超螺旋滑模控制(STA)

超螺旋滑模控制(Super Twisting Algorithm, STA)

超螺旋滑模控制又称超扭滑模控制,可以说是二阶系统中最好用的滑模控制方法。

系统模型

对于二阶系统可以建立具有标准柯西形式的微分方程组
{ x ˙ 1 = x 2 x ˙ 2 = f + g ⋅ u \begin{cases} \dot x_1 = x_2 \\ \dot x_2 = f + g \cdot u \end{cases} {x˙1=x2x˙2=f+gu
与传统滑模相比,超螺旋滑模,使用积分来获取实际控制量,不含高频切换量,所以系统中没有抖振。

令滑模面为s,只要满足以下的方程,即为稳定
{ s ˙ = − λ ∣ s ∣ 1 2 ⋅ s i g n ( s ) + ν ν ˙ = − α ⋅ s i g n ( s ) \begin{cases} \dot s = -\lambda |s| ^ {\frac {1} {2}} \cdot sign(s) + \nu \\ \dot \nu = - \alpha \cdot sign(s) \\ \end{cases} {s˙=λs21sign(s)+νν˙=αsign(s)

控制器设计

设状态 x 1 x_1 x1 的期望值为 x d x_d xd ,则跟踪误差为
{ e 1 = x 1 − x d e 2 = e ˙ 1 = x ˙ 1 − x ˙ d = x 2 − x ˙ d \begin{cases} e_1 = x_1 - x_d \\ e_2 = \dot e_1 = \dot x_1 - \dot x_d = x_2 - \dot x_d \end{cases} {e1=x1xde2=e˙1=x˙1x˙d=x2x˙d
设计滑模面为
s = c e 1 + e 2 s = ce_1 + e_2 s=ce1+e2
则滑模面的导数为
s ˙ = c e ˙ 1 + e ˙ 2 = c e ˙ 2 + f + g ⋅ u − x ¨ d = − λ ∣ s ∣ 1 2 ⋅ s i g n ( s ) + ν = − λ ∣ s ∣ 1 2 ⋅ s i g n ( s ) − α ⋅ s i g n ( s ) \begin{align} \dot s & = c \dot e_1 + \dot e_2 \nonumber \\ & = c \dot e_2 + f + g \cdot u - \ddot x_d \nonumber\\ & = -\lambda |s| ^ {\frac {1} {2}} \cdot sign(s) + \nu \nonumber\\ & = -\lambda |s| ^ {\frac {1} {2}} \cdot sign(s) - \alpha \cdot sign(s) \nonumber\\ \end{align} s˙=ce˙1+e˙2=ce˙2+f+gux¨d=λs21sign(s)+ν=λs21sign(s)αsign(s)
可以得到控制量
u = g − 1 ( − f + x ¨ d − c 1 e 2 − λ ∣ s ∣ 1 2 s i g n ( s ) − α ⋅ s i g n ( s ) ) u = g ^ {-1} (-f + \ddot x_d - c_1e_2 - \lambda |s| ^ {\frac {1} {2}}sign(s) - \alpha \cdot sign(s)) u=g1(f+x¨dc1e2λs21sign(s)αsign(s))
参数设定为
λ ˙ = ω 1 γ 1 2 α = λ ε + 1 2 ( β + 4 ε 2 ) \begin{align} \dot \lambda &= \omega _ 1 \sqrt{\frac {\gamma_1} {2}} \nonumber\\ \alpha &= \lambda \varepsilon + \frac{1}{2}(\beta+4\varepsilon ^ {2}) \nonumber \end{align} λ˙α=ω12γ1 =λε+21(β+4ε2)
式中, α , β , ε , ω 1 , γ 1 \alpha , \beta , \varepsilon , \omega_1 , \gamma_1 α,β,ε,ω1,γ1 均大于0。

