（十 五）二阶张量的分解——加法分解（和分解）
本文主要内容如下：
- 1. 加法分解的存在性与唯一性
- 2. 仿射量(反)对称部分主不变量与仿射量主不变量之间的联系
- 3. 球形张量与偏斜张量及其特征值与特征方向
- 4. 对称仿射量球(偏)量部分主不变量与对称仿射量主不变量之间的联系
1. 加法分解的存在性与唯一性
对于任意二阶张量（仿射量） T \bold{T} T可构造出二阶对称张量 N \bold N N与二阶反对称张量 Ω \bold{\Omega} Ω：
N = T + T T 2 ( a ) Ω = T − T T 2 ( b ) \bold N=\frac{\bold{T}+\bold{T}^T}{2}\qquad(a)\\\ \\ \bold{\Omega}=\frac{\bold{T}-\bold{T}^T}{2}\qquad(b) N=2T+TT(a) Ω=2T−TT(b)
且
T = N + Ω \bold{T}=\bold{N}+\bold{\Omega} T=N+Ω
这说明任意二阶张量可分解为对称部分与反对称部分的和，下面继续证明这种和分解的唯一性：
假设
T = N 1 + Ω 1 = N 2 + Ω 2 \bold{T}=\bold{N}_1+\bold{\Omega}_1=\bold{N}_2+\bold{\Omega}_2 T=N1+Ω1=N2+Ω2
则
( N 1 − N 2 ) + ( Ω 1 − Ω 2 ) = 0 ( 1 ) (\bold{N}_1-\bold{N}_2)+(\bold{\Omega}_1-\bold{\Omega}_2)=0 \qquad(1) (N1−N2)+(Ω1−Ω2)=0(1)
对上式左右同时转置，得到：
( N 1 − N 2 ) − ( Ω 1 − Ω 2 ) = 0 ( 2 ) (\bold{N}_1-\bold{N}_2)-(\bold{\Omega}_1-\bold{\Omega}_2)=0 \qquad(2) (N1−N2)−(Ω1−Ω2)=0(2)
由 ( 1 ) ( 2 ) (1)(2) (1)(2)两式可得：
{ N 1 − N 2 = 0 ⟹ N 1 = N 2 Ω 1 − Ω 2 = 0 ⟹ Ω 1 = Ω 2 \begin{cases} \bold{N}_1-\bold{N}_2=0&\Longrightarrow & \bold{N}_1=\bold{N}_2&\\\\ \bold{\Omega}_1-\bold{\Omega}_2=0&\Longrightarrow&\bold{\Omega}_1=\bold{\Omega}_2 \end{cases} ⎩ ⎨ ⎧N1−N2=0Ω1−Ω2=0⟹⟹N1=N2Ω1=Ω2
综上所述，任意仿射量可唯一地分解为 ( a ) (a) (a)确定的对称部分与 ( b ) (b) (b)确定的反对称部分之和。
2. 仿射量(反)对称部分主不变量与仿射量主不变量之间的联系
结合反对称二阶张量的主不变量的性质，可推知：
C 1 T = t r ( T ) = T i i = N i i + Ω i i = t r ( N ) + t r ( Ω ) = C 1 N + C 1 Ω = C 1 N C 2 T = 1 2 [ t r ( T ) 2 − t r ( T ∙ T ) ] = 1 2 [ t r ( N ) 2 − t r ( N ∙ N + Ω ∙ Ω + Ω ∙ N + N ∙ Ω ) ] = 1 2 [ t r ( N ) 2 − t r ( N ∙ N ) ] − 1 2 t r ( Ω ∙ Ω ) = C 2 N + C 2 Ω C 3 T = 1 6 t r ( T ) 3 − 1 2 t r ( T ) t r ( T ∙ T ) + 1 3 t r ( T ∙ T ∙ T ) = 1 6 t r ( N ) 3 − 1 2 t r ( N ) [ t r ( N ∙ N ) + t r ( Ω ∙ Ω ) + 2 t r ( N ∙ Ω ) ] + 1 3 [ t r ( N ∙ N ∙ N ) + t r ( Ω ∙ Ω ∙ Ω ) + 3 t r ( N ∙ N ∙ Ω ) + 3 t r ( Ω ∙ Ω ∙ N ) ] = [ 