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Injectivity of the Cauchy-stress tensor along rank-one connected lines under strict rank-one convexity condition

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Abstract

In this note, we show that the Cauchy stress tensor \(\sigma\) in nonlinear elasticity is injective along rank-one connected lines provided that the constitutive law is strictly rank-one convex. This means that \(\sigma(F+\xi\otimes\eta)=\sigma(F)\) implies \(\xi \otimes\eta=0\) under strict rank-one convexity. As a consequence of this seemingly unnoticed observation, it follows that rank-one convexity and a homogeneous Cauchy stress imply that the left Cauchy-Green strain is homogeneous, as is shown in Mihai and Neff (Int. J. Non-Linear Mech., 2016, to appear).

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Notes

  1. The following alternative proof, which uses the identity \(\mathrm{Cof}(F+\xi\otimes\eta).\eta= \mathrm {Cof}F.\eta\), see [17, eq. 1.1.18], was kindly suggested by the reviewer:

    $$\begin{aligned} &\bigl\langle S_{1}(F+\xi\otimes\eta) - S_{1}(F),\, \xi \otimes \eta\bigr\rangle = \bigl\langle \sigma(F+\xi\otimes\eta) \cdot \mathrm {Cof}(F+\xi \otimes\eta) - \sigma(F) \cdot \mathrm {Cof}F,\, \xi\otimes\eta\bigr\rangle \\ &\quad = \bigl\langle \sigma(F+\xi\otimes\eta),\, (\xi\otimes\eta) \bigl(\mathrm {Cof}(F+ \xi\otimes \eta) \bigr)^{T}\bigr\rangle - \bigl\langle \sigma(F),\, (\xi \otimes \eta) (\mathrm {Cof}F)^{T}\bigr\rangle \\ &\quad = \bigl\langle \sigma(F+\xi\otimes\eta),\; \xi \otimes \bigl(\mathrm {Cof}(F+\xi \otimes\eta). \eta \bigr)\bigr\rangle - \bigl\langle \sigma(F),\, \xi \otimes(\mathrm {Cof}F.\eta)\bigr\rangle \\ &\quad = \bigl\langle \sigma(F+\xi\otimes\eta),\; \xi\otimes \bigl(\mathrm {Cof}(F).\eta \bigr)\bigr\rangle - \bigl\langle \sigma(F),\, \xi\otimes(\mathrm {Cof}F.\eta)\bigr\rangle \\ &\quad = \bigl\langle \sigma(F+\xi\otimes\eta) - \sigma(F),\, \xi \otimes \bigl(( \mathrm {Cof}F).\eta \bigr)\bigr\rangle . \end{aligned}$$

    If the stored energy density function is strictly rank-one convex, the latter identity implies that if \(\sigma(F+\xi\otimes\eta)=\sigma (F)\), then \(\xi\otimes\eta=0\).

  2. The following alternative proof was kindly suggested by the reviewer: Rewriting (4.5) as

    $$\bigl(F\eta+ \lVert \eta \rVert ^{2}\xi \bigr)\otimes\xi= - \xi\otimes F\eta $$

    and recalling that \(a\otimes b = c\otimes d \neq0\) if and only if there is \(\lambda\in\mathbb{R}\setminus\{0\}\) such that \(a=\lambda \,c\) and \(b=\frac{d}{\lambda}\), then the assumption \(\xi\otimes\eta \neq0\) implies

    $$F\eta+ \lVert \eta \rVert ^{2}\xi= \lambda\,\xi\,,\quad F\eta= -\lambda\, \xi \,, $$

    and thus \(\lambda=\frac{1}{2}\,\lVert \eta \rVert ^{2}\). But then

    $$\det(F+\xi\otimes\eta) = \det F + \bigl\langle \mathrm {Cof}(F)\eta,\,\xi\bigr\rangle = \bigl(1+ \bigl\langle \eta,\, F^{-1}\xi\bigr\rangle \bigr)\,\det F = \Bigl(1-\Bigl\langle \eta,\, \frac{2}{\lVert \eta \rVert ^{2}}\,\eta\Bigr\rangle \Bigr)\,\det F = -\det F\,, $$

    which contradicts the assumption \(F,F+\xi\otimes\eta\in{\mathrm {GL}}^{+}(3)\).

