From: Plasticity - mathematical theory and numerical analysis, Second Edition, by Weimin Han, B. Daya Reddy

The function $\mathbf y$ will have to satisfy certain conditions if it is to be used to model adequately the motion of the body. First, we must ensure that no two points get mapped to a single point by $\mathbf y$; in other words, $\mathbf y$ must be one-to-one, and hence invertible. Second, we must ensure that the motion is orientation-preserving; that is, the Jacobian $J$, defined by $$J=det(\nabla \mathbf y)$$ must be positive.

I know that Jacobian must be positive, because it is equal to $\frac{dV}{dV_0}$ where $V$ and $V_0$ are current and reference volumes respectively; and they are positive. So, Jacobian must be positive. But, what does orientation-preserving mean? Which orientation does it refer? Is it orientation of a vector or a set of vectors? If so, which vector/vectors? And finally, how this orientation-preserving is related to the Jacobian?


1 Answer 1


The Jacobian is automatically positive under the hypotheses of smoothness and that no two points get mapped to a single point by ${\bf y}$, no special requirement is necessary. (Your argument however is wrong because $dV/dV_0$ is the absolute value of the Jacobian and not the Jacobian itself...) We have ${\bf y}={\bf y}(t, {\bf x})$ where ${\bf x}$ is the initial position of a particle and ${\bf y}$ is the position of that particle at time $t$. Since no particles superpose along their motion, for fixed $t$, the map ${\bf y}={\bf y}(t, {\bf x})$ where ${\bf x}$ must be invertible producing ${\bf x}={\bf x}(t, {\bf y})$. This way we have $$\delta^i_j = \frac{\partial y_i}{\partial y_j}= \sum_k \frac{\partial y_i}{\partial x_k}\frac{\partial x_k}{\partial y_j}$$ that means $$I= \nabla {\bf y}_t \nabla {\bf x}_t$$ so that $$1= \det(\nabla {\bf y}_t) \det(\nabla {\bf x}_t)$$ so that $\det(\nabla {\bf y})\neq 0$ in particular for every $t$. Finally, for fixed ${\bf x}$, the map $$t \mapsto \det(\nabla {\bf y}_t)$$ is continuous and its value at $t=0$ is positive ($ \det(\nabla {\bf y}_0)=1$) and therefore it cannot change its sign otherwise it should vanish somewhere which is not permitted.

Regarding orientation, assume $n=3$ and consider a closed a surface $\Sigma$ of particles at time $t=0$. Assume that it is orientable, i.e., there is a well defined outwards pointing normal ${\bf n}$. Now consider the evolution in time of this surface: $$\Sigma_t =\{ {\bf y}(t, {\bf x})\:|\: {\bf x} \in \Sigma\}$$ It is possible to prove that the transformed normal $${\bf n}_t := \nabla {\bf y}_t {\bf n}$$ is still outwards pointing for every $t$, just in view of the fact that $det(\nabla {\bf y}_t)$ is always positive.

However this condition is automatically fulfilled and it does not need a further explicit requirement as I showed above.


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