1 6 t r ( N ) 3 − 1 2 t r ( N ) t r ( N ∙ N ) + 1 3 t r ( N ∙ N ∙ N ) ] + [ 1 3 t r ( Ω ∙ Ω ∙ Ω ) ] − 1 2 t r ( N ) t r ( Ω ∙ Ω ) − t r ( N ) t r ( N ∙ Ω ) + t r ( N ∙ N ∙ Ω ) + t r ( Ω ∙ Ω ∙ N ) = C 3 N + C 3 Ω − 1 2 C 1 N C 2 ∗ Ω + t r ( Ω ∙ N ∙ Ω ) = C 3 N + C 2 Ω + C 1 Ω ∙ N ∙ Ω \begin{aligned} & \mathscr{C}^T_1 =tr(\bold T)=T^i_i =N^i_i+\Omega^i_i =tr(\bold N)+tr(\bold{\Omega}) =\mathscr{C}^N_1+\mathscr{C}^{\Omega}_1 =\mathscr{C}^N_1\\ \\ &\mathscr{C}^T_2=\frac{1}{2}[tr(\bold{T})^2-tr(\bold{T}\bullet\bold{T})]\\\\ &\quad\ =\frac{1}{2}[tr(\bold{N})^2-tr(\bold{N}\bullet\bold{N}+\bold{\Omega}\bullet\bold{\Omega}+\bold{\Omega}\bullet\bold{N}+\bold{N}\bullet\bold{\Omega})]\\ \\ &\quad\ =\frac{1}{2}[tr(\bold{N})^2-tr(\bold{N}\bullet\bold{N})]-\frac{1}{2}tr(\bold{\Omega}\bullet\bold{\Omega})\\ \\ &\quad\ =\mathscr{C}^N_2+\mathscr{C}^{\bold{\Omega}}_2\\ \\ &\mathscr{C}^T_3 =\frac{1}{6}tr(\bold{T})^3-\frac{1}{2}tr(\bold{T})tr(\bold{T}\bullet\bold{T})+\frac{1}{3}tr(\bold{T}\bullet\bold{T}\bullet\bold{T})\\ \\ &\quad\ \ =\frac{1}{6}tr(\bold{N})^3-\frac{1}{2}tr(\bold{N})[tr(\bold{N}\bullet\bold{N})+tr(\bold{\Omega}\bullet\bold{\Omega})+2tr(\bold{N}\bullet\bold{\Omega})]+\frac{1}{3}[tr(\bold{N}\bullet\bold{N}\bullet\bold{N})+tr(\bold{\Omega}\bullet\bold{\Omega}\bullet\bold{\Omega})+3tr(\bold{N}\bullet\bold{N}\bullet\bold{\Omega})+3tr(\bold{\Omega}\bullet\bold{\Omega}\bullet\bold{N})]\\\ \\ &\quad\ \ =[\frac{1}{6}tr(\bold{N})^3-\frac{1}{2}tr(\bold{N})tr(\bold{N}\bullet\bold{N})+\frac{1}{3}tr(\bold{N}\bullet\bold{N}\bullet\bold{N})]+[\frac{1}{3}tr(\bold{\Omega}\bullet\bold{\Omega}\bullet\bold{\Omega})]-\frac{1}{2}tr(\bold{N})tr(\bold{\Omega}\bullet\bold{\Omega})-tr(\bold{N})tr(\bold{N}\bullet\bold{\Omega})+tr(\bold{N}\bullet\bold{N}\bullet\bold{\Omega})+tr(\bold{\Omega}\bullet\bold{\Omega}\bullet\bold{N})\\\\ &\quad\ \ =\mathscr{C}_3^N+\mathscr{C}_3^{\bold{\Omega}}-\frac{1}{2}\mathscr{C}^N_1\mathscr{C}^{*\Omega}_2+tr(\bold{\Omega}\bullet\bold{N}\bullet\bold{\Omega})\\\\ &\quad\ \ =\mathscr{C}_3^N+\mathscr{C}^{\Omega}_2+\mathscr{C}^{\bold{\Omega}\bullet\bold{N}\bullet\bold{\Omega}}_1 \end{aligned} C1T=tr(T)=Tii=Nii+Ωii=tr(N)+tr(Ω)=C1N+C1Ω=C1NC2T=21[tr(T)2−tr(T∙T)] =21[tr(N)2−tr(N∙N+Ω∙Ω+Ω∙N+N∙Ω)] =21[tr(N)2−tr(N∙N)]−21tr(Ω∙Ω) =C2N+C2ΩC3T=61tr(T)3−21tr(T)tr(T∙T)+31tr(T∙T∙T) =61tr(N)3−21tr(N)[tr(N∙N)+tr(Ω∙Ω)+2tr(N∙Ω)]+31[tr(N∙N∙N)+tr(Ω∙Ω∙Ω)+3tr(N∙N∙Ω)+3tr(Ω∙Ω∙N)] =[61tr(N)3−21tr(N)tr(N∙N)+31tr(N∙N∙N)]+[31tr(Ω∙Ω∙Ω)]−21tr(N)tr(Ω∙Ω)−tr(N)tr(N∙Ω)+tr(N∙N∙Ω)+tr(Ω∙Ω∙N) =C3N+C3Ω−21C1NC2∗Ω+tr(Ω∙N∙Ω) =C3N+C2Ω+C1Ω∙N∙Ω
Remark： t r ( N ∙ Ω ) = t r ( Ω ∙ N ) = N i j Ω j i = N j i Ω i j = − N i j Ω j i ⟹ t r ( N ∙ Ω ) = t r ( Ω ∙ N ) = 0 t r ( N ∙ N ∙ Ω ) = N ∙ j i N ∙ k j Ω ∙ i k t r ( N ∙ Ω ∙ N ) = N ∙ j i Ω ∙ k j N ∙ i k = N ∙ k j Ω ∙ i k N ∙ j i t r ( Ω ∙ N ∙ N ) = Ω ∙ j i N ∙ k j N ∙ i k = Ω ∙ i k N ∙ j i N ∙ k j } ⟹ t r ( N ∙ N ∙ Ω ) = t r ( N ∙ Ω ∙ N ) = t r ( Ω ∙ N ∙ N ) t r ( N ∙ Ω ∙ Ω ) = t r ( Ω ∙ N ∙ Ω ) = t r ( Ω ∙ Ω ∙ N ) ( N ∙ Ω ∙ N ) T = N T ∙ Ω T ∙ N T = − ( N ∙ Ω ∙ N ) ⟹ t r ( N ∙ Ω ∙ N ) = C 1 N ∙ Ω ∙ N = 0 \begin{aligned} &tr(\bold{N}\bullet\bold{\Omega})=tr(\bold{\Omega}\bullet\bold{N})=N^{ij}\Omega_{ji}=N^{ji}\Omega_{ij}=-N^{ij}\Omega_{ji}\Longrightarrow tr(\bold{N}\bullet\bold{\Omega})=tr(\bold{\Omega}\bullet\bold{N})=0\\\\ &\left.\begin{aligned} &tr(\bold{N}\bullet\bold{N}\bullet\bold{\Omega})=N^i_{\bullet j}N^j_{\bullet k}{\Omega}^k_{\bullet i}\\\\ &tr(\bold{N}\bullet\bold{\Omega}\bullet\bold{N})=N^i_{\bullet j}{\Omega}^j_{\bullet k}N^k_{\bullet i}=N^j_{\bullet k}{\Omega}^k_{\bullet i}N^i_{\bullet j}\\\\ &tr(\bold{\Omega}\bullet\bold{N}\bullet\bold{N})={\Omega}^i_{\bullet j}N^j_{\bullet k}N^k_{\bullet