4階等方tensorの逆tensorとその存在条件

$ {\cal\pmb{C}}^{-1}=-\frac{\alpha}{\beta+\gamma}\frac{1}{(\mathrm{tr}\bm I)\alpha+\beta+\gamma}\bm{I}\bm{I}+\frac{\beta}{\beta^2-\gamma^2}{\cal\pmb{I}}+\frac{-\gamma}{\beta^2-\gamma^2}\tilde{\cal\pmb{I}}

$ =-\frac{\alpha}{2s}\frac{1}{(\mathrm{tr}\bm I)\alpha+2s}\bm{I}\bm{I}+\frac{1}{2s}{\cal\pmb{S}}+\frac{1}{2w}{\cal\pmb{W}}

$ s:=\frac12(\beta+\gamma)

$ w:=\frac12(\beta-\gamma)

$ {\cal\pmb{C}}=\alpha\bm{I}\bm{I}+2s({\cal\pmb D}+\frac1{\mathrm{tr}\bm I}\bm I\bm I):{\cal\pmb S}+2w{\cal\pmb W}

$ = ((\mathrm{tr}\bm I)\alpha+2s)\frac1{\mathrm{tr}\bm I}\bm I\bm I+2s{\cal\pmb D}:{\cal\pmb S}+2w{\cal\pmb W}

$ {\cal\pmb C}^{-1}=-\frac{\alpha}{2s}\frac{1}{(\mathrm{tr}\bm I)\alpha+2s}\bm{I}\bm{I}+\frac{1}{2s}({\cal\pmb D}+\frac1{\mathrm{tr}\bm I}\bm I\bm I):{\cal\pmb{S}}+\frac{1}{2w}{\cal\pmb{W}}

$ =\frac{1}{2s}\frac1{\mathrm{tr}\bm I}\frac{(\mathrm{tr}\bm I)\alpha+2s-(\mathrm{tr}\bm I)\alpha}{(\mathrm{tr}\bm I)\alpha+2s}\bm{I}\bm{I}+\frac{1}{2s}{\cal\pmb D}:{\cal\pmb{S}}+\frac{1}{2w}{\cal\pmb{W}}

$ =\frac{1}{(\mathrm{tr}\bm I)\alpha+2s}\frac1{\mathrm{tr}\bm I}\bm{I}\bm{I}+\frac{1}{2s}{\cal\pmb D}:{\cal\pmb{S}}+\frac{1}{2w}{\cal\pmb{W}}

この計算あってるのかなあ？

$ {\cal\pmb C}:{\cal\pmb C}^{-1}=(\alpha\bm{I}\bm{I}+2s{\cal\pmb S}+2w{\cal\pmb W}):\left(-\frac{\alpha}{2s}\frac{1}{(\mathrm{tr}\bm I)\alpha+2s}\bm{I}\bm{I}+\frac{1}{2s}{\cal\pmb{S}}+\frac{1}{2w}{\cal\pmb{W}}\right)

$ =\left(-\frac{\alpha}{2s}\frac{(\mathrm{tr}\bm I)\alpha}{(\mathrm{tr}\bm I)\alpha+2s}-\frac{\alpha}{(\mathrm{tr}\bm I)\alpha+2s}+\frac{\alpha}{2s}\right)\bm{I}\bm{I}+ {\cal\pmb S}+{\cal\pmb W}

$ =\frac{\alpha}{2s}\left(-\frac{(\mathrm{tr}\bm I)\alpha}{(\mathrm{tr}\bm I)\alpha+2s}-\frac{2s}{(\mathrm{tr}\bm I)\alpha+2s}+1\right)\bm{I}\bm{I}+ {\cal\pmb I}

$ ={\cal\pmb I}

あってるみたい

方針

導出

$ \pmb\varepsilon=\frac{1}{2\mu}(\pmb{\sigma}-\lambda\mathrm{tr}(\pmb\varepsilon)\pmb I)

$ =\frac{1}{2\mu}\left(\pmb{\sigma}-\frac{\lambda}{3\lambda+2\mu}\mathrm{tr}(\pmb{\sigma})\pmb I\right)

