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Compatibility Equations of Nonlinear Elasticity for Non-Simply-Connected Bodies
Compatibility equations of elasticity are almost 150 years old. Interestingly they do not seem to have been rigorously studied for non-simply-connected bodies to this date. In this paper we derive necessary and sufficient compatibility equations of nonlinear elasticity for arbitrary non-simply-connected bodies when the ambient space is Euclidean. For a non-simply-connected body, a measure of strain may not be compatible even if the standard compatibility equations ("bulk" compatibility equations) are satisfied. It turns out that there may be topological obstructions to compatibility and this paper aims to understand them for both deformation gradient F and the right Cauchy-Green strain C. We show that the necessary and sufficient conditions for compatibility of deformation gradient F are vanishing of its exterior derivative and all its periods, i.e. its integral over generators of the first homology group of the material manifold. We will show that not every non-null-homotopic path requires supplementary compatibility equations for F and linearized strain e. We then find both necessary and sufficient compatibility conditions for the right Cauchy-Green strain tensor C for arbitrary non-simply-connected bodies when the material and ambient space manifolds have the same dimensions. We discuss the well-known necessary compatibility equations in the linearized setting and the Cesaro-Volterra path integral. We then obtain the sufficient conditions of compatibility for the linearized strain when the body is not simply-connected.
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Comments
Congradulations
Dear Arash,
It's a geat work. Congradulations!!!
Mohsen
Re: Compatibility
Great stuff. Made my day.
-- Biswajit
Can dealing with the rotation group be avoided?
Hi Arash,
Nice work!
Question: I found the last two sentences of your abstract intriguing. As stated, you seem to suggest that because the target space for your (2.24) is the (non-trivial) manifold of rotations, which is non-commutative one has a problem.
However, if you look at your eqn.s 2.18 and 2.21 the compatibility question you pose can as well be posed in terms of determining an invertible tensor field, F, given C on a possibly multiple connected domain. 2.18 + 2.21 translate to a linear system of equations for F.
Since the set of all invertible tensors is an open set in the space of second order tensors rather than being a manifold, does this ease your burden?
For instance, from the proof of C compatibility in a simply connected domain, one can show that it is possible to construct "global solutions" of invertible F fields, given C (having vanishing curvature) and invertible 'initial' data at a point of the domain (this uses the fact that parallel transport preserves angles between vector fields)
In a multiply-connected domain if one could impose an analog of your second ii) in Proposition 27, then I think the proof for simply-connected domains would go through in the general case, except
could you explain how second ii) of Proposition 27 is a condition on given data (i.e. the domain and the C field) for proving sufficiency, even when you deal with C compatibility through R.
Keep up the good work,
- Amit
Re: Can dealing with the rotation group be avoided?
Dear Amit:
Thanks for the excellent questions.
"I found the last two sentences of your abstract intriguing. As stated, you seem to suggest that because the target space for your (2.24) is the (non-trivial) manifold of rotations, which is non-commutative one has a problem."
When you look at compatibility equations for F, the nice thing is that you can write the topological equations in terms of the generators of the first homology group. The first homology group is Abelian and easy to calculate compared to fundamental group (which is non-Abelian). Here the "problem" is working with a non-Abelian group that is hard to calculate.
"However, if you look at your eqn.s 2.18 and 2.21 the compatibility question you pose can as well be posed in terms of determining an invertible tensor field, F, given C on a possibly multiple connected domain. 2.18 + 2.21 translate to a linear system of equations for F. Since the set of all invertible tensors is an open set in the space of second order tensors rather than being a manifold, does this ease your burden?"
You are correct about posing the compatibility equations in terms of F. What you would get is similar to (2.24); you replace R by F and K by L=\Gamma \dot{X}. You will still have a system of linear ODEs with variable coefficients. The solution is written similar to (2.27); you replace K by L. However, this would not resolve the issue as the general linear group is not commutative either. So, you will still have your topological compatibility equations written on the generators of the fundamental group and not the first homology group.
"For instance, from the proof of C compatibility in a simply connected domain, one can show that it is possible to construct "global solutions" of invertible F fields, given C (having vanishing curvature) and invertible 'initial' data at a point of the domain (this uses the fact that parallel transport preserves angles between vector fields).
