Representation of a Solution of the Cauchy Problem for an Oscillating System with Multiple Delays and Pairwise Permutable Matrices

Theorem 1. Let τ₁, τ₂ > 0, τ : = max {τ₁, τ₂}, and φ ∈ C¹([−τ, 0], ℝ^N). Let B₁,  B₂ be N × N permutable matrices; that is, B₁B₂ = B₂B₁, and let f : [0, ∞) → ℝ^N be a given function. Solution x(t) of

satisfying initial condition has the form where

Lemma 2. Let n ≥ 2 be fixed. Then

for any i₁, …, i_n ≥ 1.

Proof. If n = 2, then the statement follows from the property of binomial coefficients:

Let the statement be true for n − 1. Next, we use the property of multinomial coefficient with inductive hypothesis to derive Clearly, from (16), we get Applying the case n = 2 (property of binomial coefficient) and (16), we get Putting (18) and (19) in (17), we obtain that the statement holds for n and the proof is complete.

Lemma 3. Let n ≥ 1 and τ₁, …, τ_n > 0. Let B₁, …, B_n be N × N pairwise permutable matrices, that is, B_iB_j = B_jB_i for each i, j ∈ {1, …, n}. Then for any t ∈ ℝ,

where the sums are taken over all subsets of {1, …, n} including the trivial ones, and

Proof. Denote ℕ₀, ℕ the set of all nonnegative, positive integers, respectively; that is, ℕ₀ = {0} ∪ ℕ. Thus, we have the trivial identity

Analogically, for any t ∈ ℝ each n-tuple j₁, …, j_n ≥ 0 such that $$ can be divided in two distinct sets of i-s so that j_i ≥ 1 if i ∈ M ⊂ {1, …, n} and j_i = 0 if i ∈ {1, …, n}∖M. That is, M denotes the set of all indices i such that j_i = 0. Moreover, $$. Accordingly, we can write

where the union is taken over all subsets of {1, …, n} including the trivial ones. So, in the view of definition (20), the statement for $$ follows.

Statement for $$ can be proved in a similar way.

Lemma 4. Let n ≥ 3 and τ₁, …, τ_n > 0. Let B₁, …, B_n be N × N pairwise permutable matrices; that is, B_iB_j = B_jB_i for each i, j ∈ {1, …, n}. Then the following holds for any t ∈ ℝ:

• (1)

  if B_i = Θ for some i ∈ {1, …, n}, then

  $$ ()

• (2)

  if τ_i = τ_k for i < k, i, k ∈ {1, …, n}, then

  $$ ()

• (3)

  for any bijective mapping σ : {1, …, n}→{1, …, n} we get

  $$ ()

• (4)

  taking the one-sided derivatives at 0, τ₁, …, τ_n, then

  $$ ()

• (5)

  considering the one-sided derivatives at 0 (they both equal Θ), then

  $$ ()

Statements (1)–(4) hold with 𝒴 instead of 𝒳.

Proof. Statement (1) follows easily from definition of $$, because Θ²ⁱ = E if i = 0 and Θ²ⁱ = Θ whenever i > 0. Next, if τ_i = τ_k, then

for any matrix function F. Thus, using the property of multinomial coefficient (see (16)) for (2), we obtain Property (3) is trivial.

Now, we prove the statement (4). If τ : = τ₁ = ⋯ = τ_n, then

by (2) and from the property of $$ (see (22)).

Hence, without any loss of generality, we assume that τ_i ≠ τ_j for each i ≠ j, i, j ∈ {1, …, n} (in the other case, we collect matrices as stated in (2)). Note the case n = 2 was proved in [16, Lemma 2.3.] Now, assume that $$ solves

that is, that the statement is fulfilled for n − 1 different delays.

Let τ_k : = max _i=1,…,nτ_i. If t < τ_k, then t − τ_k < 0, that is,

and from definition (20) it holds for such t. Consequently, by the inductive hypothesis.

