I have a $2D$ fluid parcel with coordinates $(0.5,-0.5), (-0.5,-0.5), (0.5,0.5)$ and $(-0.5,0.5)$ and this parcel is deformed by a steady flow field of $u=ay$ and $v=0$, defined on the basis ${(1,0), (0,1)}$. I tried to calculate the velocity gradient tensor $\tau_{ij}$ given by the matrix $ \left( \begin{array}{ccc} 0 & a \\ 0 & 0 \end{array} \right) $. I now need to decompose this into the symmetric strain rate tensor and the antisymmetric rotation tensor. However, this matrix isn't diagonalizable. Am I missing something here?

I need to use this part to "solve" the transformation equation of a point that is given by $$x_i(t+dt)=x_i+(u_i+du_i)dt$$where $du_i = \tau_{ij}dx_j$. Then, I must remove the rotation rate tensor from the velocity gradient tensor (which basically means I have the strain rate tensor left, if I am not mistaken) and use that to show that the transformation equation then becomes $$x_i(t+dt)=x_i+(u_i+\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})\partial x_j)dt$$ and finally determine the transformation equations for $x(t+dt), y(t+dt)$.


Any tensor $A_{ij}$ can be decomposed into symmetric and antisymmetric parts, regardless of whether or not it is diagonalizable.

$$ \begin{align} S_{ij} & = \frac{1}{2}\left(A_{ij} + A_{ji}\right) \\ \Omega_{ij} & = \frac{1}{2}\left(A_{ij} - A_{ji}\right) \end{align} $$ so that $$ A_{ij} = S_{ij} + \Omega_{ij} $$

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  • $\begingroup$ But is my velocity gradient tensor correct? I tried proceeding but it made no sense. $\endgroup$ – Artemisia Sep 3 '14 at 6:03
  • $\begingroup$ Your velocity gradient tensor looks consistent with the velocity fields you gave. Where did it start not making sense? $\endgroup$ – Darwin Sep 3 '14 at 7:41
  • $\begingroup$ Haha oops well I tried it once again and I think I have it this time :) thank you. $\endgroup$ – Artemisia Sep 3 '14 at 10:29

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