In a multiply-connected domain if one could impose an analog of your second ii) in Proposition 27, then I think the proof for simply-connected domains would go through in the general case, except could you explain how second ii) of Proposition 27 is a condition on given data (i.e. the domain and the C field) for proving sufficiency, even when you deal with C compatibility through R."
Not sure if I understand your question here. ii) in Proposition 27 guarantees uniqueness of the rotation field R given a C field.
Regards,
Arash
Arash, W.r.t. your
Arash,
W.r.t. your remark
****
You are correct about posing the compatibility equations in terms of F. What you would get is similar to (2.24); you replace R by F and K by L=\Gamma \dot{X}.
You will still have a system of linear ODEs with variable coefficients.
The solution is written similar to (2.27); you replace K by L.
However, this would not resolve the issue as the general linear group
is not commutative either. So, you will still have your topological
compatibility equations written on the generators of the fundamental
group and not the first homology group.
****
is it correct to think that this issue would remain in your way of doing the proof even if it were a simply connected domain?
For my other question, you had two ii)s in Proposition 27. I was talkng about the last item there, and the question is that when you are doing sufficiency, you only have the C field and the domain B in hand and the goal is to construct a deformation and its F. So it was not clear to me how one could pose as a hypothesis a condition on F which is an object that has to be constructed...
- Amit
Re: Amit's Question
Dear Amit:
Answer to your first question is no. If the domain is simply connected, every closed path can be shrunk to any point and hence the fundamental group (and consequently the first homology group) would be trivial. So, both "ii)"'s would be trivially satisfied in that case. By the way, thanks to you I see a typo in Proposition 27, the second "ii)" should be "iii)".
Regarding your second question, I see your point. If the first "ii)" is satisfied, then F can be uniquely constructed. This F then has to satisfy the second "ii)". I'm not sure if this can be written any more explicitly than this in the nonlinear setting. However, in the linearized case things can be explicitly written in terms of strain as you can see.
Regards,
Arash
Arash, Could you explain
Arash,
Could you explain the connection between seeking solutions to ODEs with values in the Linear Group or Orthogonal Group and certain equivalence classes of curves in the domain that characterize the connectedness of the domain.
My first question had more to do with whether some of the difficulties you mentioned had to do with the specific product integration based formula you were using or not. Because there are other ways of getting solutions in the Linear group related to the compatibility problem that get away without even recognizing you are seeking solutions in the linear group, but the property is recovered after the fact of having obtained solutions in the space of second order tensors, merely by virtue of the eqn that has actually been solved.
- Amit
Re: Amit's Question
Dear Amit:
You have a system of ODEs that you need to solve on an arbitrary curve. Now, compatibility is equivalent to path independence of the solution. When the domain is not simply connected, in addition to curvature of C vanishing, you need to make sure starting from a given point coming back to it along any closed curve you recover the original (in my solution) rotation tensor. This you only need to enforce on the generators of the fundamental group (i.e. all the "independent" closed paths). Of course, if any closed path can be continuously shrunk to a point (simply-conected domain) then these conditions would be trivially satisfied.
I don't claim that solving for dR/ds=RK product integration is the only approach. However, shouldn't we be looking at path independence of the solution regardless of the method of solution? Again, let me emphasize that for C there may be a more explicit way of writing the topological equations. Certainly worth exploring.
Regards,
Arash
Path independence/Compatibility
Arash,
I completely agree that path independence is the crux of the issue of any compatibility question, regardless of the connectedness of the domain. I was just curious as to whether seeking solutions in the rotation group or linear group is really a serious problem or not. This is because in a simple instance of a non-simply connected domain I think I can do the proof just by knowing some standard ODE theory. I haven't thought about this non-simply connected case at all, so what I say below may have a gap (I don't think so), but then again I am not writing a paper :) - great thing about imechanica, that such discussions are possible.
So let me tell you how I would do the compatibility proof for C in the simplest non-simply connected domain. Take a right-circular thick-walled hollow cylinder. Make a cut along an imaginary plane to obtain a simply-connected domain (with two more boundary surfaces than the original cylinder). Assume RC tensor of given C field vanishes on the cylinder. Then using a beautiful paper of T.Y.Thomas (1934) - this is reference 12 of Shield's paper - where he gives a completely elementary - and in my opinion very clever - proof of the Froebenius theorem, one can show that there exists global solutions (because the system is linear) to your eqn 2.18 in this simply connected domain. Moreover, using parallel transport of vector fields, one can show that if you start with an invertible tensor specified at a point, then the whole constructed field is invertible (this is in my posted compatibility notes on imechanica).
