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Derivative of F with Respect to C
Hi All,
The derivative of C (Right Cauchy-Green Tensor) with respect to F (Deformation Gradient) is computed very easily, but I'm not sure if there is an explicit way to compute the derivative of F with respect to C.
Mohsen
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Derivative of F with Respect to C
Hi Mohsen,
Your question is of course not a simple one. I think it looks very similar to the famous old problem: how to compute the derivative of the square root of a tensor (or a matrix)? There is a huge literature on that. For the beginning I recommend the paper of Anne Hoger and Donald Carlson “On the derivative of the square root of a tensor and Guo's rate theorems”, where (at least one) explicit formula was established. You could directly apply this formula for your purposes after performing the corresponding polar decomposition of the deformation gradient. But I’m not sure this is the best trick. Nevertheless I wouldn’t expect very nice formula here, since with C known we may find F only up to some multiplicative rotation tensor.
Svyatoslav
Derivative of F with Respect to C
Hi Svyatoslav,
Thanks for your comment. As you said by polar decomposition we can write
F=RU. U can be expressed as an explicit function of C and invariants of U and
therefore computing the derivative of U with respect to C although complex
(involving eigenvalues and eigenprojections of U/C) but is feasible. It remians
to claim that C is not a function of R (C=FtF=URtRU=U2) and vice versa and thus
R has no derivative with respect to C. Therefore we can write dF/dC=RdU/dC and
the result follows.
Mohsen
R is not a function of C
R is not a function of C does not mean R is the constant function of C. It means you cannot define the derivative of something that is no a function to begin with - or think of R as a 'multi-valued function' of C.
R is not a function of C
Dear Professor Acharya,
Why not to take the derivative? The question is a somewhat delicate one. Consider, for example, a function of two variables f(x,y)=sqrt(x)y and differentiate it with respect to x.
I think the terminology which uses the word 'multi-valued' should be very concrete. Like in complex analysis when we deal with Riemann surfaces. Then it is clear 'multi-valued' here means 'the lack of injectivity' that results in a countable number of sheets of the surface. How to understand properly "R is a multi-valued function of C" is in general not easy. Yes, of course, I remember your idea related to phase transitions and g-disclinations. By the way, have you received my reply?
Svyatoslav
Svyatoslav, My use of
Svyatoslav,
My use of 'multivalued function' was just to loosely motivate the problem to show the contradiction in terms.
I would not take the derivative for the following reason: as you know, in set theory a function requires a well-defined association rule that assigns a unique element in the co-domain of a function to each element of its domain (and this is of course not the same as injectivity). Once one has a function, say on a vector space, and if it has the smoothness to have a (Frechet) derivative, then one can take a derivative. But one needs a function, in the above sense, to begin with. R cannot be defined as a function in this sense on the vector space of symmetric positive definite tensors. If smoothness was the issue, one could start thinking about distributions, but distributions are definitely not functions in the strict sense and do not admit classical derivatives - at any rate the 'R as as fn. of C' problem does not fit this question also.
In the example you give, f(x,y) is at least a function - it is not differentiable (in the classical sense).
I agree 'multuvaluedness' should be used very carefully, if at all. I do not personally like it, especially for using in theory or applications where one has to do computing with it.
Yes, I did receive your reply - thanks.
Our g-disclination paper avoids the use of multivaluedness completely but takes care of the topological content of the objects without ambiguity through the use of smooth (but augmented) fields.
- Amit
R is not a Function of C
Dear Prof. Acharya,
You are looking to the problem from a pure mathematical standpoint while I'm looking from mechanical one. C and U are measures of stretch and any superposed rotation (or more specifically rigid body rotation) does not affect the components of C or U. In this regard C or U do not change with R and therefore do not have derivative with respect to R. In other words, in this problem we are investigating the rate of change of F with respect to its stretching part not its rotational part.
Being that said, from your comment I understand that R is in some sense dependent on C (not of course in the same way that a function depends on its variable). Then how can we show this dependence?
Mohsen