3次元2階tensorの固有方程式の解

3次元2階tensorの固有方程式を解いて2階tensorの固有値を具体的に求める

$ \lambda^3-({\rm tr}\bm A)\lambda^2+\frac12(({\rm tr}\bm A)^2-{\rm tr}\bm A^2)\lambda-\det\bm A=0

texはhttps://ja.wikipedia.org/wiki/三次方程式#一般解を改変

$ a_3 x^3 +a_2 x^2 +a_1 x+a_0=0

$ x_0 =\textcolor{red}{-\frac{a_2}{3a_3}} +\frac{\textcolor{blue}{\sqrt[3]{\fbox{2}}}}{3\textcolor{orange}{\sqrt[3]{2} a_3}} -\frac{\sqrt[3]{2}\textcolor{green}{\fbox4}}{3\textcolor{blue}{\sqrt[3]{\fbox2}}a_3}

$ x_1 =\textcolor{red}{-\frac{a_2}{3a_3}} -\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}{6\textcolor{orange}{\sqrt[3]{2} a_3}} +\frac{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}a_3}

$ x_2 =\textcolor{red}{-\frac{a_2}{3a_3}} -\frac{{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}}{6\textcolor{orange}{\sqrt[3]{2} a_3}} +\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}a_3}

$ \fbox1=-2{a_2}^3 +9a_1 a_2 a_3 -27a_0 {a_3}^2

$ \fbox2=\fbox 1+\sqrt{\fbox3}

$ \fbox 3=\fbox1^2+4\fbox4^3

$ \fbox4=-{a_2}^2 +3a_1 a_3

$ a_3=1を代入

$ x^3 +a_2 x^2 +a_1 x+a_0=0

$ x_0 =\textcolor{red}{-\frac{a_2}{3}} +\frac{\textcolor{blue}{\sqrt[3]{\fbox{2}}}}{3\textcolor{orange}{\sqrt[3]{2}}} -\frac{\sqrt[3]{2}\textcolor{green}{\fbox4}}{3\textcolor{blue}{\sqrt[3]{\fbox2}}}

$ x_1 =\textcolor{red}{-\frac{a_2}{3}} -\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}{6\textcolor{orange}{\sqrt[3]{2}}} +\frac{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}}

$ x_2 =\textcolor{red}{-\frac{a_2}{3}} -\frac{{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}}{6\textcolor{orange}{\sqrt[3]{2}}} +\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \fbox1=-2{a_2}^3 +9a_1 a_2-27a_0

$ \fbox2=\fbox 1+\sqrt{\fbox3}

$ \fbox 3=\fbox1^2+4\fbox4^3

$ \fbox4=-{a_2}^2 +3a_1

$ xを$ \lambdaに置換

$ \lambda^3 +a_2 \lambda^2 +a_1 \lambda+a_0=0

$ \lambda_0 =\textcolor{red}{-\frac{a_2}{3}} +\frac{\textcolor{blue}{\sqrt[3]{\fbox{2}}}}{3\textcolor{orange}{\sqrt[3]{2}}} -\frac{\sqrt[3]{2}\textcolor{green}{\fbox4}}{3\textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \lambda_1 =\textcolor{red}{-\frac{a_2}{3}} -\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}{6\textcolor{orange}{\sqrt[3]{2}}} +\frac{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \lambda_2 =\textcolor{red}{-\frac{a_2}{3}} -\frac{{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}}{6\textcolor{orange}{\sqrt[3]{2}}} +\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \fbox1=-2{a_2}^3 +9a_1 a_2-27a_0

$ \fbox2=\fbox 1+\sqrt{\fbox3}

$ \fbox 3=\fbox1^2+4\fbox4^3

$ \fbox4=-{a_2}^2 +3a_1

$ a_2=-{\rm tr}\bm Aを代入

$ \lambda^3-({\rm tr}\bm A)\lambda^2 +a_1 \lambda+a_0=0

$ \lambda_0 =\textcolor{red}{A_m} +\frac{\textcolor{blue}{\sqrt[3]{\fbox{2}}}}{3\textcolor{orange}{\sqrt[3]{2}}} -\frac{\sqrt[3]{2}\textcolor{green}{\fbox4}}{3\textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \lambda_1 =\textcolor{red}{A_m} -\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}{6\textcolor{orange}{\sqrt[3]{2}}} +\frac{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \lambda_2 =\textcolor{red}{A_m} -\frac{{\textcolor{midnightblue}{\left(1+i\sqrt3\right)}\textcolor{blue}{\sqrt[3]{\fbox2}}}}{6\textcolor{orange}{\sqrt[3]{2}}} +\frac{\textcolor{violet}{\left(1-i\sqrt3\right)}\textcolor{green}{\fbox4}}{3\cdot2^{\frac{2}{3}} \textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \fbox1=54{A_m}^3-9a_1A_m-27a_0

$ \fbox2=\fbox 1+\sqrt{\fbox3}

$ \fbox 3=\fbox1^2+4\fbox4^3

$ \fbox4=-9{A_m}^2 +3a_1

$ A_m:=\frac13{\rm tr}\bm A($ \bm Aの等方成分)とした

$ a_0を展開する

第3不変量を第1不変量と第2不変量で表すより、$ a_0=-\det\bm A=-J_1J_2+\frac13(J_1^3-{\rm tr}(\bm T^3))=-3A_ma_1+9A_m^3-\frac13{\rm tr}\bm A^3

$ \lambda^3-({\rm tr}\bm A)\lambda^2 +a_1 \lambda-\det\bm A=0

$ \lambda_0 =\textcolor{red}{A_m} +\frac{\textcolor{blue}{\sqrt[3]{\fbox{2}}}}{3\textcolor{orange}{\sqrt[3]{2}}} -\frac{\sqrt[3]{2}\textcolor{green}{\fbox4}}{3\textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \fbox1=-189{A_m}^3 +72a_1A_m+9{\rm tr}\bm A^3

$ 54-27*9

$ \fbox2=\fbox 1+\sqrt{\fbox3}

$ \fbox 3=\fbox1^2+4\fbox4^3

$ \fbox4=-9{A_m}^2 +3a_1

$ a_1=\frac12(({\rm tr}\bm A)^2-{\rm tr}\bm A^2)=\frac92A_m^2-\frac12{\rm tr}\bm A^2を代入

$ \lambda^3-({\rm tr}\bm A)\lambda^2 +\frac12(({\rm tr}\bm A)^2-{\rm tr}\bm A^2)\lambda-\det\bm A=0

$ \lambda_0 =\textcolor{red}{A_m} +\frac{\textcolor{blue}{\sqrt[3]{\fbox{2}}}}{3\textcolor{orange}{\sqrt[3]{2}}} -\frac{\sqrt[3]{2}\textcolor{green}{\fbox4}}{3\textcolor{blue}{\sqrt[3]{\fbox2}}}

$ \fbox1=135{A_m}^3-\frac12{\rm tr}\bm A^2+9{\rm tr}\bm A^3

$ 72*9/2-189

$ \fbox2=\fbox 1+\sqrt{\fbox3}

$ \fbox 3=\fbox1^2+4\fbox4^3

$ \fbox4=\frac92{A_m}^2-\frac32{\rm tr}\bm A^2

導出できたけど、複雑すぎてすぐにわかることが出てきなさそう

#2024-01-03 09:09:32