Nonautonomous Differential Equations in Banach Space and Nonrectifiable Attractivity in Two-Dimensional Linear Differential Systems

Theorem 1. For each x₀ ∈ 𝕏, there exists a unique solution x of (1).

Theorem 2. Let there exist a solution x of (1) such that ∥x(t)∥_𝕏 → 0 as t → 0. If ∥dx/dt∥_𝕏 ∉ L¹(0, t₀), then one has:

Theorem 3. Let the hypotheses (H₁)-(H₂) be fulfilled. Then, for every x₀ ∈ 𝕏, the corresponding solution x of (1) satisfies

• (i)

  ∥x(t)∥_𝕏 → 0 as t → 0;

• (ii)

  x is an injective function on (0, t₀] if and only if x₀ is not an eigenvector of U⁻¹(s)U(t) with an eigenvalue 1 for all s, t ∈ (0, t₀], s ≠ t;

• (iii)

  ∥dx/dt∥_𝕏 ∉ L¹(0, t₀) if and only if ωa ∉ L¹(0, t₀).

Definition 4. The zero solution of system (8) is rectifiable attractive (resp., nonrectifiable attractive) if it is attractive and the curve Γ_x of every solution x of (8) is a rectifiable (resp., nonrectifiable) Jordan curve in ℝ².

Definition 5. Let ℳ₂ denote the linear space of all 2 × 2 matrix with the elements in ℂ. We say that (8) is a linear integrable system if there exists an invertible matrix T ∈ ℳ₂ such that for every t ∈ (0, t₀] the matrix Λ(t) = T⁻¹A(t)T ∈ ℳ₂ is a diagonal matrix for every t ∈ (0, t₀].

Theorem 6. Let μ² − 4νρ < 0. One supposes (15) and

The zero solution of an integrable system (8) with A(t) of form (12) is nonrectifiable (resp., rectifiable) attractive if and only if

Remark 7. Under the assumptions of Theorem 6, we conclude that if g(t) ≡ 0, then the zero solution of integrable system (8) is rectifiable attractive.

Corollary 8. Let h₀ > 0 and b ∈ ℝ. One has that:

• (i)

  if a = 1, the zero solution of (18) is rectifiable attractive provided b < h₀ + 1 and nonrectifiable attractive provided b ≥ h₀ + 1;

• (ii)

  if a > 1, then the zero solution of (18) is rectifiable attractive for all h₀ > 0 and b ∈ ℝ.

Definition 9. Let (F, | · |_n) be a Fréchet space. A mapping T : F → F is said to be an α-contraction on F if for each n ∈ ℕ there exists α_n ∈ [0,1) such that

Lemma 10. Let (F, | · |_n) be a Fréchet space and T : F → F be an α-contraction on F with respect to a family of seminorms ∥·∥_n equivalent to | · |_n. Then, T has a fixed point on F.

Proof of Theorem 1. Let F = C((0, t₀]; 𝕏), x ∈ F, and

It is known that | · |_n is a family of seminorms on F and (F, | · |_n) is a Fréchet space.

Let T : F → F be an operator defined in (20). Since A(t) may be singular at t = 0, it is easy to check that in general T is not an α-contraction with respect to the family of seminorms |x|_n given in (22).

Next, let ω_n and h_n be two sequences of real numbers determined by

Let us remark that if ∥A(t)∥ satisfies (3), then obviously ω_n → ∞ as n → ∞. In respect to these two sequences ω_n and h_n, we define Such sequence ∥·∥_n is a family of seminorms ∥·∥_n which are equivalent with respect to | · |_n because

Next, according to (20) and (23), for t ∈ [1/n, t₀], we calculate

Multiplying previous inequality by $$, where h_n is defined in (24), we obtain Taking the maximum in the previous inequality over the interval [1/n, t₀] and using (24) and (25), we conclude that Now, applying Lemma 10, we obtain the existence of x ∈ F satisfying problem (19)-(20). Since x ∈ F = C((0, t₀]; 𝕏) and A ∈ C((0, t₀]; L(𝕏)), from (19)-(20) follows x ∈ C¹((0, t₀]; 𝕏). Hence, by differentiating (19), we conclude that x = x(t) is also a solution of equation (1). Thus, the existence of a solution of (1) is proved.

Now, we will show the uniqueness of solution of equation (1). Let u(t) and v(t) be two solutions of (1) such that u(t₀) = v(t₀) = x₀. Integrating (1) with respect to u(t) and v(t), we obtain u = Tu and v = Tv, which yields ∥Tu−Tv∥_n = ∥u−v∥_n. Puting this equality into (29), we conclude that

On the other hand, if the solutions u(t) and v(t) are different, then there is t^* ∈ (0, t₀] such that u(t^*) ≠ v(t^*), which implies that $$. In particular, there is a big enough m ∈ ℕ such that t^* ∈ [1/m, t₀], and, hence, that is, Putting (32) into (30), we obtain that α_m ≥ 1. But it is not possible since α_m ∈ [0,1). Hence, the assumption about the existence of t^* ∈ (0, t₀] such that u(t^*) ≠ v(t^*) is not possible. Therefore, we conclude that u(t) = v(t) for all t ∈ (0, t₀], and, thus, the uniqueness of solution of (1) is shown.

