Copyright © by SIAM. Unauthorized reproduction of this article is ...
Copyright © by SIAM. Unauthorized reproduction of this article is ... Copyright © by SIAM. Unauthorized reproduction of this article is ...
HOMOGENEOUS OBSERVER DESIGN 1841Appendix C. Proof of Lemma 2.13. The proof of this lemma is divided intothree parts.1. We first show, by contradiction, that there exists a real number c 0 satisfyingη 0 (θ) − cγ 0 (θ) < 0 ∀ θ ∈ S r0 , ∀ c ≥ c 0 .Suppose there is no such c 0 . This means there is a sequence (θ i ) i∈N in S r0which satisfiesη 0 (θ i ) − iγ 0 (θ i ) ≥ 0 ∀ i ∈ N .The sequence (θ i ) i∈N lives in a compact set. Thus we can extract a convergentsubsequence (θ il ) l∈N which converges to a point denoted θ ∞ .As the functions η 0 and γ 0 are bounded on S r0 and γ 0 takes nonnegativevalues, 8 γ 0 (θ il ) must go to 0 as i l goes to infinity. Since the functions η 0 andγ 0 are continuous, we get γ 0 (θ ∞ ) = 0 and η 0 (θ ∞ ) ≥ 0, which is impossible.Consequently, there exist c 0 and ε 0 > 0 such that(C.1) η 0 (θ) − cγ 0 (θ) ≤ −ε 0 < 0 ∀ θ ∈ S r0 , ∀ c ≥ c 0 .Moreover, since the functions η 0 and γ 0 are homogeneous in the standardsense (see Remark 2.6), we have the second inequality in (2.4).Following the same argument, we can find positive real numbers c ∞ and ε ∞such that(C.2) η ∞ (θ) − cγ ∞ (θ) < −ε ∞ ∀ θ ∈ S r∞ , ∀ c ≥ c ∞ ,and the third inequality in (2.4) holds.In the rest of the proof, letc 1 = max{c 0 ,c ∞ }, ε 1 = min{ε 0 ,ε ∞ } .2. Since η and γ are homogeneous in the 0-limit, there exists λ 0 such that, forall λ ∈ (0,λ 0 ] and all θ ∈ S r0 , we haveη(λ r0 ⋄ θ) ≤ λ d0 η 0 (θ) +λ d0 ε 14 , λd0 γ 0 (θ) − λ d0 ε 14c 1≤ γ(λ r0 ⋄ θ) ,which readily givesη(λ r0 ⋄ θ) − c 1 γ(λ r0 ⋄ θ) ≤ λ d0 η 0 (θ) +λ d0 ε 12 − c 1λ d0 γ 0 (θ) .Using (C.1), we getη(λ r0 ⋄ θ) − c 1 γ(λ r0 ⋄ θ) ≤ −λ d0 ε 12and therefore, since γ takes nonnegative values,∀ λ ∈ (0,λ 0 ] , ∀ θ ∈ S r0 ,η(λ r0 ⋄ θ) − cγ(λ r0 ⋄ θ) ≤−λ d0 ε 12∀ λ ∈ (0,λ 0 ] , ∀ θ ∈ S r0 , ∀ c ≥ c 1 .8 Indeed, if we had γ 0 (x) < 0 for some x in R n \{0}, by letting ɛ = − γ 0(x), the homogeneity in2the 0-limit of γ would give a real number λ> 0 satisfying γ(λr 0 ⋄x)λ d ≤ γ0 0 (x) +ɛ = γ 0(x)< 0. This2contradicts the fact that γ takes nonnegative values only. Also by continuity we have γ 0 (0) ≥ 0.Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.
1842 V. ANDRIEU, L. PRALY, AND A. ASTOLFISimilarly, there exists λ ∞ satisfyingη(λ r∞ ⋄θ)−cγ(λ r∞ ⋄θ) ≤−λ d∞ ε 12Consequently, for each c ≥ c 1 , the setif not empty, must be a subset of∀ λ ∈ [λ ∞ , +∞) , ∀ θ ∈ S r∞ , ∀ c ≥ c 1 .{x ∈ R n \{0} |η(x) − cγ(x) ≥ 0} ,C = {x ∈ R n : |x| r0 ≥ λ 0 } ⋃ {x ∈ R n : |x| r∞ ≤ λ ∞ } ,which is compact and does not contain the origin.3. Suppose now that for all c the first inequality in (2.4) is not true. This meansthat, for all integers c larger than c 1 , there exists x c in R n satisfyingη(x c ) − cγ(x c ) ≥ 0 ,and therefore x c is in C. Since C is a compact set, there is a convergentsubsequence (x cl ) l∈N which converges to a point denoted x ∗ different fromzero. Also as above, we must have γ(x ∗ ) = 0 and η(x ∗ ) ≥ 0. But thiscontradicts the assumption, namely,{ x ∈ R n \{0} , γ(x) =0} ⇒ η(x) < 0 .Appendix D. Proof of Proposition 2.18. Because the vector field f is homogeneousin the ∞-limit, its approximating vector field f ∞ is homogeneous in thestandard sense (see Remark 2.6). Let d V∞ be a positive real number larger thanr ∞,i for all i in {1,...,n}. Following Rosier [29], there exists a C 1 , positive definite,proper, and homogeneous function V ∞ : R n → R + , with weight r ∞ and degree d V∞ ,satisfying(D.1)∂V ∞∂x (x)f ∞(x) < 0 ∀ x ≠ 0 .From P1 in section 2.2, we know that the function x ↦→ ∂V∞∂x(x)f(x) is homogeneousin the ∞-limit with associated triple ( r ∞ , d ∞ + d V∞ , ∂V∞∂x (x)f ∞(x) ) . Letɛ ∞ = − 1 2 maxθ ∈ S r∞{ ∂V∞∂x (θ)f ∞(θ)and note that, by inequality (D.1), ɛ ∞ is a strictly positive real number. By thedefinition of homogeneity in the ∞-limit, there exists λ ∞ such that∂V ∞∂x(λ r∞ ⋄ θ)f(λ r∞ ⋄ θ)∣ λ d − ∂V ∞V∞ +d∞∂x (θ)f ∞(θ)∣ ≤ ɛ ∞ ∀ θ ∈ S r∞ , ∀ λ ≥ λ ∞ .This yields∂V ∞∂x (λr∞ ⋄ θ)f(λ r∞ ⋄ θ) ≤ λ d V∞ +d∞ ( ∂V∞∂x (θ)f ∞(θ) +ɛ ∞)≤−λ d V∞ +d∞ ɛ ∞ ∀ θ ∈ S r∞ , ∀ λ ≥ λ ∞ ,},Copyright © by SIAM. Unauthorized reproduction of this article is prohibited.
