多智能体模型
{ x ˙ i ( t ) = v i ( t ) v ˙ i ( t ) = u i ( t ) + f ( x i , v i ) + d i ( t ) \begin{equation} \begin{cases} {\dot x}_i(t) = v_i(t) \\ {\dot v}_i(t) = u_i(t) + f(x_i,v_i) + d_i(t) \end{cases} \end{equation} {x˙i(t)=vi(t)v˙i(t)=ui(t)+f(xi,vi)+di(t)
引理1 假设向量场在相关三元组 ( r ∞ , σ ∞ , ψ ∞ ) (r_{\infty},\sigma_{\infty},\psi_{\infty}) (r∞,σ∞,ψ∞) 与 ( r 0 , σ 0 , ψ 0 ) (r_{0},\sigma_{0},\psi_{0}) (r0,σ0,ψ0) 的双极限上是齐次的。如果系统的原点是全局渐进稳定的,并且存在 σ ∞ > 0 > σ 0 \sigma_{\infty} > 0 > \sigma_{0} σ∞>0>σ0,那么系统的所有解可以在固定时间内收敛到系统的原点。
引理2 L + B L+B L+B 为正定矩阵
引理3 ∑ i = 1 N ∑ j = 1 N q i j y i ϕ ( z i − z j ) = 1 2 ∑ i = 1 N ∑ j = 1 N q i j ( y i − y j ) ϕ ( z i − z j ) \sum_{i=1}^N\sum_{j=1}^N q_{ij} y_i \phi(z_i - z_j) = \frac{1}{2}\sum_{i=1}^N\sum_{j=1}^N q_{ij} (y_i-y_j) \phi(z_i - z_j) ∑i=1N∑j=1Nqijyiϕ(zi−zj)=21∑i=1N∑j=1Nqij(yi−yj)ϕ(zi−zj) 其中 Q = [ q i j ] Q = [q_{ij}] Q=[qij] 是一个对称矩阵, ϕ ( ⋅ ) \phi(\cdot) ϕ(⋅) 是一个奇函数
引理4 x , y ∈ R x,y \in \mathbb{R} x,y∈R, k 1 , k 2 > 0 k_1,k_2 > 0 k1,k2>0,有 ∣ x ∣ k 1 ∣ x ∣ k 2 ≤ k 1 / ( k 1 + k 2 ) ∣ x ∣ k 1 + k 2 + k 2 / ( k 1 + k 2 ) ∣ y ∣ k 1 + k 2 |x|^{k_1} |x|^{k_2} \le k_1 / (k_1 + k_2) |x|^{k_1 + k_2} + k_2/(k_1+k_2) |y|^{k_1+k_2} ∣x∣k1∣x∣k2≤k1/(k1+k2)∣x∣k1+k2+k2/(k1+k2)∣y∣k1+k2 。
引理5 $\frac{1}{2} \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij} |x_i - x_j|^2 = x^T \hat L x $
引理6 $x^T \hat L x \ge K x^T x $ 其中 K = ∣ ∣ p ∣ ∣ 1 2 N ∣ ∣ p ∣ ∣ 2 2 λ 2 ( L ^ ) K = \frac{||p||_1^2}{N ||p||_2^2 \lambda_2(\hat L)} K=N∣∣p∣∣22λ2(L^)∣∣p∣∣12 p = ( p 1 , … , p N ) T p = (p_1,\dots,p_N)^T p=(p1,…,pN)T 是图 G \mathcal{G} G 权重向量, L L L 是拉普拉斯矩阵, L ^ = d i a g ( p ) L \hat L = diag(p) L L^=diag(p)L
假设1 非线性函数 f f f 满足下面的条件
∣ f ( z 1 , z 2 ) − f ( z 1 ′ , z 2 ′ ) ∣ ≤ ρ ( ∣ z 1 − z 1 ′ ∣ + ∣ z 2 − z 2 ′ ∣ ) |f(z_1,z_2) - f(z_1',z_2')| \le \rho(|z_1 - z_1'| + |z_2 - z_2'|) ∣f(z1,z2)−f(z1′,z2′)∣≤ρ(∣z1−z1′∣+∣z2−z2′∣)