稳定性证明

可以看出,控制量中含有的不再是滑模面,而是多项式 ∣ s ∣ 1 2 |s| ^ {\frac {1} {2}} s21 。除此之外,在 s ˙ \dot s s˙ 中还出现了另一个参数 ν \nu ν ,不妨把这两者定义为新的状态变量,在此基础上设成李雅普诺夫函数。
{ z 1 = ∣ s ∣ 1 2 z 2 = ν ⇒ { z ˙ 1 = 1 2 ∣ s ∣ − 1 2 s ˙ = 1 2 ∣ s ∣ − 1 2 ( − λ ∣ s ∣ 1 2 ⋅ s i g n ( s ) − α ⋅ s i g n ( s ) ) z ˙ 2 = ν ˙ = − α ⋅ s i g n ( s ) \begin{cases} z_1 = |s| ^ {\frac {1} {2}} \nonumber\\ z_2 = \nu \\ \end{cases} \Rightarrow \begin{cases} \dot z_1 = {\frac {1} {2}} |s| ^ {-\frac {1} {2}} \dot s = {\frac {1} {2}} |s| ^ {-\frac {1} {2}}(-\lambda |s| ^ {\frac {1} {2}} \cdot sign(s) - \alpha \cdot sign(s)) \\ \dot z_2 = \dot \nu = -\alpha \cdot sign(s) \\ \end{cases} {z1=s21z2=ν{z˙1=21s21s˙=21s21(λs21sign(s)αsign(s))z˙2=ν˙=αsign(s)
将第一项带入第二项
{ z ˙ 1 = 1 2 ∣ z 1 ∣ ( − λ z 1 + z 2 ) z ˙ 2 = ν ˙ = − α ⋅ s i g n ( s ) = − α ⋅ s i g n ( s ) ∣ s ∣ 1 2 ∣ s ∣ − 1 2 = − α z 1 ∣ z 1 ∣ ⇒ { z ˙ 1 = 1 2 ∣ z 1 ∣ ( − λ z 1 + z 2 ) z ˙ 2 = − α z 1 ∣ z 1 ∣ \begin{align} &\begin{cases} \dot z_1 = \frac {1} {2|z_1|}(-\lambda z_1 + z_2) \\ \dot z_2 = \dot \nu = -\alpha \cdot sign(s) = -\alpha \cdot sign(s) |s| ^ {\frac {1}{2}} |s| ^ {-\frac {1}{2}} = -\alpha {\frac {z_1}{|z_1|}} \nonumber \end{cases} \\ \nonumber & \Rightarrow \\ \nonumber &\begin{cases} \dot z_1 = \frac {1} {2|z_1|}(-\lambda z_1 + z_2) \\ \dot z_2 = -\alpha {\frac {z_1}{|z_1|}} \\ \end{cases} \\ \end{align} \nonumber {z˙1=2∣z11(λz1+z2)z˙2=ν˙=αsign(s)=αsign(s)s21s21=αz1z1{z˙1=2∣z11(λz1+z2)z˙2=αz1z1
设置新的状态变量为
Z = [ z 1 z 2 ] Z = \begin{bmatrix} z_1 \\ z_2 \\ \end{bmatrix} Z=[z1z2]
设置李雅普诺夫函数为
V 0 = Z T P Z = ( β + 4 ε 2 ) z 1 2 + z 2 2 − 4 ε z 1 z 2 V_0 = Z^TPZ = (\beta+4\varepsilon^2)z_1^2 + z_2^2 - 4\varepsilon z_1 z_2 V0=ZTPZ=(β+4ε2)z12+z224εz1z2
其中 P P P
P = [ β + 4 ε 2 − 2 ε − 2 ε 1 ] P=\begin{bmatrix} \beta+4\varepsilon^2 & -2\varepsilon \\ -2\varepsilon & 1 \\ \end{bmatrix} P=[β+4ε22ε2ε1]