Proof of Theorem 2. It is enough to show that if the statement (3) does not hold, then ∥dx/dt∥_𝕏 ∈ L¹(0, t₀) for every x₀ ∈ 𝕏 and the corresponding solution x of (1) such that x(t₀) = x₀. In fact, if we suppose contrary to (3), then there exists c₀ > 0 such that ∥A(t)∥ ≤ c₀, t ∈ (0, t₀]. Also, because of ∥x(t)∥_𝕏 → 0 as t → 0, there exists c₁ > 0 such that ∥x(t)∥_𝕏 ≤ c₁, t ∈ (0, t₀]. Hence, from (1), we obtain

which shows that ∥dx/dt∥_𝕏 ∈ L¹(0, t₀).

Proof of Theorem 3. The conclusion (i) immediately follows from the hypothesis (H₂) because $$ as t → 0.

Next, we give the proof of the conclusion (ii). Let s, t ∈ (0, t₀], s < t, be such that x₀ is an eigenvector of U⁻¹(s)U(t) with an eigenvalue 1. Then,

Applying U(s) to the pervious formula, we obtain the expression U(t)x₀ = U(s)x₀, which by definition of x(t) equals to x(t) = x(s). Hence, x = x(t) is not an injective function. Note that, since U(t) is postulated invertible for all t ∈ (0, t₀], all the steps of this proof are reversible. Thus, the required equivalence holds.

Finally, the conclusion (iii) follows from the hypotheses (H₁)-(H₂) because

Hence, the required equivalence in (iii) holds.

Lemma 11. Suppose that system (8) is integrable. Then, there exist functions g, h : (0, t₀] → ℝ and constants μ, ν, ρ ∈ ℝ such that

The eigenvalues λ_1,2 : (0, t₀] → ℂ of A(t) are given by the formula

Proof. Let λ_1,2 : (0, t₀] → ℂ be two eigenvalues of the matrix-valued function A(t). Since system (8) is supposed to be an integrable system, from Definition 5, there is a regular matrix T ∈ ℳ₂ such that

If det (T) ∈ ℝ and det (T) < 0, then we can change the columns in T which causes det (T) > 0 and diag [λ₂(t), λ₁(t)] instead of diag [λ₁(t), λ₂(t)]. Also, if det (T) ∈ ℝ and det (T) > 0 or det (T) ∈ ℂ∖ℝ, then Hence, we may suppose that det (T) = 1; that is, for some complex numbers α, β, γ, δ with αδ − βγ = 1, we may set Now from (38) and (40), we obtain where the matrix elements of A(t) are real numbers for all t ∈ (0, t₀].

We denote that

and if λ_1,2(t) ∈ ℝ for all t ∈ (0, t₀], then let and if λ_1,2(t) form a complex-conjugate pair for all t ∈ (0, t₀], then let

Putting previous notations in (41), we get the desired conclusion (36). Note that all of these functions and constants are real, since h is defined as a matrix-element of a real matrix, g is defined real in both cases, and the rest of the constants are necessarily real by virtue of satisfying (36). Now, the conclusion (37) immediately follows from (36).

Lemma 12. Let x be a solution of linear integrable system (8) and let T ∈ ℳ₂ be a matrix determined in Definition 5. Then, there exists a matrix-valued function U = U(t), U : (0, t₀] → ℳ₂, such that x(t) = U(t)x₀ and satisfies the following properties:

• (1)

  U(t₀) is the identity matrix;

• (2)

  U(t) is invertible for every t ∈ (0, t₀];

• (3)

  T⁻¹U(t)T is a diagonal matrix for every t ∈ (0, t₀];

• (4)

  if λ(t) is an eigenvalue of A(t), then $$ is an eigenvalue of U(t);

• (5)

  U^′(t) = A(t)U(t) for every t ∈ (0, t₀].

Proof. Let T be the matrix that diagonalizes A(t). Let Λ_A(t) = T⁻¹A(t)T be of form

We define Now, Λ_U(t₀) is obviously the identity matrix, and thus U(t₀) = TIT⁻¹ = I. Since Λ_U(t) is invertible for all t ∈ (0, t₀], so is U(t). Next, T⁻¹U(t)T = Λ_U(t) is diagonal by construction. For an eigenvalue λ(t) of A(t), by construction, $$ is a diagonal element of Λ_U(t); so, it is an eigenvalue of U(t). Also, And finally, the function x(t) = U(t)x₀ solves the integrable system (8) because and has the property x(t₀) = U(t₀)x₀ = x₀.