其中 z 1 , z 2 , z 1 ′ , z 2 ′ ∈ R z_1,z_2,z_1',z_2'\in \mathbb{R} z1,z2,z1′,z2′∈R, ρ \rho ρ 是正常数
设计控制协议如下
u i ( t ) = u 1 i ( t ) + u 1 i ( t ) + u 1 i ( t ) u 1 i ( t ) = − l 1 ∑ j = 0 N a i j ⌈ x i − x j ⌋ α 1 − l 1 ∑ j = 0 N a i j ⌈ v i − v j ⌋ α 1 ′ u 2 i ( t ) = − l 2 ∑ j = 0 N a i j ⌈ x i − x j ⌋ α 2 − l 2 ∑ j = 0 N a i j ⌈ v i − v j ⌋ α 2 ′ u 3 i ( t ) = − k ∑ j = 0 N a i j ( x i − x j ) − k ∑ j = 0 N a i j ( v i − v j ) \begin{aligned} u_i(t) &= u_{1i}(t) + u_{1i}(t) + u_{1i}(t) \\ u_{1i}(t) &= -l_1 \sum_{j=0}^N a_{ij} \lceil x_i - x_j \rfloor^{\alpha_{1}} - l_1 \sum_{j=0}^N a_{ij} \lceil v_i - v_j \rfloor^{\alpha_{1}'} \\ u_{2i}(t) &= -l_2 \sum_{j=0}^N a_{ij} \lceil x_i - x_j \rfloor^{\alpha_{2}} - l_2 \sum_{j=0}^N a_{ij} \lceil v_i - v_j \rfloor^{\alpha_{2}'} \\ u_{3i}(t) &= -k \sum_{j=0}^N a_{ij} (x_i - x_j) -k \sum_{j=0}^N a_{ij} (v_i - v_j) \end{aligned} ui(t)u1i(t)u2i(t)u3i(t)=u1i(t)+u1i(t)+u1i(t)=−l1j=0∑Naij⌈xi−xj⌋α1−l1j=0∑Naij⌈vi−vj⌋α1′=−l2j=0∑Naij⌈xi−xj⌋α2−l2j=0∑Naij⌈vi−vj⌋α2′=−kj=0∑Naij(xi−xj)−kj=0∑Naij(vi−vj)
其中 q 1 > 0 , l 2 > 0 , k > 0 , 0 < α 1 < 1 , α 1 ′ = 2 α 1 / ( 1 + α 1 ) , α 2 > 1 , α 2 ′ = 2 α 2 / ( 1 + α 2 ) q_1>0,l_2>0,k>0,0<\alpha_1<1,\alpha_1'=2\alpha_1/(1+\alpha_1),\alpha_2>1,\alpha_2'=2\alpha_2/(1+\alpha_2) q1>0,l2>0,k>0,0<α1<1,α1′=2α1/(1+α1),α2>1,α2′=2α2/(1+α2), 以及 a i 0 > 0 a_{i0}>0 ai0>0 如果智能体 i i i 可以获取到leader的信息,定义 B = d i a g ( a 10 , a 20 , ⋯ , a N 0 ) B = diag(a_{10},a_{20},\cdots,a_{N0}) B=diag(a10,a20,⋯,aN0)。
k ≥ c / 2 + ( ( 3 + 4 ρ ) p m a x + l 4 ) / ( 2 K ) k \ge c/2 + ((3+4 \rho)p_{max} + l_4)/(2K) k≥c/2+((3+4ρ)pmax+l4)/(2K)
c ≥ 2 l 1 / ( 1 + α 1 ′ ) + 2 l 2 / ( 1 + α 2 ′ ) + ( ( 2 ρ + 1 ) p m a x + l 3 ) / K c \ge 2l_1 / (1 + \alpha_1') + 2l_2 / (1 + \alpha_2') + ((2\rho + 1)p_{max} + l_3)/K c≥2l1/(1+α1′)+2l2/(1+α2′)+((2ρ+1)pmax+l3)/K
K = ∣ ∣ p ∣ ∣ 1 2 N ∣ ∣ p ∣ ∣ 2 2 λ 2 ( L ^ ) K = \frac{||p||_1^2}{N ||p||_2^2 \lambda_2(\hat L)} K=N∣∣p∣∣22λ2(L^)∣∣p∣∣12