李雅普诺夫函数的导数

对李雅普诺夫函数进行求导
V ˙ 0 = 2 ( β + 4 ε 2 ) z 1 z ˙ 1 + 2 z 2 z ˙ 2 − 4 ε z ˙ 1 z 2 − 4 ε z 1 z ˙ 2 = 2 ( β + 4 ε 2 ) z 1 ( 1 2 ∣ z 1 ∣ ( − λ z 1 + z 2 ) ) + 2 z 2 ( − α z 1 ∣ z 1 ∣ ) − 4 ε ( 1 2 ∣ z 1 ∣ ( − λ z 1 + z 2 ) ) z 2 − 4 ε z 1 ( − λ z 1 + z 2 ) = − 1 ∣ z 1 ∣ Z T Q Z \begin{align} \dot V_0 &= 2(\beta+4\varepsilon^2)z_1 \dot z_1 + 2z_2 \dot z_2 - 4\varepsilon \dot z_1 z_2 - 4\varepsilon z_1 \dot z_2 \nonumber\\ &= 2(\beta+4\varepsilon^2)z_1 (\frac {1} {2|z_1|}(-\lambda z_1 + z_2)) + 2z_2(-\alpha {\frac {z_1}{|z_1|}}) - 4\varepsilon (\frac {1} {2|z_1|}(-\lambda z_1 + z_2)) z_2 - 4\varepsilon z_1 (-\lambda z_1 + z_2) \nonumber\\ &= - \frac {1} {|z_1|} Z^T Q Z \nonumber \end{align} V˙0=2(β+4ε2)z1z˙1+2z2z˙24εz˙1z24εz1z˙2=2(β+4ε2)z1(2∣z11(λz1+z2))+2z2(αz1z1)4ε(2∣z11(λz1+z2))z24εz1(λz1+z2)=z11ZTQZ
其中 Q Q Q
Q = [ − 4 α ε + λ ( β + 4 ε 2 ) − 1 2 ( β + 4 ε 2 ) + α − λ ε − 1 2 ( β + 4 ε 2 ) + α − λ ε 2 ε ] = [ A B C D ] Q = \begin{bmatrix} -4\alpha \varepsilon + \lambda(\beta+4 \varepsilon^2) & -\frac{1}{2}(\beta+4\varepsilon^2) + \alpha-\lambda \varepsilon \\ -\frac{1}{2} (\beta+4\varepsilon^2) + \alpha-\lambda \varepsilon & 2\varepsilon \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} Q=[4αε+λ(β+4ε2)21(β+4ε2)+αλε21(β+4ε2)+αλε2ε]=[ACBD]
这样我们得到李雅普诺夫函数
V ˙ 0 = − 1 ∣ z 1 ∣ Z T Q Z \dot V_0 = - \frac {1} {|z_1|} Z^T Q Z V˙0=z11ZTQZ
Q Q Q 的特征根
∣ p I − Q ∣ = ∣ p − A B C p − D ∣ = p 2 − ( A + D ) p + A D − B C = 0 |pI -Q| = \begin{vmatrix} p-A & B \\ C & p - D \end{vmatrix} = p^2-(A+D)p + AD - BC = 0 pIQ= pACBpD =p2(A+D)p+ADBC=0
解方程组解得特征根为
{ p m a x ( Q ) = A + D + ( A − D ) 2 + 4 B C 2 p m i n ( Q ) = A + D − ( A − D ) 2 + 4 B C 2 \begin{cases} p_{max}(Q) = \frac {A+D + \sqrt{(A-D)^2+4BC}} {2}\\ p_{min}(Q) = \frac {A+D - \sqrt{(A-D)^2+4BC}} {2} \end{cases} pmax(Q)=2A+D+(AD)2+4BC pmin(Q)=2A+D(AD)2+4BC
所以
p m i n ( Q ) Z T Z = A + D + ( A − D ) 2 + 4 B C 2 ( z 1 2 + z 2 2 ) p_{min}(Q) Z^T Z = \frac {A+D + \sqrt{(A-D)^2+4BC}} {2} (z_1^2 + z_2^2) pmin(Q)ZTZ=2A+D+(AD)2+4BC (z12+z22)

