By Xiaoxin Liao, Pei Yu
Following the hot advancements within the box of absolute balance, Prof. Xiaoxin Liao, along side Prof. Pei Yu, has created a moment variation of his seminal paintings at the topic. Liao starts with an advent to the Lurie challenge and Lurie keep an eye on process, sooner than relocating directly to the straightforward algebraic enough stipulations for absolutely the balance of self reliant and non-autonomous ODE structures, in addition to a number of distinctive sessions of Lurie-type platforms. the point of interest of the e-book then shifts towards the recent effects and study that experience seemed within the decade because the first version was once released. This ebook is aimed for use through undergraduates within the components of utilized arithmetic, nonlinear keep watch over structures, and chaos keep an eye on and synchronisation, yet can also be precious as a reference for researchers and engineers. The ebook is self-contained, although a uncomplicated wisdom of calculus, linear approach and matrix thought, and usual differential equations is a prerequisite.
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Additional resources for Absolute Stability of Nonlinear Control Systems, 2nd Edition (Mathematical Modelling: Theory and Applications)
Therefore, cT b + d TB−1 d ≥ 0. 11 following the order (1) =⇒ (2) =⇒ (3) =⇒ (1). Proof. 11 holds. 15) that ∀x0 = 0, U(x0 ) > 0 when cT x0 > 0. 4 Several Equivalent SANC for Negative Deﬁnite Derivatives 51 where m is a constant. Thus, cT x¯ = cT B−1 d − cT B−1 d − m T −1 c B c = m. cT B−1 c So for all m ∈ (0, +∞), U(x) ¯ > 0. However, on the other hand, U(x) ¯ = x¯T B x¯ − 2 d T x¯ − cT b, cT B−1 d − m T −1 T c (B ) cT B−1 c cT B−1 d − m −1 B c × B B−1 d − cT B−1 c cT B−1 d − m −1 B c − cTb. 21) Because B is a symmetric positive deﬁnite matrix, so is B−1 , and (B−1 )T = B−1 .
X are ⎧ T −1 ⎨ c B d ≤ 0, 1 ⎩ (cT B−1 d)2 − d T B−1 d − cT b ≥ 0. 4) to prove the following equivalent result. 11. t. x. 50 3 Sufﬁcient Conditions of Absolute Stability: Classical Methods (1) U(x) = xT B x − 2 d Tx − cT b ≥ 0 (cT x = 0), 1. 7; 2. cT B−1 d + cT B−1 c (d T B−1 d + cTb) ≤ 0; ⎧ T −1 2 ⎨ (1) cT b + η TB−1 d − (c B d) ≤ 0, cT B−1 c where d = 12 (AT c + 2 Pb). 3. 11, we need the following lemma. 12. For any symmetric matrix B, we have cT b + d TB−1 d ≥ 0, where d= 1 T (A c + 2 Pb), 2 Proof.
T. 13) where d = P b + 12 AT c. Proof. Since B is a symmetric positive deﬁnite, so is B−1 , and there exists matrix H such that B = H T H, B−1 = H −1 (H −1 )T . 14) ⎪ ⎩ 2 U f (σ ) when f (σ ) = 0, where U= H x − (H −1 )d f (σ ) T H x − (H −1)d − (d T B−1 d + cTb). 15) 48 3 Sufﬁcient Conditions of Absolute Stability: Classical Methods 1 we only need to prove that for any cT x ≥ , we have U > 0. To achieve this, k introduce the nonsingular transform y = H x − (H −1)T d under which U becomes U = yT y − ρ , where ρ = d T B−1 d + cTb.
Absolute Stability of Nonlinear Control Systems, 2nd Edition (Mathematical Modelling: Theory and Applications) by Xiaoxin Liao, Pei Yu