i}={\Omega}^k_{\bullet i}N^i_{\bullet j}N^j_{\bullet k} \end{aligned}\right\}\Longrightarrow tr(\bold{N}\bullet\bold{N}\bullet\bold{\Omega})=tr(\bold{N}\bullet\bold{\Omega}\bullet\bold{N})=tr(\bold{\Omega}\bullet\bold{N}\bullet\bold{N})\\\\ &tr(\bold{N}\bullet\bold{\Omega}\bullet\bold{\Omega})=tr(\bold{\Omega}\bullet\bold{N}\bullet\bold{\Omega})=tr(\bold{\Omega}\bullet\bold{\Omega}\bullet\bold{N})\\\\ &(\bold{N}\bullet\bold{\Omega}\bullet\bold{N})^T=\bold{N}^T\bullet\bold{\Omega}^T\bullet\bold{N}^T=-(\bold{N}\bullet\bold{\Omega}\bullet\bold{N})\Longrightarrow tr(\bold{N}\bullet\bold{\Omega}\bullet\bold{N})=\mathscr{C}_1^{\bold{N}\bullet\bold{\Omega}\bullet\bold{N}}=0 \end{aligned} tr(N∙Ω)=tr(Ω∙N)=NijΩji=NjiΩij=−NijΩji⟹tr(N∙Ω)=tr(Ω∙N)=0tr(N∙N∙Ω)=N∙jiN∙kjΩ∙iktr(N∙Ω∙N)=N∙jiΩ∙kjN∙ik=N∙kjΩ∙ikN∙jitr(Ω∙N∙N)=Ω∙jiN∙kjN∙ik=Ω∙ikN∙jiN∙kj⎭ ⎬ ⎫⟹tr(N∙N∙Ω)=tr(N∙Ω∙N)=tr(Ω∙N∙N)tr(N∙Ω∙Ω)=tr(Ω∙N∙Ω)=tr(Ω∙Ω∙N)(N∙Ω∙N)T=NT∙ΩT∙NT=−(N∙Ω∙N)⟹tr(N∙Ω∙N)=C1N∙Ω∙N=0
3. 球形张量与偏斜张量及其特征值与特征方向
任意对称仿射量 N \bold N N（任意二阶张量 T \bold T T的对称部分）可分解为球形张量 P \bold P P与偏斜张量 S \bold S S的和，即：
N = P + S ( c ) \bold{N}=\bold{P}+\bold{S} \qquad (c) N=P+S(c)
其中，
P = 1 3 C 1 N G = 1 3 C 1 T G \bold{P}=\frac{1}{3}\mathscr{C}^N_1\bold{G}=\frac{1}{3}\mathscr{C}^T_1\bold{G} P=31C1NG=31C1TG
注意到，球形张量与偏斜张量均为对称仿射量。
对于球形张量，任意方向均为其特征方向，特征值为 1 3 C 1 N \frac{1}{3}\mathscr{C}^N_1 31C1N，这是由于，对于任意方向 v ⃗ \vec{v} v 均有：
P ∙ v ⃗ = 1 3 C 1 N G ∙ v ⃗ = 1 3 C 1 N v ⃗ v ⃗ ∙ P = 1 3 C 1 N v ⃗ ∙ G = 1 3 C 1 N v ⃗ \bold{P}\bullet\vec{v}=\frac{1}{3}\mathscr{C}^N_1\bold{G}\bullet\vec{v}=\frac{1}{3}\mathscr{C}^N_1\vec{v}\\\ \\ \vec{v}\bullet\bold{P}=\frac{1}{3}\mathscr{C}^N_1\vec{v}\bullet\bold{G}=\frac{1}{3}\mathscr{C}^N_1\vec{v} P∙v=31C1NG∙v=31C1Nv v∙P=31C1Nv∙G=31C1Nv
设对称张量 N \bold N N的三个特征值及其对应的线性无关的特征向量分别为:
λ 1 − u ⃗ 1 、 λ 2 − u ⃗ 2 、 λ 3 − u ⃗ 3 \lambda_1-\vec{u}_1、\lambda_2-\vec{u}_2、\lambda_3-\vec{u}_3 λ1−u1、λ2−u2、λ3−u3