定义 $\tilde{x}_i = x_i - x_0, \tilde{v}_i = v_i - v_0, i = 1,2,3,\cdots,N $, 作为共识误差,可以得到 x i − x j = x ~ i − x ~ j x_i-x_j = \tilde{x}_i - \tilde{x}_j xi−xj=x~i−x~j 。同时定义向量 x ~ = [ x ~ 1 , x ~ 2 , ⋯ , x ~ N ] T , v ~ = [ v ~ 1 , v ~ 2 , ⋯ , v ~ N ] T , z = [ x ~ T , v ~ T ] , u = [ u 1 , u 2 , u 3 , ⋯ , u N ] \tilde{x} = [\tilde{x}_1,\tilde{x}_2,\cdots,\tilde{x}_N]^T, \tilde{v} = [\tilde{v}_1,\tilde{v}_2,\cdots,\tilde{v}_N]^T,z = [\tilde{x}^T,\tilde{v}^T],u=[u_1,u_2,u_3,\cdots,u_N] x~=[x~1,x~2,⋯,x~N]T,v~=[v~1,v~2,⋯,v~N]T,z=[x~T,v~T],u=[u1,u2,u3,⋯,uN] 。
因此可以获得误差系统
{ x ~ ˙ i = v ~ i v ~ ˙ i = u i + f ( x i , v i ) − f 0 ( x 0 , v 0 ) \begin{cases} \dot{\tilde{x}}_i = \tilde{v}_i \\ \dot{\tilde{v}}_i = u_i + f(x_i,v_i) - f_0(x_0,v_0) \end{cases} {x~˙i=v~iv~˙i=ui+f(xi,vi)−f0(x0,v0)
整体的证明可以分为三部分,第一部分我们证明上述方程右侧矢量场在双极限上是齐次的,第二部分我们证明上述方程的0极限和1极限逼近系统是全局渐进稳定的,最后我们证明了整个闭环全局系统是渐进稳定的。根据引理1证明了系统可以在固定时间内收敛。
证明 ψ ( z ) = ( v ~ 1 , ⋯ , v ~ N , u 1 + f ( x 1 , v 1 ) − f ( x 0 , v 0 ) , ⋯ , u N + f ( x N , v N ) − f ( x 0 , v 0 ) ) ∈ R 2 N \psi(z) = (\tilde{v}_1,\cdots,\tilde{v}_N,u_1+f(x_1,v_1)-f(x_0,v_0), \cdots, u_N + f(x_N,v_N) - f(x_0,v_0)) \in \mathbb{R}^{2N} ψ(z)=(v~1,⋯,v~N,u1+f(x1,v1)−f(x0,v0),⋯,uN+f(xN,vN)−f(x0,v0))∈R2N 是与相关三元组 ( r ∞ , α 2 − 1 , ψ ∞ ) (r_{\infty},\alpha_2 - 1,\psi_{\infty}) (r∞,α2−1,ψ∞) 与 ( r 0 , α 1 − 1 , ψ 0 ) (r_{0},\alpha_1 - 1,\psi_{0}) (r0,α1−1,ψ0) 双极限齐次的,其中 r ∞ = ( 2 , ⋯ , 2 , 1 + α 2 , ⋯ , 1 + α 2 ) r_{\infty} = (2,\cdots,2,1+\alpha_2,\cdots,1+\alpha_2) r∞=(2,⋯,2,1+α2,⋯,1+α2) , r 0 = ( 2 , ⋯ , 2 , 1 + α 1 , ⋯ , 1 + α 1 ) r_0 = (2,\cdots,2,1+\alpha_1,\cdots,1+\alpha_1) r0=(2,⋯,2,1+α1,⋯,1+α1), ψ ∞ = ( v ~ 1 , ⋯ , v ~ N , u 21 , ⋯ , u 2 N ) \psi_{\infty} = (\tilde{v}_1,\cdots,\tilde{v}_N, u_{21},\cdots,u_{2N}) ψ∞=(v~1,⋯,v~N,u21,⋯,u2N), ψ 0 = ( v ~ 1 , ⋯ , v ~ N , u 11 , ⋯ , u 1 N ) \psi_{0} = (\tilde{v}_1,\cdots,\tilde{v}_N, u_{11},\cdots,u_{1N}) ψ0=(v~1,⋯,v~N,u11,⋯,u1N).
