I recently focus on solid mechanics and I am reading Nonlinear Solid Mechanics A Continuum Approach for Engineering by Gerhard A. Holzapfel. However, I was confused by a mathematical formula eq(2.49), it is as follows, \begin{equation} \operatorname{div} \mathbf{A}=\operatorname{Div} \mathbf{A} \mathbf{F}^{-\mathrm{T}} \tag{2.49} \end{equation} where $\mathbf{A}$ is a smooth second-order tensor field, $\mathbf{F}^{-1}$ is the inverse of the deformation gradient. I tried to prove this formula but failed, my current progress is as follows,

\begin{equation} \operatorname{div} \mathbf{A}= \frac{\partial A_{i j}}{\partial x_{j}} \mathbf{e}_{i} \\ \operatorname{Div} \mathbf{A} \mathbf{F}^{-\mathrm{T}} = \frac{\partial A_{i j}}{\partial X_{j}} \mathbf{e}_{i} \frac{\partial X_{k}}{\partial x_{l}} \mathbf{e}_{l} \otimes \mathbf{e}_{k}=\frac{\partial A_{i j}}{\partial X_{j}}\frac{\partial X_{k}}{\partial x_{l}}\delta_{il}\mathbf{e}_{k} = \frac{\partial A_{i j}}{\partial X_{j}}\frac{\partial X_{k}}{\partial x_{i}}\mathbf{e}_{k} \end{equation}

I stopped here and can't reach that equation(2.49) in the book. Can someone help me?

  • 1
    $\begingroup$ What is the difference between a div and a Div? $\endgroup$ – G. Smith Jun 22 at 4:02
  • $\begingroup$ And what is $F^{-T}$ $\endgroup$ – Eli Jun 22 at 6:20
  • $\begingroup$ $\operatorname{Div}$ is the divergence with respect to material coordinates while $\operatorname{div}$ is the divergence with respect to spatial coordinates. $F^{-T}$ is the transpose of $F^{-1}$. $\endgroup$ – John Lionel Jun 22 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.