Z T Q Z = A z 1 2 + ( B + C ) Z 1 Z 2 + D z 2 2 Z^TQZ = A z_1^2 + (B+C)Z_1Z_2 + Dz_2^2 ZTQZ=Az12+(B+C)Z1Z2+Dz22

比较 $p_{min}(Q) Z^T Z 与 与 Z^TQZ$的大小,为了简便运算,将根号项用 R R R 表示
D v a l = 2 ( Z T Q Z − p m i n ( Q ) Z T Z ) = ( A − D + R ) z 1 2 + ( D − A + R ) z 2 2 + 2 ( B + C ) z 1 z 2 = ( A − D + R ) [ z 1 2 + ( D − A + R ) ( A − D + R ) z 2 2 + 2 ( B + C ) ( A − D + R ) z 1 z 2 ] = ( A − D + R ) [ z 1 2 + ( D − A + R ) ( D + R − A ) ( A − D + R ) ( D + R − A ) z 2 2 + 2 ( B + C ) ( R + D − A ) ( A − D + R ) ( R + D − A ) z 1 z 2 ] = ( A − D + R ) [ z 1 2 + ( D + R − A ) 2 4 B C z 2 2 + 2 ( B + C ) ( R + D − A ) 4 B C z 1 z 2 ] = ( A − D + R ) [ z 1 2 + ( D + R − A ) 2 4 B C z 2 2 + ( D + R − A ) 2 4 B C z 1 z 2 ( D + R − A ) 2 4 B C z 1 z 2 + 2 ( B + C ) ( R + D − A ) 4 B C z 1 z 2 ] = ( A − D + R ) [ ( z 1 + D + R − A 2 B C z 2 ) 2 + ( 2 B + 2 C − 4 B C ) ( R + D − A ) 4 B C z 1 z 2 ] \begin{align} D_{val} &=2(Z^TQZ - p_{min}(Q) Z^T Z ) \nonumber\\ &= (A-D+R)z_1^2+(D-A+R)z_2^2+2(B+C)z_1z_2 \nonumber\\ &= (A-D+R)\left[z_1^2 + \frac{(D-A+R)}{(A-D+R)}z_2^2 + \frac{2(B+C)}{(A-D+R)}z_1z_2\right] \nonumber\\ &= (A-D+R)\left[z_1^2 + \frac{(D-A+R)(D+R-A)}{(A-D+R)(D+R-A)}z_2^2 + \frac{2(B+C)(R+D-A)}{(A-D+R)(R+D-A)}z_1z_2\right] \nonumber \\ &= (A-D+R)\left[z_1^2 + \frac{(D+R-A)^2}{4BC}z_2^2 + \frac{2(B+C)(R+D-A)}{4BC}z_1z_2\right] \nonumber\\ &= (A-D+R)\left[z_1^2 + \frac{(D+R-A)^2}{4BC}z_2^2 + \sqrt{\frac{(D+R-A)^2}{4BC}}z_1z_2 \sqrt{\frac{(D+R-A)^2}{4BC}}z_1z_2 + \frac{2(B+C)(R+D-A)}{4BC}z_1z_2\right] \nonumber\\ &= (A-D+R)\left[(z_1 + \frac{D+R-A}{2 \sqrt{BC}}z_2)^2 + \frac{(2B+2C-4\sqrt{BC})(R+D-A)}{4BC}z_1z_2\right] \nonumber\\ \end{align} Dval=2(ZTQZpmin(Q)ZTZ)=(AD+R)z12+(DA+R)z22+2(B+C)z1z2=(AD+R)[z12+(AD+R)(DA+R)z22+(AD+R)2(B+C)z1z2]=(AD+R)[z12+(AD+R)(D+RA)(DA+R)(D+RA)z22+(AD+R)(R+DA)2(B+C)(R+DA)z1z2]=(AD+R)[z12+4BC(D+RA)2z22+4BC2(B+C)(R+DA)z1z2]=(AD+R)[z12+4BC(D+RA)2z22+4BC(D+RA)2 z1z24BC(D+RA)2 z1z2+4BC2(B+C)(R+DA)z1z2]=(AD+R)[(z1+2BC D+RAz2)2+4BC(2B+2C4BC )(R+DA)z1z2]
上式中
R + A − D = ( A − D ) 2 + 4 B C + ( A − D ) ≥ 0 R + A - D = \sqrt{(A-D)^2+4BC} + (A - D) \ge 0 R+AD=(AD)2+4BC +(AD)0

( z 1 + D + R − A 2 B C z 2 ) 2 ≥ 0 (z_1 + \frac{D+R-A}{2 \sqrt{BC}}z_2)^2 \ge 0 (z1+2BC D+RAz2)20

{ 2 B + 2 C − 4 B C ≥ 0 R + D − A = ( A − D ) 2 + 4 B C + ( D − A ) ≥ 0 ⇒ ( 2 B + 2 C − 4 B C ) ( R + D − A ) 4 B C ≥ 0 \begin{cases} 2B+2C-4\sqrt{BC} \ge 0 \\ R+D-A = \sqrt{(A-D)^2+4BC} + (D - A) \ge 0 \\ \end{cases} \Rightarrow \frac{(2B+2C-4\sqrt{BC})(R+D-A)}{4BC} \ge 0 {2B+2C4BC 0R+DA=(AD)2+4BC +(DA)04BC(2B+2C4BC )(R+DA)0

所以我们得到
Z T Q Z ≥ p m i n ( Q ) Z T Z Z^TQZ \ge p_{min}(Q) Z^T Z ZTQZpmin(Q)ZTZ
同理可证
Z T Q Z ≤ p m a x ( Q ) Z T Z Z^TQZ \le p_{max}(Q) Z^T Z ZTQZpmax(Q)ZTZ

李雅普诺夫函数导数的变换

上式是根据 V ˙ 0 = − 1 ∣ z 1 ∣ Z T Q Z \dot V_0 = -\frac {1} {|z_1|} Z^TQZ V˙0=z11ZTQZ 做出的,对于 V 0 = Z T P Z V_0 = Z ^ T P Z V0=ZTPZ 同样根据上式可得