Lemma 13. Let (8) be a linear integrable system such that μ² − 4νρ < 0 and g(t) ≠ 0, t ∈ (0, t₀], where the real numbers μ, ν, ρ, and g(t) are determined by Lemma 11. Then, the solution′s curve Γ_x of every solution x(t) is a Jordan curve in ℝ² if and only if for every pair s, t ∈ (0, t₀], s < t, at least one of the statements in (15) holds true.

Proof. In order to prove this lemma, we use the equivalence stated in the conclusion (ii) of Theorem 3 (we may use Theorem 3 because of Lemma 12; see also the remark at the end of Theorem 3). According to that, we need to find pairs s and t such that 1 is an eigenvalue of U⁻¹(s)U(t). Since the evolution operator U(t) can be diagonalized into the form Λ_U(t), we compute

The diagonalized form of U⁻¹(s)U(t) is then given by the formula Now, the eigenvalues of U⁻¹(s)U(t) are $$ and $$. Since $$, we only need to solve the equation This equation is solved whenever where the order of integration has been reversed by multiplying by −1.

Since by this argument, U⁻¹(s)U(t) = I if and only if 1 is an eigenvalue of U⁻¹(s)U(t), we get the full equivalence.

Proof of Theorem 6. The key point of this proof is to show that the integrable system (8) satisfies the required hypotheses (H₁)-(H₂), because all conclusions of Theorem 6 follow immediately from the conclusions (i), (ii), and (iii) of Theorem 3. Before we show that, we state the following two propositions which will be proved in Section 5.

Proposition 14. Let f : ℝ × ℝ → ℝ be defined by $$, where μ, ν, ρ are real numbers such that μ² < 4νρ. Then, there exist constants c₁, c₂ > 0 such that

Proposition 15. Let Λ : (0, t₀] → ℳ₂ be a diagonal matrix of the form

for all t ∈ (0, t₀]. Then, det Λ(t) ≥ 0, t ∈ (0, t₀] and

Proof of Corollary 8. It is enough to show that all assumptions of Theorem 6: μ² − 4νρ < 0, (15), (16), and (17) are fulfilled in particular for

where h₀ > 0, a ≥ 1, and b ∈ ℝ. In fact, the following elementary calculation is shown: Moreover, for a = 1, so, condition (17) is fulfilled if and only if b ≥ h₀ + 1. Also, for a > 1, Therefore, we may apply Theorem 6 to system (18) which proves this corollary.

Theorem 16. The set 𝒜((0, t₀]; ℳ₂) is an algebra with respect to the classic matrix operations + and ∘; that is to say, if A, B ∈ 𝒜((0, t₀]; ℳ₂), then the following properties hold:

• (1)

  αA + βB ∈ 𝒜((0, t₀]; ℳ₂) for all α, β ∈ ℝ;

• (2)

  A∘B ∈ 𝒜((0, t₀]; ℳ₂);

• (3)

  if det A(t) ≠ 0 for all t ∈ (0, t₀], then A⁻¹ ∈ 𝒜((0, t₀]; ℳ₂);

• (4)

  if f : ℝ → ℝ is a real analytic function such that f(A(t)) converges absolutely for all t ∈ J, then f(A) ∈ 𝒜((0, t₀]; ℳ₂).

Proof. We denote the diagonal matrices Λ_A = TAT¹ and Λ_B = TBT¹.

• (1)

  T(αA + βB)T⁻¹ = αTAT⁻¹ + βTBT⁻¹ = αΛ_A + βΛ_B, which is diagonal.

• (2)

  TABT⁻¹ = TAT⁻¹TBT⁻¹ = Λ_AΛ_B, which is diagonal.

• (3)

  $$, which is also diagonal.

• (4)

  $$, which is also diagonal.

Proof of Proposition 14. Denote by u ∈ ℝ² the vector u = (x, y). It is enough to show that f = f(u) is a norm as a function f : ℝ² → ℝ, since all the norms in ℝ² are equivalent [30, p. 38], which, by definition, is the claim of this lemma. Let

By direct computation, we see that $$. Since M is symmetric and positive definite, it has a unique symmetric positive definite square root N² = M, so that $$ [31, p. 231]. Hence, $$, where v = Nu; so, the function f = f(u) is a norm.

Proof of Proposition 15. Let ϕ(t):(0, t₀] → ℝ be such that λ(t) = |λ(t) | e^iϕ(t). Then, $$; so, for any z = (z₁, z₂) ∈ ℂ², we have

since |e^±iϕ | = 1. The remaining equality is true because