PS#2 - Nonlinear Systems and Control. Spring 2011 ... system. ˙x1 = x2. ˙x2 = −(
2b − g(x1))ax2 − a2x1 where a, b > 0 and g(x1) = { 0, |x1| > 1 .... [1] H.K. Khalil.
PS#2 - Nonlinear Systems and Control
Spring 2011
Problem 1 What is a relationship between k and b that excludes the existence of periodic orbits for the following system x˙ 1 = x2 x˙ 2 = −(2b − g(x1 ))ax2 − a2 x1 where a, b > 0 and g(x1 ) =
0, k,
|x1 | > 1 |x1 | ≤ 1
Solution Let us compute ∂f1 ∂f2 + = g(x1 ) − 2b ∂x1 ∂x2 By the Bendixson criterion or the “negative Pointcar´e-Bendixson criterion”, we require that the quantity above is not equal to zero and does not change sign in order to exclude the existence of periodic orbits. This is satisfied if k < 2b Problem 2 Consider the function
f (x) =
−x1 + x1 x2 x2 − x1 x2
1. Is f (x) globally Lipschitz? If so, compute the global Lipschitz constant. 2. Is f (x) locally Lipschitz? If so, compute the Lipschitz constant over the set S = {x ∈ R2 | |x1 | ≤ a1 , |x2 | ≤ a2 } Solution 1. f (x) is not globally Lipschitz since the Jacobian ∂f −1 + x2 x1 = ∂x −x2 1 − x1 is not uniformly bounded over all of R2 . (See Appendix) 2. Since f (x) is continuously differentiable on R2 , then it is locally Lipschitz. Since S is compact, then f (x) is Lipschitz over S. We have that
∂f
= max{| − 1 + x2 | + |x1 |, |x2 | + |1 − x1 |} ≤ 1 + a1 + a2
∂x ∞ where the last inequality follows from the definition of S. The Lipschitz constant is L = 1+a1 +a2 . 1
Problem 3 Given the following nonlinear ODE 2x2 , 1 + x22 2x1 = −x2 + , 1 + x21
x˙ 1 = −x1 +
x1 (0) = a
x˙ 2
x2 (0) = b.
Show that this ODE has a unique solution defined for all t ≥ 0. Solution Let z = (x1 x2 )T and z˜ = (˜ x1 x ˜2 )T . We have that
" # 2˜ x2
−x1 + 2x22 + x
˜1 − 1+(˜
x2 )2 1+x2 kf (z) − f (˜ z )k =
2x1 2˜ x1
−x2 + 1+x
˜2 − 1+(˜ 2 + x x1 )2 1
" x # x ˜2 2
1+x22 − 1+(˜x2 )2 ≤ kz − z˜k + 2 x1
x ˜1
1+x2 − 1+(˜x1 )2 1
(x2 −˜x2 )+x2 x˜2 (x2 −˜x2 )
2 (1+x2 )(1+(˜x2 )2 ) = kz − z˜k + 2
(x1 −˜x1 )+x1 x˜1 (x1 −˜x1 )
(1+x21 )(1+(˜ x1 )2 )
x2 x˜2 (x2 −˜x2 ) (x2 −˜ x2 )
(1+x2 )(1+(˜
x2 )2 ) 2
(1+x22 )(1+(˜x2 )2 ) ≤ kz − z˜k + 2 + 2 (x1 −˜ x1 )
x1 x˜1 (x1 −˜x1 )
(1+x21 )(1+(˜x1 )2 )
(1+x21 )(1+(˜x1 )2 )
x2 − x
x2 − x ˜2 ˜2
+ 2 ≤ kz − z˜k + 2 x1 − x ˜1 x1 − x ˜1 Therefore, kf (z) − f (˜ z )k ≤ 5 kz − z˜k holds globally, and the result follows. Another way to solve this problem (since the vector field f is continuously differentiable in this case) is to look at the Jacobian of the system and try to derive a uniform upper bound for it. (See [1, Lemma 3.3, p.91])
(1−x2 ) −1 2 (1+x22)2
∂f (1 − x22 )
(1 − x21 )
2 = max 1 + 2 =
∂x
(1−x21 )
(1 + x2 )2 , 1 + 2 (1 + x2 )2 ≤ 3 −1
2 (1+x2 )2 2 1 ∞ 1
∞
Therefore, we have established a global bound on the Jacobian, which implies a global Lipschitz constant. Problem 4 1. Derive the sensitivity equation for the following system x˙ 1 = x2 x˙ 2 = −c sin(x1 ) − (a + b cos(x1 ))x2 where the nominal values of the parameters are a0 = 1, b0 = 0, and c0 = 1. 2. Simulate using Matlab the behavior of the sensitivity equation starting from x10 = x20 = 1 2
Solution T Let λ0 = 1 0 1 be the vector of nominal parameters. First, we have the nominal system x˙ 1 = x2 x˙ 2 = − sin(x1 ) − x2 Then, we compute the Jacobian matrices at λ0 to obtain ∂f 0 1 = A= − cos(x1 ) −1 ∂x λ0 and
∂f 0 0 0 = B= −x2 −x2 cos(x1 ) − sin(x1 ) ∂λ λ0
Define the sensitivity function as S(t) ,
∂x1
∂a
∂x1 ∂b
∂x1 ∂c
∂a
∂x2 ∂b
∂x2 ∂c
x3 x5 x7 = x4 x6 x8 ∂x2
λ0
Accordingly, we can not solve for the nominal system and the sensitivity function in one shot using the following system x˙ 1 x2 x10 x˙ 2 − sin(x ) − x 1 2 x20 x˙ 3 x4 0 x˙ 4 −x3 cos(x1 ) − x4 − x2 = , x(0) = 0 x˙ = x˙ 5 0 x6 x˙ 6 −x5 cos(x1 ) − x6 − x2 cos(x1 ) 0 x˙ 7 0 x8 x˙ 8
−x7 cos(x1 ) − x8 − sin(x1 )
0
Use the functions odef un− processing.m and odef un.m to simulate the system above. The results are shown in the figures below. Now, change the initial condition for the nominal system and see what happens!
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solution of the nominal system 1.5 x1 x2 1
0.5
0
−0.5
−1
0
1
2
3
4
5
6
7
8
9
10
variation of x1 with respect to parameters 0.6 x3 x5
0.4
x
7
0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1.2
0
1
2
3
4
5
6
7
8
9
10
variation of x2 with respect to parameters 0.8 x4 x6
0.6
x7 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8
0
1
2
3
4
5
4
6
7
8
9
10
1
Appendix
Lemma 1 [1, Lemma 3.3] If f (t, x) and Lipschitz in x on [a, b] × Rn if and only if
h
i
∂f h ∂x i ∂f ∂x
(t, x) are continuous on [a, b] × Rn , then f is globally is uniformly bounded on [a, b] × Rn .
References [1] H.K. Khalil. Nonlinear Systems. Prentice Hall, Upper Saddle River, NJ, 3rd edition, 2002.
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