向量的0范数,向量中非零元素的个数
向量的1范数,向量中各元素绝对值的模
向量的2范数,通常意义上的模值,欧几里得范数
向量的无穷范数,向量的最大值

矩阵的1范数,列和范数,所有矩阵列向量绝对值之和的最大值
矩阵的2范数,谱范数,即 A T A A^TA ATA矩阵的最大特征值的开平方
矩阵的无穷范数,行和范数,所有矩阵行向量绝对值之和的最大值
矩阵的F范数,Forbenius范数,所有矩阵元素绝对值的平方和再开放

Z T P Z ≥ p m i n ( P ) Z T Z ⇒ ( Z T P Z ) 1 / 2 ≥ p m i n 1 / 2 ( P ) ( Z T Z ) 1 / 2 = p m i n 1 / 2 ( P ) ∥ Z ∥ ⇒ ∥ Z ∥ ≤ ( Z T P Z ) 1 / 2 p m i n 1 / 2 ( P ) = V 0 1 / 2 p m i n 1 / 2 ( P ) \begin{gather} Z^TPZ \ge p_{min}(P)Z^TZ \nonumber\\ \Rightarrow (Z^TPZ)^{1/2} \ge p_{min}^{1/2}(P)(Z^TZ)^{1/2} = p_{min}^{1/2}(P) \Vert Z \Vert \nonumber\\ \Rightarrow \Vert Z\Vert \le \frac{(Z^TPZ)^{1/2}}{p_{min}^{1/2}(P)} = \frac {V_0^{1/2}} {p_{min}^{1/2}(P)} \nonumber \end{gather} ZTPZpmin(P)ZTZ(ZTPZ)1/2pmin1/2(P)(ZTZ)1/2=pmin1/2(P)ZZpmin1/2(P)(ZTPZ)1/2=pmin1/2(P)V01/2

Z Z Z的欧几里得范数为
∥ Z ∥ = z 1 2 + z 2 2 = ( ∣ s ∣ 1 2 s i g n ( s ) ) 2 + ν 2 = ∣ s ∣ + ν ≥ ∣ s ∣ = ∣ z 1 ∣ \Vert Z \Vert = \sqrt {z_1^2 + z_2^2} = \sqrt{(|s| ^ {\frac {1} {2}}sign(s) )^2 + \nu ^ 2} = \sqrt{|s| + \nu} \ge \sqrt{|s|} = |z_1| Z=z12+z22 =(s21sign(s))2+ν2 =s+ν s =z1
所以
− 1 ∣ z 1 ∣ ≤ − 1 ∥ Z ∥ -\frac {1}{\vert z_1 \vert} \le -\frac {1}{\Vert Z \Vert} z11Z1
我们再次回到 V ˙ 0 \dot V_0 V˙0
V ˙ 0 = − 1 ∣ z 1 ∣ Z T Q Z ≤ − 1 ∣ z 1 ∣ p m i n ( Q ) Z T Z = − 1 ∣ z 1 ∣ p m i n ( Q ) ∥ Z ∥ 2 ≤ − 1 ∥ Z ∥ p m i n ( Q ) ∥ Z ∥ 2 = − p m i n ( Q ) ∥ Z ∥ ≤ − p m i n ( Q ) V 0 1 2 p m i n 1 2 ( P ) = − r V 0 1 2 \begin{align} \dot V_0 &= - \frac{1} {|z_1|} Z^TQZ \le - \frac{1} {|z_1|} p_{min}(Q)Z^TZ \nonumber \\ &= - \frac{1} {|z_1|} p_{min}(Q) \Vert Z \Vert ^ 2 \le -\frac {1}{\Vert Z \Vert} p_{min}(Q) \Vert Z \Vert ^ 2 \nonumber\\ &= -p_{min}(Q) \Vert Z \Vert \le -p_{min}(Q) \frac {V_0^{\frac{1}{2}}} {p_{min}^{\frac{1}{2}}(P)} \nonumber\\ &= -r V_0^{\frac{1}{2}} \nonumber \end{align} V˙0=z11ZTQZz11pmin(Q)ZTZ=z11pmin(Q)Z2Z1pmin(Q)Z2=pmin(Q)Zpmin(Q)pmin21(P)V021=rV021
其中
r = p m i n ( Q ) p m i n 1 / 2 ( P ) r = \frac {p_{min}(Q)} {p_{min}^{1/2}(P)} r=pmin1/2(P)pmin(Q)

若系统满足 V ˙ ≤ − r V 1 2 \dot V \le -rV^{\frac {1} {2}} V˙rV21 其中 r > 0 r>0 r>0 ,则系统可以在有限时间内稳定

