3. Conjugacy of Noninvertible Mappings
Definition 2.
Two mappings S,T:X→X are conjugate, if there exists a homeomorphism H:X→X such that
(15)S∘H(x)=H∘T(x).
Definition 3.
Two discrete semidynamical systems Sn,Tn:X→X (n∈ℕ) are conjugate, if there exists a homeomorphism H:X→X such that
(16)Sn∘H(x)=H∘Tn(x).
It is easily verified that two discrete semidynamical systems Sn and Tn, generated by mappings S and T, are conjugate if and only if the mappings S and T are conjugate.
Suppose that mapping (1) has an invariant manifold given by Lipschitzian mapping u:E→F such that
(17)supx∥u(x)∥≤δ,|u(x)-u(x′)|≤|x-x′|.
Our aim is to find a simpler mapping conjugated with (1).
Theorem 4.
If supx(∥(Dg(x))-1∥∥A(x)∥)+5ɛ supx∥(Dg(x))-1∥<1, then there exists a continuous mapping v:E×B(δ)→E which is Lipschitzian with respect to the second variable such that mappings (1) and
(18)x1=X(x,u(x)),r1=R(x+v(x,r),r)
are conjugated in a small neighbourhood of the invariant manifold r=u(x).
We will seek the mapping establishing the conjugacy of (1) and (18) in the form
(19)H(x,r)=(x+v(x,r),r).
We get the following functional equation:
(20)X(x+v(x,r),r)=X(x,u(x)) +v(X(x,u(x)),R(x+v(x,r),r))
or equivalently
(21)v(x,r)=(Dg(x))-1((X(x,u(x)),R(x+v(x,r),r))Dg(x)v(x,r)-X(x+v(x,r),r)xxxxxxxxxxx+X(x,u(x))xxxxxxxxxxx+v(X(x,u(x)),R(x+v(x,r),r))).
The proof of the theorem consists of four lemmas.
Lemma 5.
The functional equation (20) has a unique solution in ℳ1.
Proof.
The set of continuous mappings v:E×B(δ)→E,(22)ℳ={v∈C(E×B(δ),E)∣supx,r|v(x,r)||r-u(x)|<+∞}
becomes a Banach space if we use the norm ∥v∥=supx,r(|v(x,r)|/|r-u(x)|). The set
(23)ℳ1={v∈ℳ∣∥v∥≤1,|v(x,r)-v(x,r′)|≤|r-r′|}
is a closed subset of the Banach space ℳ.
Let us consider the mapping v↦ℒv, v∈ℳ1 defined by the equality
(24)ℒv(x,r) =(Dg(x))-1v(X(x,u(x)),R(x+v(x,r),r)) +(Dg(x))-1(Dg(x)v(x,r)-g(x+v(x,r))+g(x)aaaaaaaaaaaaaaaaa-Ψ(x+v(x,r),r)+Ψ(x,u(x))).
First we obtain
(25)|Lv(x,r)| ≤∥(Dg(x))-1∥|R(x+v(x,r),r)-u(X(x,u(x)))| +∥(Dg(x))-1∥|Dg(x)v(x,r)-g(x+v(x,r))+g(x)| +∥(Dg(x))-1∥|Ψ(x+v(x,r),r)-Ψ(x,u(x))| ≤∥(Dg(x))-1∥ ×(∥A(x)∥+γ|r|+2ε+ω(|r-u(x)|)+2ε) ×|r-u(x)|.
Here we used Hadamard lemma:
(26)g(x′)-g(x)=∫01Dg(x+θ(x′-x))dθ(x′-x).
Next we get
(27)|Lv(x,r)-Lv(x,r′)| ≤∥(Dg(x))-1∥|R(x+v(x,r),r)-R(x+v(x,r′),r′)| +∥(Dg(x))-1∥|Dg(x)(v(x,r)-v(x,r′))xxxxxxxxxxxxxxx-g(x+v(x,r))+g(x+v(x,r′))| +∥(Dg(x))-1∥ ×|Ψ(x+v(x,r),r)-Ψ(x+v(x,r′),r′)| ≤∥(Dg(x))-1∥(∥A(x)∥+2γmax{|r-u(x)|,|r′|}+2ε) ×|r-r′| +∥(Dg(x))-1∥ ×(ω(max{|r-u(x)|,|r′-u(x)|})+2ε)|r-r′|.