则有：
( N − P ) ∙ u ⃗ i = ( λ i − 1 3 C 1 N ) u ⃗ i = S ∙ u ⃗ i (\bold{N-P})\bullet\vec{u}_i=(\lambda_i-\frac{1}{3}\mathscr{C}^N_1)\vec{u}_i=\bold S\bullet\vec{u}_i (N−P)∙ui=(λi−31C1N)ui=S∙ui
故，对称仿射量的特征向量与其偏量部分具有相同的特征方向，而二者的特征值满足：
λ i S = λ i N − 1 3 C 1 N ， u ⃗ i S = u ⃗ i N \lambda_i^S=\lambda_i^N-\frac{1}{3}\mathscr{C}^N_1，\vec{u}_i^S=\vec{u}_i^N λiS=λiN−31C1N，uiS=uiN
4. 对称仿射量球(偏)量部分主不变量与对称仿射量主不变量之间的联系
C 1 P = t r ( P ) = P ∙ i i = 1 3 C 1 N δ i i = C 1 N C 2 P = 1 2 [ t r 2 ( P ) − t r ( P 2 ) ] = 1 2 { ( C 1 N ) 2 − t r [ 1 9 ( C 1 N ) 2 G ] } = 1 3 ( C 1 N ) 2 C 3 P = 1 6 t r 3 ( P ) − 1 2 t r ( P ) t r ( P 2 ) + 1 3 t r ( P 3 ) = 1 2 ( C 1 N ) 3 C 1 S = t r ( N ) − t r ( P ) = 0 C 2 S = − 1 2 t r ( S 2 ) = − 1 2 [ t r ( N 2 ) − 2 t r ( N ∙ 1 3 C 1 N G ) + t r ( P 2 ) ] = − 1 2 [ ( C 1 N ) 2 − 2 C 2 N − 2 3 ( C 1 N ) 2 + 1 3 ( C 1 N ) 2 ] = C 2 N − 1 3 ( C 1 N ) 2 ≤ 0 C 3 S = 1 3 t r [ ( N − P ) 3 ] = 1 3 [ t r ( N 3 ) − t r ( P 3 ) − 2 t r ( N 2 ∙ P ) + 2 t r ( N ∙ P 2 ) − t r ( N ∙ P ∙ N ) + t r ( P ∙ N ∙ P ) ] = 1 3 [ t r ( N 3 ) + 2 9 ( C 1 N ) 3 − C 1 N t r ( N 2 ) ] = C 3 N − 1 3 C 1 N C 2 N + 2 27 ( C 1 N ) 3 \begin{aligned} &\mathscr{C}^P_1=tr(\bold P)=P^i_{\bullet i}=\frac{1}{3}\mathscr{C}^N_1\delta^{i}_i=\mathscr{C}^N_1 \\\ \\ &\mathscr{C}^P_2=\frac{1}{2}[tr^2(\bold P)-tr(\bold P^2)]=\frac{1}{2}\{(\mathscr{C}^N_1)^2-tr[\frac{1}{9}(\mathscr{C}^N_1)^2\bold G]\}=\frac{1}{3}(\mathscr{C}^N_1)^2 \\\ \\ &\mathscr{C}^P_3=\frac{1}{6}tr^3(\bold{P})-\frac{1}{2}tr(\bold{P})tr(\bold{P}^2)+\frac{1}{3}tr(\bold{P^3})=\frac{1}{2}(\mathscr{C}^N_1)^3 & \\ \\ \\ &\mathscr{C}^S_1=tr(\bold N)-tr(\bold P)=0 \\\ \\ &\mathscr{C}^S_2=-\frac{1}{2}tr(\bold S^2)=-\frac{1}{2}[tr(\bold N^2)-2tr(\bold {N}\bullet\frac{1}{3}\mathscr{C}^N_1\bold{G})+tr(\bold P^2)]\\\ \\ &\quad\ =-\frac{1}{2}[(\mathscr{C}^N_1)^2-2\mathscr{C}^N_2-\frac{2}{3}(\mathscr{C}^N_1)^2+\frac{1}{3}(\mathscr{C}^N_1)^2]=\mathscr{C}^N_2-\frac{1}{3}(\mathscr{C}^N_1)^2\le0 \\\ \\ &\mathscr{C}^S_3=\frac{1}{3}tr[\bold{(N-P)^3}]\\\ \\ &\quad\ =\frac{1}{3}[tr(\bold N^3)-tr(\bold P^3)-2tr(\bold{N}^2\bullet \bold{P})+2tr(\bold{N\bullet P}^2)-tr(\bold{N\bullet P\bullet N})+tr(\bold{P\bullet N\bullet P})] \\\ \\ &\quad\ = \frac{1}{3}[tr(\bold N^3)+\frac{2}{9}(\mathscr{C}^N_1)^3-\mathscr{C}^N_1tr(\bold N^2)] \\\ \\ &\quad\ =\mathscr{C}^N_3-\frac{1}{3}\mathscr{C}^N_1\mathscr{C}^N_2+\frac{2}{27}(\mathscr{C}^N_1)^3 \end{aligned} C1P=tr(P)=P∙ii=31C1Nδii=C1NC2P=21[tr2(P)−tr(P2)]=21{(C1N)2−tr[91(C1N)2G]}=31(C1N)2C3P=61tr3(P)−21tr(P)tr(P2)+31tr(P3)=21(C1N)3C1S=tr(N)−tr(P)=0C2S=−21tr(S2)=−21[tr(N2)−2tr(N∙31C1NG)+tr(P2)] =−21[(C1N)2−2C2N−32(C1N)2+31(C1N)2]=C2N−31(C1N)2≤0C3S=31tr[(N−P)3] =31[tr(N3)−tr(P3)−2tr(N2∙P)+2tr(N∙P2)−tr(N∙P∙N)+tr(P∙N∙P)] =31[tr(N3)+92(C1N)3−C1Ntr(N2)] =C3N−31C1NC2N+272(C1N)3
偏斜张量的第二不变量恒为非正的理由如下：
设对称张量 N \bold N N的特征值分别为：
λ 1 、 λ 2 、 λ 3 ∈ R \lambda_1、\lambda_2、\lambda_3\in R λ1、λ2、λ3∈R
那么，
C 1 N = λ 1 + λ 2 + λ 3 C 2 N = λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 ⟹ C 2 S = C 2 N − 1 3 ( C 1 N ) 2 = − 1 6 [ ( λ 1 − λ 2 ) 2 + ( λ 1 − λ 3 ) 2 + ( λ 2 − λ 3 ) 2 ] ≤ 0 ( 当且仅当有三重特征值时为零 ) \begin{aligned} &\qquad\ \mathscr{C}^N_1=\lambda_1+\lambda_2+\lambda_3 \\\ \\ &\qquad\ \mathscr{C}^N_2=\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3\\\ \\ &\Longrightarrow\mathscr{C}^S_2=\mathscr{C}^N_2-\frac{1}{3}(\mathscr{C}^N_1)^2\\\\ &\qquad\quad\ \ =-\frac{1}{6}[(\lambda_1-\lambda_2)^2+(\lambda_1-\lambda_3)^2+(\lambda_2-\lambda_3)^2]\le0\\\\ &\qquad\quad\ \ (当且仅当有三重特征值时为零) \end{aligned} C1N=λ1+λ2+λ3 C2N=λ1λ2+λ1λ3+λ2λ3⟹C2S=C2N−31(C1N)2 =−61[(λ1−λ2)2+(λ1−λ3)2+(λ2−λ3)2]≤0 (当且仅当有三重特征值时为零)
（证毕）