矩阵Q正定性的保证

上面的证明保证了系统具有李雅普诺夫稳定性,但是只有在 r > 0 r > 0 r>0的情况下才能保证系统稳定,此时需要 p m i n ( Q ) {p_{min}(Q)} pmin(Q)

p m i n 1 / 2 ( P ) {p_{min}^{1/2}(P)} pmin1/2(P) 保持同号,由于矩阵 P P P为正定矩阵,所以 p m i n 1 / 2 ( P ) {p_{min}^{1/2}(P)} pmin1/2(P)必大于0,那么需要保证 p m i n ( Q ) {p_{min}(Q)} pmin(Q)也大于0。

正定矩阵的特征值都是正数

Q = [ − 4 α ε + λ ( β + 4 ε 2 ) − 1 2 ( β + 4 ε 2 ) + α − λ ε − 1 2 ( β + 4 ε 2 ) + α − λ ε 2 ε ] Q = \begin{bmatrix} -4\alpha \varepsilon + \lambda(\beta+4 \varepsilon^2) & -\frac{1}{2}(\beta+4\varepsilon^2) + \alpha-\lambda \varepsilon \\ -\frac{1}{2} (\beta+4\varepsilon^2) + \alpha-\lambda \varepsilon & 2\varepsilon \end{bmatrix} Q=[4αε+λ(β+4ε2)21(β+4ε2)+αλε21(β+4ε2)+αλε2ε]

不妨直接取
α = λ ε + 1 2 ( β + 4 ε 2 ) \alpha = \lambda \varepsilon + \frac{1}{2}(\beta+4\varepsilon^2) α=λε+21(β+4ε2)
这样的话可以简化一下
Q = [ ( λ − 2 ε ) ( β + 4 ε 2 ) − 4 λ ε 2 0 0 2 ε ] Q = \begin{bmatrix} (\lambda-2\varepsilon)(\beta+4 \varepsilon^2)-4\lambda \varepsilon^2 & 0\\ 0 & 2\varepsilon \end{bmatrix} Q=[(λ2ε)(β+4ε2)4λε2002ε]
所以 Q Q Q 的特征根为
{ p 1 = ( λ − 2 ε ) ( β + 4 ε 2 ) − 4 λ ε 2 p 2 = 2 ε \begin{cases} p_1 = (\lambda-2\varepsilon)(\beta+4 \varepsilon^2)-4\lambda \varepsilon^2 \\ p_2 = 2\varepsilon \end{cases} {p1=(λ2ε)(β+4ε2)4λε2p2=2ε
由于 ε > 0 \varepsilon > 0 ε>0 所以 p 2 > 0 p_2 > 0 p2>0非常显然,现在只需要保证 p 1 > 0 p_1>0 p1>0,则可以有
λ > 2 ε ( β + 4 ε 2 ) β \lambda > \frac{2\varepsilon(\beta+4\varepsilon^2)} {\beta} λ>β2ε(β+4ε2)