In addition,
(28)|Lv(x,r)-Lv′(x,r)| ≤∥(Dg(x))-1∥|R(x+v(x,r),r)-R(x+v′(x,r),r)| +∥(Dg(x))-1∥|v′v(X(x,u(x)),R(x+v(x,r),r))xxxxxxxxxxxxxxx-v′(X(x,u(x)),R(x+v(x,r),r))| +∥(Dg(x))-1∥|Dg(x)(v(x,r)-v′(x,r))xxxxxxxxxxxxxxx-g(x+v(x,r))+g(x+v′(x,r))| +∥(Dg(x))-1∥ ×|Ψ(x+v(x,r),r)-Ψ(x+v′(x,r),r)| ≤∥(Dg(x))-1∥(∥A(x)∥+2γ|r|+3ε)∥v-v′∥|r-u(x)| +∥(Dg(x))-1∥(ω(|r-u(x)|)+ε) ×∥v-v′∥|r-u(x)|.
We choose δ>0, where max{|r|,|r′|}=δ≤a, such that
(29)supx(∥(Dg(x))-1∥∥A(x)∥) +(5ɛ+ω(8δ)+4γδ)supx∥(Dg(x))-1∥<1.
Then ∥ℒv∥≤1, |ℒv(x,r)-ℒv(x,r′)|≤|r-r′|, the mapping ℒ is a contraction, and consequently the functional equation (20) has unique solution in ℳ1.
Next we will prove that the mapping H is a homeomorphism in the small neighbourhood of the invariant manifold r=u(x). Let us consider the functional equation
(30)X(x+v1(x,r),u(x+v1(x,r))) =X(x,r)+v1(X(x,r),R(x,r))
or equivalently
(31)v1(x,r) =(Dg(x))-1(Dg(x)v1(x,r)xxxxxxxxxxxx-X(x+v1(x,r),u(x+v1(x,r)))xxxxxxxxxxxx+X(x,r)+v1(X(x,r),R(x,r))).
Lemma 6.
The functional equation (30) has a unique solution in ℳ2.
Proof.
The set
(32)ℳ2={v∈ℳ∣∥v∥≤1}
is a closed subset of the Banach space ℳ.
Let us consider the mapping v1↦ℒv1, v1∈ℳ2 defined by the equality
(33)ℒv1(x,r) =(Dg(x))-1v1(X(x,r),R(x,r))+(Dg(x))-1 ×(Dg(x)v1(x,r)-g(x+v1(x,r))+g(x) -Ψ(x+v1(x,r),u(x+v1(x,r)))+Ψ(x,r)).
We have
(34)|Lv1(x,r)| ≤∥(Dg(x))-1∥|R(x,r)-u(X(x,r))|+∥(Dg(x))-1∥ ×|Dg(x)v1(x,r)-g(x+v1(x,r))+g(x)| +∥(Dg(x))-1∥ ×|Ψ(x+v1(x,r),u(x+v1(x,r)))-Ψ(x,r)| ≤∥(Dg(x))-1∥(∥A(x)∥+2ε+ω(|r-u(x)|)+3ε) ×|r-u(x)|.
We obtain
(35)|Lv1(x,r)-Lv1′(x,r)| ≤∥(Dg(x))-1∥ ×|v1(X(x,r),R(x,r))-v1′(X(x,r),R(x,r))| +∥(Dg(x))-1∥ ×|Dg(x)(v1(x,r)-v1′(x,r))xxxxxx-g(x+v1(x,r))+g(x+v1′(x,r))|+∥(Dg(x))-1∥ ×|v1′Ψ(x+v1(x,r),u(x+v1(x,r)))xxxxxx-Ψ(x+v1′(x,r),u(x+v1′(x,r)))| ≤∥(Dg(x))-1∥(∥A(x)∥+2ε+ω(|r-u(x)|)+2ε) ×∥v1-v1′∥|r-u(x)|.
We get that ℒ is a contraction and consequently the functional equation (30) has a unique solution in ℳ2.
Consider the mapping G defined by equality G(x,r)=(x+v1(x,r),r).
Lemma 7.
One has G∘H=id.
Proof.
Let us consider the functional equation
(36)X(x+v2(x,r),u(x+v2(x,r))) =X(x,u(x))+v2(X(x,u(x)),R(x+v(x,r),r))
or equivalently
(37)v2(x,r) =(Dg(x))-1 ×(Dg(x)v2(x,r)-X(x+v2(x,r),u(x+v2(x,r)))xxxxx+X(x,u(x))+v2(X(x,u(x)),R(x+v(x,r),r))).