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Acknowledgements

The support for L. Angela Mihai by the Engineering and Physical Sciences Research Council of Great Britain under research grant EP/M011992/1 is gratefully acknowledged. We thank the reviewer for pointing out the shorter proofs noted in Sect. 2 and in the Appendix.

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Correspondence to Patrizio Neff.

Appendix

Appendix

In this appendix we showFootnote 2 that

$$ \widehat{F}\,\widehat{F}^{T} = F\,F^{T}\,,\quad\widehat{F} = F + \xi\otimes \eta\,,\quad\det\widehat{F},\; \det F > 0\quad\implies\quad\xi \otimes\eta= 0\,. $$
(4.1)

We note that \(\widehat{F}\) and \(F\) are twins [17, Sect. 2.5] since they are rank-one connected and their principal stretches coincide. Here, not only their principal stretches coincide, but the left-stretch tensor is the same as well.

Proof

Since \(\widehat{F}\,\widehat{F}^{T} = F\,F^{T}\), we see that \((\det\widehat{F})^{2} = (\det F)^{2}\), and by assumption (4.1)3 we can conclude that \(\det\widehat{F} = \det F\). Since

det ( F + ξ η ) = det ( F ( 1 + F 1 ξ η ) ) = det F det ( 1 + F 1 ξ η ) = det F ( 1 + tr ( F 1 ξ η ) )
(4.2)

and \(\det(F + \xi\otimes\eta) = \det\widehat{F} = \det F\), by (4.1)2 we conclude from (4.2)

$$ \operatorname {tr}\bigl(F^{-1}\, \xi\otimes\eta \bigr)\ =\ \langle F^{-1}\xi \,, \,\eta \rangle \ =\ 0\,. $$
(4.3)

Assumption (4.1)1 and (4.1)2 together imply

$$ \widehat{F}\,\widehat{F}^{T}\ =\ F\,F^{T} + F\eta\otimes\xi+ \xi\otimes F\eta+ \lVert \eta \rVert ^{2}(\xi\otimes\xi)\ =\ F\,F^{T} \,, $$
(4.4)

thus we must have

$$ F\eta\otimes\xi+ \xi\otimes F \eta+ \lVert \eta \rVert ^{2}(\xi\otimes \xi)= 0 \,. $$
(4.5)

We introduce \(\widehat{\xi}= F^{-1}\xi\), \(\xi= F\widehat{\xi}\) and insert into (4.3) and (4.5) to yield

$$ F\eta\otimes F\widehat{\xi}+ F\widehat{\xi}\otimes F\eta+ \lVert \eta \rVert ^{2} \,(F\widehat{\xi}\otimes F\widehat{\xi})\ =\ 0\,,\qquad \langle \widehat{\xi}\,,\,\eta \rangle \ =\ 0\,. $$
(4.6)

This is equivalent to

$$ F\, \bigl\{ \,\eta\otimes\widehat{\xi}+ \widehat{\xi}\otimes\eta+ \lVert \eta \rVert ^{2}\,(\widehat{\xi}\otimes\widehat{\xi})\, \bigr\} F^{T}\ =\ 0\,,\qquad \langle \widehat{\xi}\,,\,\eta \rangle \ =\ 0\,. $$
(4.7)

Since \(\det F>0\) we have as well

$$ \eta\otimes\widehat{\xi}+ \widehat{\xi}\otimes\eta+ \lVert \eta \rVert ^{2}\,( \widehat{\xi}\otimes\widehat{\xi})\ =\ 0\,,\qquad \langle \widehat{\xi}\,,\,\eta \rangle \ =\ 0 \,. $$
(4.8)

Multiplying (4.8) with \(\eta\neq0\) we obtain \(\eta \underbrace{ \langle \widehat{\xi}\,,\,\eta \rangle }_{=0} + \widehat{\xi}\, \lVert \eta \rVert ^{2} + \lVert \eta \rVert ^{2}\,\widehat{\xi}\,\underbrace{\langle \widehat{\xi}\,,\,\eta \rangle }_{=0} = 0\). Hence, \(\widehat{\xi} \lVert \eta \rVert ^{2} = 0\) implies \(\widehat{\xi}=0\). □

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Neff, P., Mihai, L.A. Injectivity of the Cauchy-stress tensor along rank-one connected lines under strict rank-one convexity condition. J Elast 127, 309–315 (2017). https://doi.org/10.1007/s10659-016-9609-y

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