Mechanics of Continua and Structures
The procedure to compute the pull-back acceleration field is discussed in the section Pull-back Acceleration field. In that section, we presented the procedure for computing the tensor \(\boldsymbol{P}\) and the vector \(\boldsymbol{q}\). In this section, we present the procedure for computing the orthonormal tensor \(\boldsymbol{Q}(t)\).
From the definition of \(\boldsymbol{P}\) given in (2) in the previous section, it follows that \(\boldsymbol{Q}(t)\) is determined/characterized by the equation
\(\begin{align} \ddot{\boldsymbol{Q}}(t)=\boldsymbol{Q}(t) \boldsymbol{P}(t) \label{eq:QGoverning} \end{align}\) Equation \(\eqref{eq:QGoverning}\) is a second order, linear, homogenous, tensor differential equation with time varying coefficients. It is in general not difficult to numerically solve equations of the form \(\eqref{eq:QGoverning}\). However, in our particular case, the solution of \(\eqref{eq:QGoverning}\) is complicated by the constraint that its solution \(\boldsymbol{Q}(t)\) needs to be an orthonormal tensor. That, in addition to \(\eqref{eq:QGoverning}\), the tensor \(\boldsymbol{Q}(t)\) also needs to satisfy the constraint
\(\begin{align} \boldsymbol{Q}^{\textsf{T}}(t)\boldsymbol{Q}(t)&=\boldsymbol{I}, \label{eq:Qconstraint1} \end{align}\) or equivalently
\[\begin{align} \boldsymbol{Q}(t)\boldsymbol{Q}^{\textsf{T}}(t)&=\boldsymbol{I}, \label{eq:Qconstraint2} \end{align}\]where \(\boldsymbol{I}\) is the identity tensor. Therefore, instead of directly trying to solve \(\eqref{eq:QGoverning}\) numerically, we pursue a different strategy.
After differentiating equations \(\eqref{eq:Qconstraint1}\) and \(\eqref{eq:Qconstraint2}\) with respect to time we get \(\begin{align} \dot{\boldsymbol{Q}}^{\textsf{T}}(t)\boldsymbol{Q}(t)+\boldsymbol{Q}^{\textsf{T}}\dot{\boldsymbol{Q}}(t)&=\boldsymbol{0}, \label{eq:ConstraintDiff1} \\ \end{align}\) and \(\begin{align} \dot{\boldsymbol{Q}}(t)\boldsymbol{Q}^{\textsf{T}}(t)+\boldsymbol{Q}\dot{\boldsymbol{Q}}^{\textsf{T}}(t)&=\boldsymbol{0}. \label{eq:ConstraintDiff2} \end{align}\)
Defining \(\begin{align} \boldsymbol{\Omega}_{0}(t)&:=\boldsymbol{Q}^{\textsf{T}}\dot{\boldsymbol{Q}}(t) \label{eq:DefOmegaZero} \end{align}\)
and
\(\begin{align} \boldsymbol{\Omega}(t)&:=\dot{\boldsymbol{Q}}(t)\boldsymbol{Q}^{\textsf{T}}(t) \end{align}\) it follows from \(\eqref{eq:ConstraintDiff1}\) and \(\eqref{eq:ConstraintDiff2}\) that
\(\begin{align} \boldsymbol{\Omega}_{0}(t)&=-\left.\boldsymbol{\Omega}_{0}(t)\right.^{\textsf{T}}, \label{eq:OmegaZeroDef}\\ \end{align}\) and \(\begin{align} \boldsymbol{\Omega}(t)&=-\left.\boldsymbol{\Omega}(t)\right.^{\textsf{T}}.\\ \end{align}\) That is, that \(\boldsymbol{\Omega}_{0}\) and \(\boldsymbol{\Omega}\) are Skew-Symmetric tensors.
