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Year 4 of the mathematics course of the University of Oxford

Solid mechanics course of the University of Oxford

Past exam of the solid mechanics course of the University of Oxford

2011

Define the invariants by the <characteristic polynomial>
$$
\det(tI-B)=t^3-I_Bt^2+II_Bt-III_B.
$$
Because $B=FF^T$ has positive <eigenvalues> $\lambda_j^2$, its invariants are the <elementary symmetric functions> in those <eigenvalues>:
$$
I_B=\lambda_1^2+\lambda_2^2+\lambda_3^2,\qquad
II_B=\lambda_1^2\lambda_2^2+\lambda_1^2\lambda_3^2+\lambda_2^2\lambda_3^2,\qquad
III_B=\lambda_1^2\lambda_2^2\lambda_3^2.
$$
The <Frobenius norm> gives $|F|^2=\operatorname{tr}(F^TF)=\sum_j\lambda_j^2$. Write $J=\det F>0$. Since $\operatorname{cof}F=JF^{-T}$, its <singular values> are $J/\lambda_j$, so
$$
|\operatorname{cof}F|^2=J^2\sum_j\lambda_j^{-2}
=\lambda_2^2\lambda_3^2+\lambda_1^2\lambda_3^2+\lambda_1^2\lambda_2^2.
$$
Finally $\det B=\det F\det F^T=J^2$. Therefore
$$
\boxed{I_B=|F|^2,\qquad II_B=|\operatorname{cof}F|^2,\qquad III_B=(\det F)^2.}
$$
These identities also give $I_B=\operatorname{tr}B$ and $II_B=((\operatorname{tr}B)^2-\operatorname{tr}(B^2))/2$.

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Past exam of the solid mechanics course of the University of Oxford / 2011 / 1 / b / Solution

Past exam of the solid mechanics course of the University of Oxford / 2011 / 1 / b / Solution

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