$ =-\frac{\lambda}{2\mu}\frac{1}{3\lambda+2\mu}\mathrm{tr}(\pmb{\sigma})\pmb I+\frac{1}{2\mu}\pmb{\sigma}

$ =\left(-\frac{\lambda}{2\mu}\frac{1}{3\lambda+2\mu}\pmb{I}\pmb{I}+\frac{1}{2\mu}{\cal\pmb{I}}\right):\pmb{\sigma}

$ \underline{\therefore {{\cal\pmb{C}}_h}^{-1}=-\frac{\lambda}{2\mu}\frac{1}{3\lambda+2\mu}\pmb{I}\pmb{I}+\frac{1}{2\mu}{\cal\pmb{I}}\quad}_\blacksquare

$ \pmb\sigma=\frac13(3K-2G)\mathrm{tr}(\pmb\varepsilon)\pmb I+2G\pmb\varepsilon

$ \iff\pmb\varepsilon=\frac{1}{2G}\left(\pmb\sigma-\frac13(3K-2G)\mathrm{tr}(\pmb\varepsilon)\pmb I\right)

$ = \frac{1}{2G}\left(\pmb\sigma-\frac13\frac{3K-2G}{3K}\mathrm{tr}(\pmb\sigma)\pmb I\right)

$ \because\mathrm{tr}(\pmb\sigma)=3K\mathrm{tr}(\pmb\varepsilon)

$ = \frac{1}{2G}\left({\cal\pmb I}-\frac13\frac{3K-2G}{3K}\pmb I\pmb I\right):\pmb\sigma

$ \underline{\therefore {{\cal\pmb{C}}_h}^{-1}=\frac13\left(\frac{1}{3K}-\frac{1}{2G}\right)\pmb{I}\pmb{I}+\frac{1}{2G}{\cal\pmb{I}}\quad}_\blacksquare

$ = \frac13\frac1{3K}\pmb I\pmb I+\frac1{2G}{\cal\pmb D}

逆tensorも等方tensorだと仮定して解いてみる

$ {\cal\pmb{C}}^{-1}=\delta\pmb{I}\pmb{I}+\epsilon{\cal\pmb{I}}+\zeta\tilde{\cal\pmb{I}}

$ \implies {\cal\pmb{C}}^{-1}:{\cal\pmb{C}}={\cal\pmb{I}}

$ \iff (\delta\pmb{I}\pmb{I}+\epsilon{\cal\pmb{I}}+\zeta\tilde{\cal\pmb{I}}):(\alpha\pmb{I}\pmb{I}+\beta{\cal\pmb{I}}+\gamma\tilde{\cal\pmb{I}})={\cal\pmb{I}}

$ \iff 3\delta\alpha\pmb{I}\pmb{I}+(\epsilon\beta+\zeta\gamma){\cal\pmb{I}}+\delta(\beta+\gamma)\pmb{I}\pmb{I}+\epsilon\alpha\pmb{I}\pmb{I}+\epsilon\gamma\tilde{\cal\pmb{I}}+\zeta\alpha\pmb{I}\pmb{I}+\zeta\beta\tilde{\cal\pmb{I}}={\cal\pmb{I}}

$ \iff (\delta(3\alpha+\beta+\gamma)+\alpha(\epsilon+\zeta))\pmb{I}\pmb{I}+(\epsilon\beta+\zeta\gamma){\cal\pmb{I}}+(\epsilon\gamma+\zeta\beta)\tilde{\cal\pmb{I}}={\cal\pmb{I}}

$ \iff \begin{pmatrix}3\alpha+\beta+\gamma&\alpha&\alpha\\0&\beta&\gamma\\0&\gamma&\beta\end{pmatrix}\begin{pmatrix}\delta\\\epsilon\\\zeta\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}

$ \iff \begin{dcases}(3\alpha+\beta+\gamma)\delta+\alpha(\epsilon+\zeta)=0\\\begin{pmatrix}\beta&\gamma\\\gamma&\beta\end{pmatrix}\begin{pmatrix}\epsilon\\\zeta\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}\end{dcases}