首先证明无穷极限
ψ ( x ~ 1 , ⋯ , x ~ N , v ~ 1 , ⋯ , v ~ N ) = ( v ~ 1 , ⋯ , v ~ N , u 1 + f ( x 1 , v 1 ) − f ( x 0 , v 0 ) , ⋯ , u N + f ( x N , v N ) − f ( x 0 , v 0 ) ) \psi(\tilde{x}_1,\cdots,\tilde{x}_N,\tilde{v}_1,\cdots,\tilde{v}_N) = (\tilde{v}_1,\cdots,\tilde{v}_N,u_1+f(x_1,v_1)-f(x_0,v_0), \cdots, u_N + f(x_N,v_N) - f(x_0,v_0)) ψ(x~1,⋯,x~N,v~1,⋯,v~N)=(v~1,⋯,v~N,u1+f(x1,v1)−f(x0,v0),⋯,uN+f(xN,vN)−f(x0,v0))
ψ ( λ r ∞ ) = ψ ( λ 2 x ~ 1 , ⋯ , λ 2 x ~ N , λ α 2 + 1 v ~ 1 , ⋯ , λ α 2 + 1 v ~ N ) \psi(\lambda^{r_{\infty}}) = \psi(\lambda^2\tilde{x}_1,\cdots,\lambda^2\tilde{x}_N,\lambda^{\alpha_2+1}\tilde{v}_1,\cdots,\lambda^{\alpha_2+1}\tilde{v}_N) ψ(λr∞)=ψ(λ2x~1,⋯,λ2x~N,λα2+1v~1,⋯,λα2+1v~N)
d ∞ = ( α 2 + 1 , ⋯ , α 2 + 1 , 2 α 2 , ⋯ , 2 α 2 ) d_{\infty} = (\alpha_2+1,\cdots,\alpha_2+1,2\alpha_2,\cdots,2\alpha_2) d∞=(α2+1,⋯,α2+1,2α2,⋯,2α2)
先验证前 N N N 项,之前是需要 v ~ 1 \tilde{v}_1 v~1 ,现在经过一系列变换已经变成了 λ α 2 + 1 v ~ 1 \lambda^{\alpha_2+1}\tilde{v}_1 λα2+1v~1 ,所以
lim λ → 0 max z ∈ C ∣ λ α 2 + 1 v ~ 1 λ α 2 + 1 − v ~ 1 ∣ = 0 \lim_{\lambda \to 0} \max_{z \in C} \left| \frac{\lambda^{\alpha_2+1}\tilde{v}_1}{\lambda^{\alpha_2 + 1}} - \tilde{v}_1\right| = 0 λ→0limz∈Cmax λα2+1λα2+1v~1−v~1 =0
恒成立,第一个三元组也就验证成功了,也就是 ( 2 , α 2 + 1 , v ~ 1 ) (2,\alpha_2 + 1, \tilde{v}_1) (2,α2+1,v~1),同理可得前 N N N 个都是这样
再验证后 N N N 项,其中
u 1 i ( λ 2 x ~ 1 , ⋯ , λ 2 x ~ N , λ α 2 + 1 v ~ 1 , ⋯ , λ α 2 + 1 v ~ N ) = − l 1 λ 2 α 1 ∑ j = 0 N a i j ⌈ x i − x j ⌋ α 1 − l 1 λ ( α 2 + 1 ) α 1 ′ ∑ j = 0 N a i j ⌈ v i − v j ⌋ α 1 ′ u_{1i}(\lambda^2\tilde{x}_1,\cdots,\lambda^2\tilde{x}_N,\lambda^{\alpha_2+1}\tilde{v}_1,\cdots,\lambda^{\alpha_2+1}\tilde{v}_N) = -l_1 \lambda^{2\alpha_1} \sum_{j=0}^N a_{ij} \lceil x_i - x_j \rfloor^{\alpha_{1}} - l_1 \lambda^{(\alpha_2 + 1)\alpha_1'}\sum_{j=0}^N a_{ij} \lceil v_i - v_j \rfloor^{\alpha_{1}'} u1i(λ2x~1,⋯,λ2x~N,λα2+1v~1,⋯,λα2+1v~N)=−l1λ2α1j=0∑Naij⌈xi−xj⌋α1−l1λ(α2+1)α1′j=0∑Naij⌈vi−vj⌋α1′
