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
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\begin{cases} \dot x_1 = x_2 \\ \dot x_2 = f + g \cdot u \end{cases}
{x˙1=x2x˙2=f+g⋅u 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
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\begin{cases} \dot s = -\lambda |s| ^ {\frac {1} {2}} \cdot sign(s) + \nu \\ \dot \nu = - \alpha \cdot sign(s) \\ \end{cases}
{s˙=−λ∣s∣21⋅sign(s)+νν˙=−α⋅sign(s)
控制器设计
设状态
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x1 的期望值为
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x_d
xd ，则跟踪误差为
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\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=x1−xde2=e˙1=x˙1−x˙d=x2−x˙d 设计滑模面为
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s = ce_1 + e_2
s=ce1+e2 则滑模面的导数为
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\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+g⋅u−x¨d=−λ∣s∣21⋅sign(s)+ν=−λ∣s∣21⋅sign(s)−α⋅sign(s) 可以得到控制量
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u = g ^ {-1} (-f + \ddot x_d - c_1e_2 - \lambda |s| ^ {\frac {1} {2}}sign(s) - \alpha \cdot sign(s))
u=g−1(−f+x¨d−c1e2−λ∣s∣21sign(s)−α⋅sign(s)) 参数设定为
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\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，
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\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
∣
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|s| ^ {\frac {1} {2}}
∣s∣21 。除此之外，在
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\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。
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\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=∣s∣21z2=ν⇒{z˙1=21∣s∣−21s˙=21∣s∣−21(−λ∣s∣21⋅sign(s)−α⋅sign(s))z˙2=ν˙=−α⋅sign(s) 将第一项带入第二项
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\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∣z1∣1(−λz1+z2)z˙2=ν˙=−α⋅sign(s)=−α⋅sign(s)∣s∣21∣s∣−21=−α∣z1∣z1⇒{z˙1=2∣z1∣1(−λz1+z2)z˙2=−α∣z1∣z1 设置新的状态变量为
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Z = \begin{bmatrix} z_1 \\ z_2 \\ \end{bmatrix}
Z=[z1z2] 设置李雅普诺夫函数为
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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+z22−4εz1z2 其中
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P=\begin{bmatrix} \beta+4\varepsilon^2 & -2\varepsilon \\ -2\varepsilon & 1 \\ \end{bmatrix}
P=[β+4ε2−2ε−2ε1]
李雅普诺夫函数的导数
对李雅普诺夫函数进行求导
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\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˙2−4εz˙1z2−4εz1z˙2=2(β+4ε2)z1(2∣z1∣1(−λz1+z2))+2z2(−α∣z1∣z1)−4ε(2∣z1∣1(−λz1+z2))z2−4εz1(−λz1+z2)=−∣z1∣1ZTQZ 其中
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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)+α−λε−21(β+4ε2)+α−λε2ε]=[ACBD] 这样我们得到李雅普诺夫函数
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\dot V_0 = - \frac {1} {|z_1|} Z^T Q Z
V˙0=−∣z1∣1ZTQZ 求
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|pI -Q| = \begin{vmatrix} p-A & B \\ C & p - D \end{vmatrix} = p^2-(A+D)p + AD - BC = 0
∣pI−Q∣=
p−ACBp−D
=p2−(A+D)p+AD−BC=0 解方程组解得特征根为
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\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+(A−D)2+4BC
pmin(Q)=2A+D−(A−D)2+4BC
所以
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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+(A−D)2+4BC
(z12+z22)
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Z^TQZ = A z_1^2 + (B+C)Z_1Z_2 + Dz_2^2
ZTQZ=Az12+(B+C)Z1Z2+Dz22