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1842 V. ANDRIEU, L. PRALY, AND A. ASTOLFISimilarly, there ex<strong>is</strong>ts λ ∞ sat<strong>is</strong>fyingη(λ r∞ ⋄θ)−cγ(λ r∞ ⋄θ) ≤−λ d∞ ε 12Consequently, for each c ≥ c 1 , the setif not empty, must be a subset <strong>of</strong>∀ λ ∈ [λ ∞ , +∞) , ∀ θ ∈ S r∞ , ∀ c ≥ c 1 .{x ∈ R n \{0} |η(x) − cγ(x) ≥ 0} ,C = {x ∈ R n : |x| r0 ≥ λ 0 } ⋃ {x ∈ R n : |x| r∞ ≤ λ ∞ } ,which <strong>is</strong> compact and does not contain the origin.3. Suppose now that for all c the first inequality in (2.4) <strong>is</strong> not true. Th<strong>is</strong> meansthat, for all integers c larger than c 1 , there ex<strong>is</strong>ts x c in R n sat<strong>is</strong>fyingη(x c ) − cγ(x c ) ≥ 0 ,and therefore x c <strong>is</strong> in C. Since C <strong>is</strong> a compact set, there <strong>is</strong> a convergentsubsequence (x cl ) l∈N which converges to a point denoted x ∗ different fromzero. Also as above, we must have γ(x ∗ ) = 0 and η(x ∗ ) ≥ 0. But <strong>th<strong>is</strong></strong>contradicts the assumption, namely,{ x ∈ R n \{0} , γ(x) =0} ⇒ η(x) < 0 .Appendix D. Pro<strong>of</strong> <strong>of</strong> Proposition 2.18. Because the vector field f <strong>is</strong> homogeneousin the ∞-limit, its approximating vector field f ∞ <strong>is</strong> homogeneous in thestandard sense (see Remark 2.6). Let d V∞ be a positive real number larger thanr ∞,i for all i in {1,...,n}. Following Rosier [29], there ex<strong>is</strong>ts a C 1 , positive definite,proper, and homogeneous function V ∞ : R n → R + , with weight r ∞ and degree d V∞ ,sat<strong>is</strong>fying(D.1)∂V ∞∂x (x)f ∞(x) < 0 ∀ x ≠ 0 .From P1 in section 2.2, we know that the function x ↦→ ∂V∞∂x(x)f(x) <strong>is</strong> homogeneousin the ∞-limit with associated triple ( r ∞ , d ∞ + d V∞ , ∂V∞∂x (x)f ∞(x) ) . Letɛ ∞ = − 1 2 maxθ ∈ S r∞{ ∂V∞∂x (θ)f ∞(θ)and note that, <strong>by</strong> inequality (D.1), ɛ ∞ <strong>is</strong> a strictly positive real number. By thedefinition <strong>of</strong> homogeneity in the ∞-limit, there ex<strong>is</strong>ts λ ∞ such that∂V ∞∂x(λ r∞ ⋄ θ)f(λ r∞ ⋄ θ)∣ λ d − ∂V ∞V∞ +d∞∂x (θ)f ∞(θ)∣ ≤ ɛ ∞ ∀ θ ∈ S r∞ , ∀ λ ≥ λ ∞ .Th<strong>is</strong> yields∂V ∞∂x (λr∞ ⋄ θ)f(λ r∞ ⋄ θ) ≤ λ d V∞ +d∞ ( ∂V∞∂x (θ)f ∞(θ) +ɛ ∞)≤−λ d V∞ +d∞ ɛ ∞ ∀ θ ∈ S r∞ , ∀ λ ≥ λ ∞ ,},<strong>Copyright</strong> © <strong>by</strong> <strong>SIAM</strong>. <strong>Unauthorized</strong> <strong>reproduction</strong> <strong>of</strong> <strong>th<strong>is</strong></strong> <strong>article</strong> <strong>is</strong> prohibited.
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