Differentiating equation \(\eqref{eq:Qconstraint1}\) twice with respect to time we get \(\begin{align} \ddot{\boldsymbol{Q}}^{\textsf{T}}(t)\boldsymbol{Q}(t)+ \dot{\boldsymbol{Q}}^{\textsf{T}}(t)\dot{\boldsymbol{Q}}(t)+ \dot{\boldsymbol{Q}}^{\textsf{T}}(t)\dot{\boldsymbol{Q}}(t)+ \boldsymbol{Q}^{\textsf{T}}\ddot{\boldsymbol{Q}}(t)&=\boldsymbol{0}, \label{eq:ConstraintDiff3a} \end{align}\) which can be re-arranged to read \(\begin{align} \boldsymbol{Q}^{\textsf{T}}\ddot{\boldsymbol{Q}}(t)+ \ddot{\boldsymbol{Q}}^{\textsf{T}}(t)\boldsymbol{Q}(t) &=-2\dot{\boldsymbol{Q}}^{\textsf{T}}(t)\dot{\boldsymbol{Q}}(t). \label{eq:ConstraintDiff3b} \end{align}\)
Recalling from the definition of \(\boldsymbol{P}(t)\) that the term \(\boldsymbol{Q}^{\textsf{T}}\ddot{\boldsymbol{Q}}(t)\), which is the the first term on the left hand side of \(\eqref{eq:ConstraintDiff3b}\), is in fact \(\boldsymbol{P}\), and the term \(\ddot{\boldsymbol{Q}}^{\textsf{T}}(t)\boldsymbol{Q}(t)\), which is the the second term on the left hand side of \(\eqref{eq:ConstraintDiff3b}\), is the transpose of \(\boldsymbol{P}\), we get from \(\eqref{eq:ConstraintDiff3b}\) that \(\begin{align} \textsf{sym}\left(\boldsymbol{P}\right)(t) &= -\dot{\boldsymbol{Q}}^{\textsf{T}}(t)\dot{\boldsymbol{Q}}(t), \label{eq:ConstraintDiff3c} \end{align}\) where the operator \(\textsf{sym}: \textsf{Lin}\mapsto \text{Sym}\), where \(\textsf{Sym}\) is the set of all symmetric tensors, is defined as \(\text{sym}(\cdot):=\left((\cdot)+(\cdot)^{\textsf{T}}\right)/2\). It follows from \(\eqref{eq:OmegaZeroDef}\), the definition of \(\boldsymbol{\Omega}_0\), that the right hand side of \(\eqref{eq:ConstraintDiff3c}\) is equal to \(\boldsymbol{\Omega}_0^{\textsf{T}}(t) \boldsymbol{\Omega}_0(t)\). Thus, we get from \(\eqref{eq:ConstraintDiff3c}\) that \(\begin{align} \textsf{sym}\left(\boldsymbol{P}\right)(t) &= - \boldsymbol{\Omega}_{0}^{\textsf{T}} \boldsymbol{\Omega}_{0}(t). \label{eq:ConstraintDiff3d} \end{align}\) Using the fact that \(\boldsymbol{\Omega}_0(t)\) is a skew-symmetric tensor, equation \(\eqref{eq:ConstraintDiff3d}\) can be simplified to read \(\begin{align} \textsf{sym}\boldsymbol{P}(t) &= \boldsymbol{\Omega}_{0}^{2}(t). \label{eq:ConstraintDiff3e} \end{align}\)
Noting \(\textsf{sym}\left(\boldsymbol{P}\right)(t)\) and that it is possible to define the square root of all symmetric tensors, we can write \(\eqref{eq:ConstraintDiff3e}\) as \(\begin{align} \boldsymbol{\Omega}_{0}(t) &= \left(\textsf{sym}\left(\boldsymbol{P}\right)\right)^{1/2}(t) \label{eq:ConstraintDiff3f} \end{align}\) The steps to compute \(\boldsymbol{\Omega}_0(t)\) from \(\boldsymbol{P}(t)\) using equation \(\eqref{eq:ConstraintDiff3f}\) are detailed in the section Skew-symmetric square root of a symmetric tensor.
If follows from \(\eqref{eq:DefOmegaZero}\) and \(\eqref{eq:ConstraintDiff3f}\) that \(\begin{align} \dot{\boldsymbol{Q}}(t)&= \boldsymbol{Q}(t) \left(\textsf{sym}\left(\boldsymbol{P}\right)\right)^{1/2}(t) \label{eq:FirstOrderODE} \end{align}\)
Equation \(\eqref{eq:FirstOrderODE}\) is a first-order, homogenous, ordinary tensor differential equation with time varying coefficients. It can be shown that its unique solution is \(\begin{align} \boldsymbol{Q}(t)&=e^{\boldsymbol{W}(t)} \label{eq:SolQ} \end{align}\) where, \(\begin{align} \boldsymbol{W}(t)&:=\int_{0}^{t} \left(\textsf{sym}\left(\boldsymbol{P}\right)\right)^{1/2}(\tau)\, d\tau, \label{eq:DefW} \end{align}\) and the exponential of a tensor is a tensor defined as \(\begin{align} e^{\boldsymbol{W}(t)}&= \boldsymbol{I}+\boldsymbol{W}(t)+\frac{1}{2!}\boldsymbol{W}^2(t)+\frac{1}{3!}\boldsymbol{W}^3(t)+\cdots \end{align}\)
It follows from \(\eqref{eq:DefW}\) that \(\boldsymbol{W}(t)\) is a skew-symmetric tensor. Using this fact, it can be shown that \(\begin{align} e^{\boldsymbol{W}(t)}&=\boldsymbol{I}+ \frac{\sin(w(t))}{w(t)}\boldsymbol{W}(t)+ \frac{\left(1-\cos(w(t))\right)}{w(t)^2}\boldsymbol{W}^2(t), \end{align}\) where \(w(t)\) is the magnitude of \(\boldsymbol{W}(t)\)’s axial vector.