重写李雅普诺夫函数

上一节中给出了保证 Q Q Q 正定性的条件,但是 α \alpha α λ \lambda λ 这两个参数值是人为给出的,因此需要把这两个参数加入到李雅普诺夫函数中来
V = V 0 + 1 2 γ 1 ( λ − λ ∗ ) 2 + 1 2 γ 2 ( α − α ∗ ) 2 V = V_0 + \frac {1} {2\gamma_1} (\lambda-\lambda^{*})^2 + \frac{1}{2\gamma_2} (\alpha-\alpha^{*})^2 V=V0+2γ11(λλ)2+2γ21(αα)2
其中 λ ∗   α ∗ \lambda^{*} \ \alpha^{*} λ α 为未知常数,对其求导
V ˙ = V ˙ 0 + 1 γ 1 ( λ − λ ∗ ) λ ˙ + 1 γ 2 ( α − α ∗ ) α ˙ ≤ − r V 0 1 2 + 1 γ 1 ( λ − λ ∗ ) λ ˙ + 1 γ 2 ( α − α ∗ ) α ˙ = − r V 0 1 2 + 1 γ 1 ( λ − λ ∗ ) λ ˙ + 1 γ 2 ( α − α ∗ ) α ˙ − ω 1 2 γ 1 ∣ λ − λ ∗ ∣ + ω 1 2 γ 1 ∣ λ − λ ∗ ∣ − ω 2 2 γ 2 ∣ α − α ∗ ∣ + ω 2 2 γ 2 ∣ α − α ∗ ∣ \begin{align} \dot V &= \dot V_0 + \frac {1} {\gamma_1} (\lambda-\lambda^{*})\dot \lambda + \frac{1}{\gamma_2} (\alpha-\alpha^{*})\dot \alpha \le -r V_0^{\frac{1}{2}} + \frac {1} {\gamma_1} (\lambda-\lambda^{*})\dot \lambda + \frac{1}{\gamma_2} (\alpha-\alpha^{*})\dot \alpha \nonumber\\ &= -r V_0^{\frac{1}{2}} + \frac {1} {\gamma_1} (\lambda-\lambda^{*})\dot \lambda + \frac{1}{\gamma_2} (\alpha-\alpha^{*})\dot \alpha -\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}|+\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}| -\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}|+\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}| \nonumber \end{align} V˙=V˙0+γ11(λλ)λ˙+γ21(αα)α˙rV021+γ11(λλ)λ˙+γ21(αα)α˙=rV021+γ11(λλ)λ˙+γ21(αα)α˙2γ1 ω1λλ+2γ1 ω1λλ2γ2 ω2αα+2γ2 ω2αα
根据 ( x 2 + y 2 + z 2 ) ≤ ∣ x ∣ + ∣ y ∣ + ∣ z ∣ (x^2 + y^2 + z^2) \le |x| + |y| + |z| (x2+y2+z2)x+y+z
− r V 0 1 2 − ω 1 2 γ 1 ∣ λ − λ ∗ ∣ − ω 2 2 γ 2 ∣ α − α ∗ ∣ ≤ − [ r 2 V 0 1 2 + ω 1 2 2 γ 1 ∣ λ − λ ∗ ∣ 2 + ω 2 2 2 γ 2 ∣ α − α ∗ ∣ 2 ] 1 2 -r V_0^{\frac{1}{2}} - \frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}|-\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}| \le - \left[r^2V_0^{\frac{1}{2}}+ \frac {\omega_1^2} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}|^2 + \frac {\omega_2^2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}|^2\right]^{\frac{1}{2}} rV0212γ1 ω1λλ2γ2 ω2αα[r2V021+2γ1 ω12λλ2+2γ2 ω22αα2]21
r , ω 1 , ω 2 r,\omega_1,\omega_2 r,ω1,ω2 中最小的数为 n n n,则上式为
[ r 2 V 0 1 2 + ω 1 2 2 γ 1 ∣ λ − λ ∗ ∣ 2 + ω 2 2 2 γ 2 ∣ α − α ∗ ∣ 2 ] 1 2 ≤ − n [ V 0 1 2 + 1 2 γ 1 ∣ λ − λ ∗ ∣ 2 + 1 2 γ 2 ∣ α − α ∗ ∣ 2 ] = − n V 1 2 \left[r^2V_0^{\frac{1}{2}}+ \frac {\omega_1^2} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}|^2 + \frac {\omega_2^2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}|^2\right]^{\frac{1}{2}} \le-n \left[V_0^{\frac{1}{2}}+ \frac {1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}|^2 + \frac {1} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}|^2 \right] = -nV^{\frac{1}{2}} [r2V021+2γ1 ω12λλ2+2γ2 ω22αα2]21n[V021+2γ1 1λλ2+2γ2 1αα2]=nV21
带入 V ˙ \dot V V˙