比较 $p_{min}(Q) Z^T Z
与
与
与Z^TQZ$的大小，为了简便运算，将根号项用
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\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(ZTQZ−pmin(Q)ZTZ)=(A−D+R)z12+(D−A+R)z22+2(B+C)z1z2=(A−D+R)[z12+(A−D+R)(D−A+R)z22+(A−D+R)2(B+C)z1z2]=(A−D+R)[z12+(A−D+R)(D+R−A)(D−A+R)(D+R−A)z22+(A−D+R)(R+D−A)2(B+C)(R+D−A)z1z2]=(A−D+R)[z12+4BC(D+R−A)2z22+4BC2(B+C)(R+D−A)z1z2]=(A−D+R)[z12+4BC(D+R−A)2z22+4BC(D+R−A)2
z1z24BC(D+R−A)2
z1z2+4BC2(B+C)(R+D−A)z1z2]=(A−D+R)[(z1+2BC
D+R−Az2)2+4BC(2B+2C−4BC
)(R+D−A)z1z2] 上式中
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R + A - D = \sqrt{(A-D)^2+4BC} + (A - D) \ge 0
R+A−D=(A−D)2+4BC
+(A−D)≥0
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(z_1 + \frac{D+R-A}{2 \sqrt{BC}}z_2)^2 \ge 0
(z1+2BC
D+R−Az2)2≥0
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\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+2C−4BC
≥0R+D−A=(A−D)2+4BC
+(D−A)≥0⇒4BC(2B+2C−4BC
)(R+D−A)≥0
所以我们得到
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Z^TQZ \ge p_{min}(Q) Z^T Z
ZTQZ≥pmin(Q)ZTZ 同理可证
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a
x
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Z
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Z^TQZ \le p_{max}(Q) Z^T Z
ZTQZ≤pmax(Q)ZTZ
李雅普诺夫函数导数的变换
上式是根据
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Z
\dot V_0 = -\frac {1} {|z_1|} Z^TQZ
V˙0=−∣z1∣1ZTQZ 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}
ZTPZ≥pmin(P)ZTZ⇒(ZTPZ)1/2≥pmin1/2(P)(ZTZ)1/2=pmin1/2(P)∥Z∥⇒∥Z∥≤pmin1/2(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
=(∣s∣21sign(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}
−∣z1∣1≤−∥Z∥1 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=−∣z1∣1ZTQZ≤−∣z1∣1pmin(Q)ZTZ=−∣z1∣1pmin(Q)∥Z∥2≤−∥Z∥1pmin(Q)∥Z∥2=−pmin(Q)∥Z∥≤−pmin(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)
与
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)+α−λε−21(β+4ε2)+α−λε2ε]
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=[(λ−2ε)(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=(λ−2ε)(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 和
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(λ−λ∗)2+2γ21(α−α∗)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(λ−λ∗)λ˙+γ21(α−α∗)α˙≤−rV021+γ11(λ−λ∗)λ˙+γ21(α−α∗)α˙=−rV021+γ11(λ−λ∗)λ˙+γ21(α−α∗)α˙−2γ1
ω1∣λ−λ∗∣+2γ1
ω1∣λ−λ∗∣−2γ2
ω2∣α−α∗∣+2γ2
ω2∣α−α∗∣ 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∣ 有
−
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}}
−rV021−2γ1
ω1∣λ−λ∗∣−2γ2
ω2∣α−α∗∣≤−[r2V021+2γ1
ω12∣λ−λ∗∣2+2γ2
ω22∣α−α∗∣2]21 设
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
ω12∣λ−λ∗∣2+2γ2
ω22∣α−α∗∣2]21≤−n[V021+2γ1
1∣λ−λ∗∣2+2γ2
1∣α−α∗∣2]=−nV21 带入
V
˙
\dot V
V˙ 有
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(λ−λ∗)λ˙+γ21(α−α∗)α˙−2γ1
ω1∣λ−λ∗∣+2γ1
ω1∣λ−λ∗∣−2γ2
ω2∣α−α∗∣+2γ2
ω2∣α−α∗∣≤−nV21+γ11(λ−λ∗)λ˙+γ21(α−α∗)α˙+2γ1
ω1∣λ−λ∗∣+2γ2
ω2∣α−α∗∣ 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∗>λ，α∗>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(λ−λ∗)λ˙+γ21(α−α∗)α˙+2γ1
ω1∣λ−λ∗∣+2γ2
ω2∣α−α∗∣=−nV21−γ11∣λ−λ∗∣λ˙−γ21∣α−α∗∣α˙+2γ1
ω1∣λ−λ∗∣+2γ2
ω2∣α−α∗∣=−nV21+∣λ−λ∗∣(2γ1
ω1−γ1λ˙)+∣α−α∗∣(2γ2
ω2−γ2λ˙) 此时若令
λ
˙
=
ω
1
γ
1
2
\dot \lambda = \omega_1 \sqrt{\frac{\gamma_1}{2}}
λ˙=ω12γ1
则
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+∣λ−λ∗∣(2γ2
ω2−γ2α˙)=−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=∣λ−λ∗∣(2γ2
ω2−γ2α˙) 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˙≤−nV21≤0 而不是传统的
V
˙
≤
0
\dot V \le 0
V˙≤0
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