lim λ → ∞ u 1 i ( λ 2 x ~ 1 , ⋯ , λ 2 x ~ N , λ α 2 + 1 v ~ 1 , ⋯ , λ α 2 + 1 v ~ N ) λ 2 α 2 → 0 \lim_{\lambda \to \infty} \frac{u_{1i}(\lambda^2\tilde{x}_1,\cdots,\lambda^2\tilde{x}_N,\lambda^{\alpha_2+1}\tilde{v}_1,\cdots,\lambda^{\alpha_2+1}\tilde{v}_N)}{\lambda^{2\alpha_2}} \to 0 λ→∞limλ2α2u1i(λ2x~1,⋯,λ2x~N,λα2+1v~1,⋯,λα2+1v~N)→0
同理可得
lim λ → ∞ u 2 i ( λ 2 x ~ 1 , ⋯ , λ 2 x ~ N , λ α 2 + 1 v ~ 1 , ⋯ , λ α 2 + 1 v ~ N ) λ 2 α 2 = u 2 i ( z ) \lim_{\lambda \to \infty} \frac{u_{2i}(\lambda^2\tilde{x}_1,\cdots,\lambda^2\tilde{x}_N,\lambda^{\alpha_2+1}\tilde{v}_1,\cdots,\lambda^{\alpha_2+1}\tilde{v}_N)}{\lambda^{2\alpha_2}} = u_{2i}(z) λ→∞limλ2α2u2i(λ2x~1,⋯,λ2x~N,λα2+1v~1,⋯,λα2+1v~N)=u2i(z)
lim λ → ∞ u 3 i ( λ 2 x ~ 1 , ⋯ , λ 2 x ~ N , λ α 2 + 1 v ~ 1 , ⋯ , λ α 2 + 1 v ~ N ) λ 2 α 2 → 0 \lim_{\lambda \to \infty} \frac{u_{3i}(\lambda^2\tilde{x}_1,\cdots,\lambda^2\tilde{x}_N,\lambda^{\alpha_2+1}\tilde{v}_1,\cdots,\lambda^{\alpha_2+1}\tilde{v}_N)}{\lambda^{2\alpha_2}} \to 0 λ→∞limλ2α2u3i(λ2x~1,⋯,λ2x~N,λα2+1v~1,⋯,λα2+1v~N)→0
∣ f ( λ 2 x i , λ 1 + α 2 v i ) − ∑ j = 1 N p j f ( λ 2 x j , λ 1 + α 2 v j ) ∣ λ 2 α 2 ≤ ∑ j = 1 N p j ∣ f ( λ 2 x i , λ 1 + α 2 v i ) − f ( λ 2 x j , λ 1 + α 2 v j ) ∣ λ 2 α 2 ≤ ρ ∑ j = 1 N p j ( λ 2 ∣ x i − x j ∣ − λ 1 + α 2 ∣ v i − v j ∣ ) λ 2 α 2 → 0 \begin{aligned} & \frac{\left|f(\lambda^2 x_i, \lambda^{1 + \alpha_2} v_i) - \sum_{j=1}^{N}p_j f(\lambda^2 x_j, \lambda^{1 + \alpha_2} v_j) \right|}{\lambda^{2 \alpha_2}} \\ & \le \frac{\sum_{j=1}^{N}p_j |f(\lambda^2 x_i, \lambda^{1 + \alpha_2} v_i) - f(\lambda^2 x_j, \lambda^{1 + \alpha_2} v_j) | }{\lambda^{2 \alpha_2}} \\ & \le \frac{\rho\sum_{j=1}^{N}p_j (\lambda^2|x_i - x_j| - \lambda^{1 + \alpha_2} |v_i - v_j|)}{\lambda^{2 \alpha_2}} \to 0 \end{aligned} λ2α2 f(λ2xi,λ1+α2vi)−∑j=1Npjf(λ2xj,λ1+α2vj) ≤λ2α2∑j=1Npj∣f(λ2xi,λ1+α2vi)−f(λ2xj,λ1+α2vj)∣≤λ2α2ρ∑j=1Npj(λ2∣xi−xj∣−λ1+α2∣vi−vj∣)→0