V ˙ ≤ − r V 0 1 2 + 1 γ 1 ( λ − λ ∗ ) λ ˙ + 1 γ 2 ( α − α ∗ ) α ˙ − ω 1 2 γ 1 ∣ λ − λ ∗ ∣ + ω 1 2 γ 1 ∣ λ − λ ∗ ∣ − ω 2 2 γ 2 ∣ α − α ∗ ∣ + ω 2 2 γ 2 ∣ α − α ∗ ∣ ≤ − n V 1 2 + 1 γ 1 ( λ − λ ∗ ) λ ˙ + 1 γ 2 ( α − α ∗ ) α ˙ + ω 1 2 γ 1 ∣ λ − λ ∗ ∣ + ω 2 2 γ 2 ∣ α − α ∗ ∣ \begin {align} \dot V &\le -r V_0^{\frac{1}{2}} + \frac {1} {\gamma_1} (\lambda-\lambda^{*})\dot \lambda + \frac{1}{\gamma_2} (\alpha-\alpha^{*})\dot \alpha -\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}|+\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}| -\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}|+\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}| \nonumber\\ &\le -nV^{\frac{1}{2}} + \frac {1} {\gamma_1} (\lambda-\lambda^{*})\dot \lambda + \frac{1}{\gamma_2} (\alpha-\alpha^{*})\dot \alpha +\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}| +\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}| \nonumber \end {align} V˙rV021+γ11(λλ)λ˙+γ21(αα)α˙2γ1 ω1λλ+2γ1 ω1λλ2γ2 ω2αα+2γ2 ω2ααnV21+γ11(λλ)λ˙+γ21(αα)α˙+2γ1 ω1λλ+2γ2 ω2αα
由于 λ ∗   α ∗ \lambda^{*} \ \alpha^{*} λ α 为未知常数,那我们假设 λ ∗ > λ , α ∗ > α \lambda^{*}>\lambda , \alpha^{*} > \alpha λ>λα>α ,总能找到两个常数满足这两个条件
V ˙ ≤ − n V 1 2 + 1 γ 1 ( λ − λ ∗ ) λ ˙ + 1 γ 2 ( α − α ∗ ) α ˙ + ω 1 2 γ 1 ∣ λ − λ ∗ ∣ + ω 2 2 γ 2 ∣ α − α ∗ ∣ = − n V 1 2 − 1 γ 1 ∣ λ − λ ∗ ∣ λ ˙ − 1 γ 2 ∣ α − α ∗ ∣ α ˙ + ω 1 2 γ 1 ∣ λ − λ ∗ ∣ + ω 2 2 γ 2 ∣ α − α ∗ ∣ = − n V 1 2 + ∣ λ − λ ∗ ∣ ( ω 1 2 γ 1 − λ ˙ γ 1 ) + ∣ α − α ∗ ∣ ( ω 2 2 γ 2 − λ ˙ γ 2 ) \begin{align} \dot V &\le -nV^{\frac{1}{2}} + \frac {1} {\gamma_1} (\lambda-\lambda^{*})\dot \lambda + \frac{1}{\gamma_2} (\alpha-\alpha^{*})\dot \alpha +\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}| +\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}| \nonumber\\ &= -nV^{\frac{1}{2}} - \frac {1} {\gamma_1} |\lambda-\lambda^{*}|\dot \lambda - \frac{1}{\gamma_2} |\alpha-\alpha^{*}|\dot \alpha +\frac {\omega_1} {\sqrt{2 \gamma_1}} |\lambda - \lambda^{*}| +\frac {\omega_2} {\sqrt{2 \gamma_2}} |\alpha - \alpha^{*}| \nonumber\\ &= -nV^{\frac{1}{2}} + |\lambda-\lambda^{*}|(\frac {\omega_1} {\sqrt{2 \gamma_1}} - \frac{\dot \lambda} {\gamma_1}) + |\alpha-\alpha^{*}|(\frac {\omega_2} {\sqrt{2 \gamma_2}} - \frac{\dot \lambda} {\gamma_2}) \nonumber \end{align} V˙nV21+γ11(λλ)λ˙+γ21(αα)α˙+2γ1 ω1λλ+2γ2 ω2αα=nV21γ11λλλ˙γ21ααα˙+2γ1 ω1λλ+2γ2 ω2αα=nV21+λλ(2γ1 ω1γ1λ˙)+αα(2γ2 ω2γ2λ˙)
此时若令
λ ˙ = ω 1 γ 1 2 \dot \lambda = \omega_1 \sqrt{\frac{\gamma_1}{2}} λ˙=ω12γ1

V ˙ ≤ − n V 1 2 + ∣ λ − λ ∗ ∣ ( ω 2 2 γ 2 − α ˙ γ 2 ) = − n V 1 2 + η \dot V \le -nV^{\frac{1}{2}} + |\lambda-\lambda^{*}|(\frac {\omega_2} {\sqrt{2 \gamma_2}} - \frac{\dot \alpha} {\gamma_2}) = -nV^{\frac{1}{2}} + \eta V˙nV21+λλ(2γ2 ω2γ2α˙)=nV21+η
其中
η = ∣ λ − λ ∗ ∣ ( ω 2 2 γ 2 − α ˙ γ 2 ) \eta = |\lambda-\lambda^{*}|(\frac {\omega_2} {\sqrt{2 \gamma_2}} - \frac{\dot \alpha} {\gamma_2}) η=λλ(2γ2 ω2γ2α˙)
所以此系统具有李雅普诺夫稳定性,尽管有 η \eta η 存在,系统仍然可以在一定程度上保持稳定,原因在于我们证明了 V ˙ ≤ − n V 1 2 ≤ 0 \dot V \le -nV^{\frac{1}{2}} \le 0 V˙nV210 而不是传统的 V ˙ ≤ 0 \dot V \le 0 V˙0

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