F}}^T = (\boldsymbol{R}\cdot\boldsymbol{U}^{-1})\cdot (\boldsymbol{U}\cdot\dot{\boldsymbol{R}}^T + \dot{\boldsymbol{U}}\cdot\boldsymbol{R}^T) = \boldsymbol{R}\cdot\dot{\boldsymbol{R}}^T + \boldsymbol{R}\cdot\boldsymbol{U}^{-1}\cdot\dot{\boldsymbol{U}}\cdot\boldsymbol{R}^T $$ So we have

\boldsymbol{w} = \frac{1}{2}~(\dot{\boldsymbol{R}}\cdot\boldsymbol{R}^T + \boldsymbol{R}\cdot\dot{\boldsymbol{U}}\cdot\boldsymbol{U}^{-1}\cdot\boldsymbol{R}^T        - \boldsymbol{R}\cdot\dot{\boldsymbol{R}}^T - \boldsymbol{R}\cdot\boldsymbol{U}^{-1}\cdot\dot{\boldsymbol{U}}\cdot\boldsymbol{R}^T) $$ Now

\boldsymbol{R}\cdot\boldsymbol{R}^T = \boldsymbol{\mathit{1}} \quad \implies \quad \dot{\boldsymbol{R}}\cdot\boldsymbol{R}^T + \boldsymbol{R}\cdot\dot{\boldsymbol{R}}^T = \boldsymbol{\mathit{0}} $$ Therefore

\boldsymbol{w} = \dot{\boldsymbol{R}}\cdot\boldsymbol{R}^T + \frac{1}{2}~\boldsymbol{R}\cdot(\dot{\boldsymbol{U}}\cdot\boldsymbol{U}^{-1} -             \boldsymbol{U}^{-1}\cdot\dot{\boldsymbol{U}})\cdot\boldsymbol{R}^T $$ The second term above is invariant for rigid body motions and zero for an uniaxial stretch. Hence, we are left with just a rotation term. This is why the quantity $$\boldsymbol{w}$$ is called a spin.

The spin tensor is a skew-symmetric tensor and has an associated axial vector $$\boldsymbol{\omega}$$ (also called the angular velocity vector) whose components are given by

\boldsymbol{\omega} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} $$ where

\mathbf{w} = \begin{bmatrix} 0 & -w_3 & w_2 \\ w_3 & 0 & -w_1 \\ -w_2 & w_1 & 0 \end{bmatrix} $$ The spin tensor and its associated axial vector appear in a number of modern numerical algorithms.

Rate of change of volume
Recall that

dv = J~dV \qquad \text{where}~ J = \det\boldsymbol{F} $$ Therefore, taking the material time derivative of $$dv$$ (keeping $$\mathbf{X}$$ fixed), we have

\cfrac{d}{dt}(dv) = \dot{J}~dV = \cfrac{\dot{J}}{J}~dv $$

At this stage we invoke the following result from tensor calculus:

If $$\boldsymbol{A}$$ is an invertible tensor which depends on $$t$$ then

\cfrac{d}{dt}(\det\boldsymbol{A}) = (\det\boldsymbol{A})~\text{tr}\left(\cfrac{d\boldsymbol{A}}{dt}\cdot\boldsymbol{A}^{-1}\right) $$

In the case where $$\boldsymbol{A} = \boldsymbol{F},~ J = \det\boldsymbol{F}$$ we have

\cfrac{d}{dt}(J) = J~\text{tr}\left(\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}\right) $$ or,

\dot{J} = J~\text{tr}(\boldsymbol{l}) = J~\text{tr}(\mathbf{d}) $$ Therefore,

\cfrac{d}{dt}(dv) = \text{tr}(\mathbf{d})~dv $$ Alternatively, we can also write

\dot{J} = \frac{1}{2}~J~\boldsymbol{C}^{-1}:\dot{\boldsymbol{C}} $$ These relations are of immense use in numerical algorithms - particularly those which involved incompressible behavior, i.e., when $$\dot{J} = 0$$.