It is easily verified that the functional equation (36) has the trivial solution. Let us prove the uniqueness of the solution in ℳ3, where
(38)ℳ3={v2∈ℳ∣∥v2∥≤3}
is a closed subset of the Banach space ℳ. We get
(39)|v2(x,r)| ≤∥(Dg(x))-1∥∥v2∥|R(x+v(x,r),r)-u(X(x,u(x)))| ×∥(Dg(x))-1∥ ×|Dg(x)v2(x,r)-g(x+v2(x,r))+g(x)| +∥(Dg(x))-1∥ ×|Ψ(x+v2(x,r),u(x+v2(x,r)))-Ψ(x,u(x))| ≤∥(Dg(x))-1∥ ×(∥A(x)∥+γ|r|+2ε+ω(3|r-u(x)|)+2ε) ×∥v2∥|r-u(x)|.
It follows that v2(x,r)≡0. The mapping w1, where
(40)w1(x,r)=v(x,r)+v1(x+v(x,r),r),
also satisfies the functional equation (36). Using the change of variables x↦x+v(x,r) in (30) we get
(41)X(x+w1(x,r),u(x+w1(x,r))) =X(x+v(x,r),r) +v1(X(x+v(x,r),r),R(x+v(x,r),r)).
Using (20), we obtain
(42)X(x+w1(x,r),u(x+w1(x,r))) =X(x,u(x))+v(X(x,u(x)),R(x+v(x,r),r)) +v1(X(x,u(x))+v(X(x,u(x)),R(x+v(x,r),r)),xxxxxxxxR(x+v(x,r),r)) =X(x,u(x))+w1(X(x,u(x)),R(x+v(x,r),r)).
Let us note that
(43)|w1(x,r)|≤|r-u(x)|+|r-u(x+v(x,r))|≤3|r-u(x)|.
Therefore ∥w1∥≤3 and we have
(44)v(x,r)+v1(x+v(x,r),r)=0.
We obtain that G∘H=id.
Lemma 8.
One has H∘G=id.
Proof.
The set of continuous mappings v3:E×B(δ)×B(δ)→E,(45)𝒩={|v3(x,r,z)|max(|r-u(x)|,|z-r|)v3∈C(E×B(δ)×B(δ),E)∣xxxxxsupx,r,z |v3(x,r,z)| max(|r-u(x)|,|z-r|)<∞}
becomes a Banach space if we use the norm ∥v3∥=supx,r,z(|v3(x,r,z)|/max (|r-u(x)|,|z-r|)). The set
(46)𝒩1={z′v3∈𝒩∣∥v3∥≤1, xxxxx|v3(x,r,z)-v3(x,r,z′)|≤|z-z′|}
is a closed subset of the Banach space 𝒩.
Let us consider the functional equation
(47)X(x,r)+v3(X(x,r),R(x,r),R(x+v3(x,r,z),z)) =X(x+v3(x,r,z),z)
or equivalently
(48)v3(x,r,z) =(Dg(x))-1 ×(Dg(x)v3(x,r,z)-g(x+v3(x,r,z))+g(x)xxxxxx+Ψ(x,r)-Ψ(x+v3(x,r,z),z)xxxxxx+v3(X(x,r),R(x,r),R(x+v3(x,r,z),z))).
Let us consider the mapping v3↦ℒv3, v3∈𝒩1 defined by the equality
(49)ℒv3(x,r,z) =(Dg(x))-1 ×(Dg(x)v3(x,r,z)-g(x+v3(x,r,z))+g(x)xxxxxx-Ψ(x+v3(x,r,z),z)+Ψ(x,r)xxxxxx+v3(X(x,r),R(x,r),R(x+v3(x,r,z),z))).
We obtain
(50)|Lv3(x,r,z)| ≤∥(Dg(x))-1∥max{v3|R(x,r)-u(X(x,r))|,xxxxxxxxxxxxxxxxx|R(x+v3(x,r,z),z)-R(x,r)|} +∥(Dg(x))-1∥ ×|Dg(x)v3(x,r,z)-g(x+v3(x,r,z))+g(x)| +∥(Dg(x))-1∥|Ψ(x,r)-Ψ(x+v3(x,r,z),z)| ≤∥(Dg(x))-1∥(∥A(x)∥+γ|z|+2ε) ×max{|r-u(x)|,|z-r|}+∥(Dg(x))-1∥ ×(ω(max{|r-u(x)|,|z-r|})+2ε) ×max{|r-u(x)|,|z-r|}.