2元連立方程式に帰着できる

$ \iff \begin{dcases}\delta&=-\frac{\alpha}{3\alpha+\beta+\gamma}(\epsilon+\zeta)\\\begin{pmatrix}\epsilon\\\zeta\end{pmatrix}&=\frac{1}{\beta^2-\gamma^2}\begin{pmatrix}\beta&-\gamma\\-\gamma&\beta\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\beta^2-\gamma^2}\begin{pmatrix}\beta\\-\gamma\end{pmatrix}\end{dcases}

$ \iff \begin{dcases}\delta&=-\frac{\alpha}{3\alpha+\beta+\gamma}\frac{1}{\beta+\gamma}\\\begin{pmatrix}\epsilon\\\zeta\end{pmatrix}&=\frac{1}{\beta^2-\gamma^2}\begin{pmatrix}\beta&-\gamma\\-\gamma&\beta\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\beta^2-\gamma^2}\begin{pmatrix}\beta\\-\gamma\end{pmatrix}\end{dcases}

$ \therefore {\cal\pmb{C}}^{-1}=-\frac{\alpha}{\beta+\gamma}\frac{1}{3\alpha+\beta+\gamma}\pmb{I}\pmb{I}+\frac{\beta}{\beta^2-\gamma^2}{\cal\pmb{I}}+\frac{-\gamma}{\beta^2-\gamma^2}\tilde{\cal\pmb{I}}

$ {\cal\pmb{C}}^{-1}=-\frac{\alpha}{2s}\frac{1}{3\alpha+2s}(\pmb{I}\pmb{I})+\frac{s+w}{4sw}{\cal\pmb{I}}+\frac{-(s-w)}{4sw}\tilde{\cal\pmb{I}}

$ =-\frac{\alpha}{2s}\frac{1}{3\alpha+2s}\pmb{I}\pmb{I}+\left(\frac{s+w}{4sw}+\frac{-(s-w)}{4sw}\right){\cal\pmb{S}}+\left(\frac{s+w}{4sw}-\frac{-(s-w)}{4sw}\right){\cal\pmb{W}}

$ =-\frac{\alpha}{2s}\frac{1}{3\alpha+2s}\pmb{I}\pmb{I}+\frac{1}{2s}{\cal\pmb{S}}+\frac{1}{2w}{\cal\pmb{W}}

消えた情報を復元できないため、逆tensorが存在しない

各項のtensor計算

$ (\pmb{I}\pmb{I}):(\pmb{I}\pmb{I})=\mathrm{tr}(\pmb{I})(\pmb{I}\pmb{I})=3\pmb{I}\pmb{I}

$ (\pmb{I}\pmb{I}):\tilde{\cal\pmb{I}}=\sum_{i,j}(\pmb{e}_i\pmb{e}_i\pmb{e}_j\pmb{e}_j):(\pmb{e}_j\pmb{e}_j\pmb{e}_j\pmb{e}_j)=\sum_{i,j}(\pmb{e}_i\pmb{e}_i\pmb{e}_j\pmb{e}_j)=\pmb{I}\pmb{I}

$ (\pmb{I}\pmb{I}):{\cal\pmb{I}}=\sum_{i,j}(\pmb{e}_i\pmb{e}_i\pmb{e}_j\pmb{e}_j):(\pmb{e}_j\pmb{e}_j\pmb{e}_j\pmb{e}_j)=\sum_{i,j}(\pmb{e}_i\pmb{e}_i\pmb{e}_j\pmb{e}_j)=\pmb{I}\pmb{I}

$ {\cal\pmb{I}}:{\cal\pmb{I}}=\tilde{\cal\pmb{I}}:\tilde{\cal\pmb{I}}={\cal\pmb{I}}

$ {\cal\pmb{I}}:\tilde{\cal\pmb{I}}=\tilde{\cal\pmb{I}}:{\cal\pmb{I}}=\tilde{\cal\pmb{I}}