5
$\begingroup$

The Taylor expansion of a vector field $f(x)$ to the order of one is $$f(x)=f(x_0)+Jf(x_0)\cdot\Delta x+o(\Delta x)$$ where $Jf$ is Jacobian of the vector field and $\Delta x=x-x_0$.

Suppose we decompose it into symmetric and skew-symmetric part such that $$S=\frac{Jf+(Jf)^T}{2}$$ $$A=\frac{Jf-(Jf)^T}{2}$$ Then $$f(x)\approx f(x_0)+S\Delta x+A\Delta x$$ Could we regard $f(x_0)$ as a translation, $S\Delta x$ as an expansion or a contradiction along axes of eigen directions for it resembles strain tensor and $A\Delta x$ a rotation since skew-symmetric matrix is infinitesimal of rotation matrix?

If it is correct, I'm a bit confused with, say operator $A$, what object it acts on?

Otherwise, does this decomposition have any practical meaning in terms of vector field or velocity field?

$\endgroup$
4
$\begingroup$

you ask whether the decomposition of the Jacobi tensor $J{f}$ into symmetric and antisymmetric parts $S$ and $A$ produces a Helmholtz decomposition of $\Delta f=J{f}\cdot\Delta{x}$ into a curl-free part $\chi=S\cdot\Delta x$ and a divergence-free part $\omega=A\cdot\Delta x$.

this is indeed correct, I will try to write it out explicitly (trying to avoid the errors I made in my first answer)

notation: the subscript $0$ indicates the dependence on $x_0$ and derivatives with respect to $x_0$; I abbreviate $\xi=\Delta x=x-x_0$ and denote by $\nabla_\xi$ the derivative with respect to $\xi$ (or $x$).

$${\omega}(\xi)={A}(x_0)\cdot\xi=\tfrac{1}{2}\xi\cdot(\nabla_0 f_0)-\tfrac{1}{2}(\nabla_0 f_0)\cdot\xi=\tfrac{1}{2}(\nabla_0\times{f}_0)\times\xi$$ $$\Rightarrow\nabla_\xi\cdot{\omega}(\xi)=-\tfrac{1}{2}(\nabla_0\times f_0)\cdot(\nabla_\xi\times\xi)=0$$

$$\chi(\xi)=S(x_0)\cdot\xi=\tfrac{1}{2}(\nabla_0 f_0)\cdot\xi+\tfrac{1}{2}\xi\cdot(\nabla_0 f_0)$$ $$\Rightarrow\nabla_\xi\times\chi(\xi)=-\tfrac{1}{2}\nabla_0\times f_0+\tfrac{1}{2}\nabla_0\times f_0=0$$


vector identities used:

$$a\times(b\times c)=b(a\cdot c)-c(a\cdot b),\;\; a\cdot(b\times c)=c\cdot(a\times b)$$

$\endgroup$
  • $\begingroup$ Thanks and I'm pretty interested in how you derive these vector identities. $\endgroup$ – Shuchang Sep 2 '13 at 2:45
  • $\begingroup$ I added some intermediate steps. You can of course always just verify these identies by writing the whole thing out in components. $\endgroup$ – Carlo Beenakker Sep 2 '13 at 6:15

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.