In addition,
(51)|Lv3(x,r,z)-Lv3(x,r,z′)| ≤∥(Dg(x))-1∥ ×|Dg(x)(v3(x,r,z)-v3(x,r,z′))xxxxx-g(x+v3(x,r,z))+g(x+v3(x,r,z′))| +∥(Dg(x))-1∥ ×|Ψ(x+v3(x,r,z),z)-Ψ(x+v3(x,r,z′),z′)| +∥(Dg(x))-1∥ ×|R(x+v3(x,r,z),z)-R(x+v3(x,r,z′),z′)| ≤∥(Dg(x))-1∥ ×(ω(max{|r-u(x)|,|z-r|,|z′-r|})+2ε)|z-z′| +∥(Dg(x))-1∥ ×(∥A∥+2ε+2γmax{|r-u(x)|,|z|,|z′-r|}) ×|z-z′|.
Let v3∈𝒩1 and v3′∈𝒩1∪𝒩2 where
(52)𝒩2={v3′∈𝒩∣supx,|r|≤δ,|z|≤δ|v3′(x,r,z)|≤8δ,xxxxx|v3′(x,r,z)-v3′(x,r,z′)|≤|z-z′|supx,|r|≤δ,|z|≤δ}.
We have
(53)|Lv3(x,r,z)-Lv3′(x,r,z)| ≤∥(Dg(x))-1∥ ×|Dg(x)(v3(x,r,z)-v3′(x,r,z))xxxxxx-g(x+v3(x,r,z))+g(x+v3′(x,r,z))| +∥(Dg(x))-1∥ ×|Ψ(x+v3(x,r,z),z)-Ψ(x+v3′(x,r,z),z)| +∥(Dg(x))-1∥ ×|v3′v3(X(x,r),R(x,r),R(x+v3(x,r,z),z))xxxxxx-v3′(X(x,r),R(x,r),R(x+v3(x,r,z),z))| +∥(Dg(x))-1∥ ×|v3′(X(x,r),R(x,r),R(x+v3(x,r,z),z))xxxxxx-v3′(X(x,r),R(x,r),R(x+v3′(x,r,z),z))| ≤∥(Dg(x))-1∥(ω(max{|r-u(x)|,|z-r|,8δ})+ε) ×∥v3-v3′∥max{|r-u(x)|,|z-r|}+∥(Dg(x))-1∥ ×max{v3′|R(x,r)-u(X(x,r))|,xxxxxxxxx|R(x+v3′(x,r,z),z)-R(x,r)|}∥v3-v3′∥ +∥(Dg(x))-1∥ ×|R(x+v3(x,r,z),z)-R(x+v3′(x,r,z),z)| ≤∥(Dg(x))-1∥(ω(8δ)+ε)∥v3-v3′∥ ×max{|r-u(x)|,|z-r|} +∥(Dg(x))-1∥(∥A∥+3ε+γ|z|) ×max{|r-u(x)|,|z-r|}∥v3-v3′∥ =∥(Dg(x))-1∥(∥A(x)∥+4ε+ω(8δ)+γ|z|) ×∥v3-v3′∥max{|r-u(x)|,|z-r|}.
Then ∥ℒv3∥≤1, |ℒv3(x,r,z)-ℒv3(x,r,z′)|≤|z-z′|, the mapping ℒ is a contraction, and consequently the functional equation (47) has a unique solution in 𝒩1. Moreover, this solution is also unique in the closed subset 𝒩2. Let us note that
(54)v3(x,r,r)=0.
The mapping w2, where
(55)w2(x,r,z)=v1(x,r)+v(x+v1(x,r),z),
satisfies (47). Using the change of variables (x,r)↦(x+v1(x,r),z) in (20) we get
(56)X(x+w2(x,r,z),z) =X(x+v1(x,r),0) +v(X(x+v1(x,r),0),R(x+w2(x,r,z),z)).
Using (30) we obtain
(57)X(x+w2(x,r,z),z) =X(x,r)+v1(X(x,r),R(x,r)) +v(X(x,r)+v1(X(x,r),R(x,r)), R(x+w2(x,r,z),z)) =X(x,r) +w2(X(x,r),R(x,r),R(x+w2(x,r,z),z)).
Let us note that
(58)|w2(x,r,z)-w2(x,r,z′)|≤|z-z′|,(59)|w2(x,r,z)|≤|r-u(x)|+|z-r| +|r-u(x+v1(x,r))|≤4max{|r-u(x)|,|z-r|}.
Therefore w2∈𝒩2 and we have
(60)v1(x,r)+v(x+v1(x,r),r)=0.
It follows that H∘G=id.
Finally we conclude that the mapping H is a homeomorphism establishing a conjugacy of the noninvertible mappings (1) and (18).