Now all that remains to be shown is whether there is a jump in the constructed F field across the cut or not.
If we now add the following condition beyond RC vanishing:
Take any closed curve S around the hole. Demand that the line integral
\int_S F^a_A \Gamma^A_{BC} dX_C = 0
where F^a_A is the F field constructed from the given C field on the simply-connected domain because RC vanishes. Then it is straightforward to show that the jump in F at any point on the cut surface is necessarily zero - construct a closed contour from a point on one side of the cut-surface tothe adjacent point on the other side of the cut, follow the cut to intersection of the closed contour S with the cut, follow S to the other side of the cut, and then again follow the cut back to A; because the line integral written above on this closed curve is zero, and it is also zero on S, it is easy to see that the contributions from the pieces from the two sides of the cut surface will have to vanish (even though F appears in the integrand), and hence the required jump F(A) - F(B) = 0. Done.
So, yes, path independence is what all this is about - we are in agreement on that for sure.
- Amit
Re: Amit's comment regarding compatibility equations
Dear Amit:
I think one issue is that unlike 2D, non-simply connected domains in 3-space are not that simple. Actually, there is no compete classification of the non-simply-connected embedded submanifolds of R^3. In the plane (as I mention in the paper), any non-simply-connected domain is a disk with a finite number of holes.
One advantage of using group theory is that you don't need to make cuts and then worry about continuity in the (relevant) fields. Of course, making cuts is fine and what you are suggesting for a hollow cylinder is correct. Remark 211 in the paper makes a connection between "cuts" and the relative homology groups.
I think we need to look at more non-trivial examples of non-simply-connected bodies.
Regards,
Arash
Re: Arash/compatibility
Arash,
*****
I think one issue is that unlike 2D, non-simply connected domains in 3-space are not that simple.
*******
I completely agree (I did not compliment you on your paper just like that!).
You would have realized that the example I gave is not of a 2-d body but for any 3-d body topologically equivalent to a thick walled hollow cylinder (with non-zero length along its axis, let's say finite).
For 3-d bodies, is there a theorem in topology that says that any non-simply connected 3-d body can be reduced to a simply-connected one by an appropriate system of cuts? If so, then I think my argument will work for any such body.
To me it looks like the group theoretic analysis of connectedness coupled with an 'explicit' construction for the deformation with the simplest possible tools seems like the optimal way to do the problem. Of course, I know next to nothing of group theory, so I am out of that game....
- Amit
Re: Amit's question on non-simply-connected bodies
Dear Amit:
Thanks for the encouraging compliment.
I agree that your example is a 3D body. However, it is a special 3D body in the sense that it has a "deformation retract" to a 2D problem (fundamental group is preserved under a deformation retract). I have uploaded two figures. In one you can see a rod with a few tubular holes. Your example has only one hole. In such cases it is clear what cuts one needs to use to make the body simply connected. In this figure c_i's are generators of the fundamental group and each corresponds to a cut (exactly the one you mentioned). Note also that the fundamental group happens to be Abelian in this case.
For embedded 3 submanifolds of R^3, I don't think there is any theorem that can tell you what cuts to use. I have uploaded a figure that shows a ball with a knot removed. Think of a knot as a "deformed" thick ring. What cut(s) should we use here? I don't know. But it is known how to calculate the fundamental group. There are four crossing points and each corresponds to a group generator. Note that this knot is connected (has only one component). It turns out that the first homology group has only one generator as long as the knot is connected.
Regards
Arash
Re: Arash, non-simply-connected
Thanks Arash.
***********
For embedded 3 submanifolds of R^3, I don't think there is any theorem that can tell you what cuts to use
***********
OK. Are you aware of any theorem that claims merely the existence of a system of cuts that renders the body simply-connected?
- Amit
Re: Amit's question on cuts
Dear Amit:
I'm not sure if there is such a theorem. However, I think there are ways of looking for such cuts given a triangulation of the manifold. I've seen some related works on discretization of Maxwell's equations for domains with holes. If you're interested I can send them to you.
Regards,
Arash