可以看出 ψ i ′ \psi_{i'} ψi′ 与三元组 ( r ∞ , 2 α 2 , u 2 , i ′ − N ) (r_{\infty},2\alpha_2,u_{2,i'-N}) (r∞,2α2,u2,i′−N) 是无穷齐次的,其中 i ′ = N + 1 , … , 2 N i' = N+1,\dots,2N i′=N+1,…,2N.
因此证明向量场(vector field) ψ \psi ψ 与三元组 ( r 0 , α 1 − 1 , ψ 0 ) (r_{0},\alpha_1 - 1,\psi_{0}) (r0,α1−1,ψ0) 零齐次,与三元组 ( r ∞ , α 2 − 1 , ψ ∞ ) (r_{\infty},\alpha_2 - 1,\psi_{\infty}) (r∞,α2−1,ψ∞) 无穷齐次。由于0齐次与无穷齐次证明一致,所以不需要再证明0齐次。
首先证明 z ˙ = ψ ∞ \dot z = \psi_{\infty} z˙=ψ∞ 是全局渐进稳定的,考虑系统 z ˙ = ψ ∞ \dot z = \psi_{\infty} z˙=ψ∞ 为
{ x ~ ˙ i = v ~ i v ~ ˙ i = u 2 i \begin{equation} \begin{cases} {\dot {\tilde{x}}}_i = \tilde{v}_i \\ {\dot {\tilde{v}}}_i = u_{2i} \end{cases} \end{equation} {x~˙i=v~iv~˙i=u2i
选择李雅普诺夫函数
V = ∑ i = 1 N p i v ~ i 2 + l 2 1 + α 2 ∑ i = 1 N ∑ j = 1 N p i a i j ∣ x ~ i − x ~ j ∣ 1 + α 2 V = \sum_{i=1}^N p_i \tilde{v}_i^2 + \frac{l_2}{1 + \alpha_2} \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij} |\tilde{x}_i - \tilde{x}_j|^{1 + \alpha_2} V=i=1∑Npiv~i2+1+α2l2i=1∑Nj=1∑Npiaij∣x~i−x~j∣1+α2
求导可得
V ˙ = 2 ∑ i = 1 N p i x ~ i v ~ ˙ i + l 2 ∑ i = 1 N ∑ j = 1 N p i a i j ⌈ x ~ i − x ~ j ⌋ α 2 ( v ~ i − v ~ j ) = 2 ∑ i = 1 N p i v ~ i ( − l 2 ∑ j = 0 N a i j ⌈ x ~ i − x ~ j ⌋ α 2 − l 2 ∑ j = 0 N a i j ⌈ v ~ i − v ~ j ⌋ α 2 ′ ) + l 2 ∑ i = 1 N ∑ j = 1 N p i a i j ⌈ x ~ i − x ~ j ⌋ α 2 ( v ~ i − v ~ j ) = − l 2 ∑ i = 1 N ∑ j = 1 N p i a i j ∣ v ~ i − v ~ j ∣ 1 + α 2 ′ \begin{equation} \begin{aligned} \dot V &= 2\sum_{i=1}^N p_i \tilde{x}_i {\dot {\tilde{v}}}_i + l_2 \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij} \lceil \tilde{x}_i - \tilde{x}_j\rfloor^{\alpha_2}(\tilde{v}_i - \tilde{v}_j)\\ &= 2\sum_{i=1}^N p_i \tilde{v}_i \left( -l_2 \sum_{j=0}^N a_{ij} \lceil \tilde{x}_i - \tilde{x}_j \rfloor^{\alpha_{2}} - l_2 \sum_{j=0}^N a_{ij} \lceil \tilde{v}_i - \tilde{v}_j \rfloor^{\alpha_{2}'} \right) + l_2 \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij}\lceil \tilde{x}_i - \tilde{x}_j\rfloor^{\alpha_2}(\tilde{v}_i - \tilde{v}_j)\\ &= -l_2 \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij}| \tilde{v}_i - \tilde{v}_j |^{1 + \alpha_2^{'}} \end{aligned} \end{equation} V˙=2i=1∑Npix~iv~˙i+l2i=1∑Nj=1∑Npiaij⌈x~i−x~j⌋α2(v~i−v~j)=2i=1∑Npiv~i(−l2j=0∑Naij⌈x~i−x~j⌋α2−l2j=0∑Naij⌈v~i−v~j⌋α2′)+l2i=1∑Nj=1∑Npiaij⌈x~i−x~j⌋α2(v~i−v~j)=−l2i=1∑Nj=1∑Npiaij∣v~i−v~j∣1+α2′
定义不变集 Ω = { ( x 1 , … , x N , v 1 , … , v N ) ∣ V ˙ = 0 } \Omega = \{ (x_1,\dots,x_N,v_1,\dots,v_N) | \dot V = 0 \} Ω={(x1,…,xN,v1,…,vN)∣V˙=0} ,注意到当图为强链接时, V ˙ = 0 \dot V = 0 V˙=0 暗含 v i = v j = v ˉ v_i = v_j = \bar v vi=vj=vˉ,这意味着 u 2 i = u 2 j u_{2i} = u_{2j} u2i=u2j。 u 2 i = − l 2 ∑ j = 0 N a i j ⌈ x i − x j ⌋ α 2 u_{2i} = -l_2 \sum_{j=0}^N a_{ij} \lceil x_i - x_j \rfloor^{\alpha_{2}} u2i=−l2∑j=0Naij⌈xi−xj⌋α2,更进一步,我们有 p i a i j = p j a j i p_i a_{ij} = p_j a_{ji} piaij=pjaji,因此 ∑ i = 1 N p i u 2 i = 0 \sum_{i=1}^N p_i u_{2i} = 0 ∑i=1Npiu2i=0, 那么下面的等式成立
0 = ∑ i = 1 N p i x i u 2 i = − ∑ i = 1 N ∑ j = 1 N p i a i j x i ⌈ x i − x j ⌋ α 2 = − 1 2 ∑ i = 1 N ∑ j = 1 N p i a i j ⌈ x i − x j ⌋ 1 + α 2 0 = \sum_{i=1}^N p_i x_i u_{2i} = - \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij} x_i \lceil x_i - x_j\rfloor^{\alpha_2} = -\frac{1}{2} \sum_{i=1}^N \sum_{j=1}^N p_i a_{ij} \lceil x_i - x_j\rfloor^{1 + \alpha_2} 0=i=1∑Npixiu2i=−i=1∑Nj=1∑Npiaijxi⌈xi−xj⌋α2=−21i=1∑Nj=1∑Npiaij⌈xi−xj⌋1+α2
因此我们有 x i = x j = x ˉ x_i = x_j = \bar x xi=xj=xˉ, 根据拉塞尔不变性原理, x i − x j → 0 x_i - x_j \to 0 xi−xj→0 v i − v j = 0 v_i - v_j = 0 vi−vj=0 当 t → ∞ t \to \infty t→∞
由于 z ˙ = ψ 0 \dot z = \psi_{0} z˙=ψ0 的证明与上面类似,这里不在赘述
由于本文过长,拆分为两篇文章