Now, let t ≥ max _i=1,…,nτ_i. Applying Lemma 3, we get

with S_M(t) given by (24) and the sum taken over all subsets of {1, …, n} including the trivial ones. Note that with a characteristic function $$ of a set $$ given by Since each M ⊂ {1, …, n} is a finite set, Lemma 2 yields We apply this identity to derive a formula for the second derivative of S_M for any ∅ ≠ M ⊂ {1, …, n}: Next, for any fixed i ∈ {1, …, n} we split the second sum to j_i = 1 and j_i ≥ 2, that is, and use the equality since So we obtain for each ∅ ≠ M ⊂ {1, …, n}. Obviously, $$. Consequently, Now, we add and subtract to the right-hand side of (50) to get and apply M = M∖{i} whenever i ∉ M: Denoting #M the number of elements of the set M, we split the last two terms of the right-hand side of the latter equality with respect to Hence, we have Now, we show that Let M ⊂ {1, …, n}, and let i ∉ M be arbitrary and fixed such that 1 ≤ #M ≤ n − 1. Then, clearly, and 2 ≤ #(M ∪ {i}) ≤ n, i ∈ M ∪ {i}. Moreover, if M₁, M₂ ⊂ {1, …, n}, i ∉ M_1,2 are such that M₁ ≠ M₂, 1 ≤ #M_1,2 ≤ n − 1, then M₁ ∪ {i} ≠ M₂ ∪ {i}.

On the other side, if M ⊂ {1, …, n}, i ∈ M are arbitrary and fixed such that 2 ≤ #M ≤ n, then

and 1 ≤ #(M∖{i}) ≤ n − 1, i ∉ M∖{i}. Furthermore, if M₁, M₂ ⊂ {1, …, n}, i ∈ M_1,2 are such that M₁ ≠ M₂, 2 ≤ #M_1,2 ≤ n, then, M₁∖{i} ≠ M₂∖{i}. In conclusion, there is 1 − 1 correspondence between the terms on the left-hand side of (56) and the terms on the right-hand side. So (56) is valid.

Putting (56) in (55) we obtain

Next, by the property of empty sum, we get Moreover, it holds Therefore, putting (60) and (61) in (59) and the result in (53), we obtain Hence, $$ solves (31) for all t ≥ 0. Clearly, the same is true for t < 0.

For $$, statements (1)–(3) can be proved as for $$. Next, if τ : = τ₁ = ⋯ = τ_n, we apply the point (2) of this lemma and property (22) for $$ to see that

So, $$ is a solution of (31) when all delays are the same.

Again, the case n = 2 with different delays was proved in [16]; thus, we assume that the statement is fulfilled for n − 1, n ≥ 3 and that τ_i ≠ τ_j for each i ≠ j,    i,   j ∈ {1, …, n}. As before, if t < τ_k and τ_k : = max _i=1,…,nτ_i, then

by definition (20), and the statement follows from the inductive hypothesis. For t ≥ max _i=1,…,nτ_i, we apply Lemma 3 to see that with $$ given by (25). This time and $$. The rest proceeds analogically to $$.

The final statement follows directly from definition (20).

Remark 5. Another proof of statements (1)–(3) of the previous lemma can be made with the aid of statement (4) of the same lemma and uses the uniqueness of a solution of the corresponding initial value problem. For instance in statement (1) of the lemma, both

solve with initial condition and τ = max _i=1,…,nτ_i.

Definition 6. Let τ₁, …, τ_n > 0, τ : = max _i=1,…,nτ_i, and φ ∈ C¹([−τ, 0], ℝ^N), and let B₁, …, B_n be N × N matrices, and let f : [0, ∞) → ℝ^N be a given function. Function x : [−τ, ∞) → ℝ^N is a solution of (11) and initial condition (8), if x ∈ C¹([−τ, ∞), ℝ^N)∩C²([0, ∞), ℝ^N) (taken the second right-hand derivative at 0) satisfies (11) on [0, ∞) and condition (8) on [−τ, 0].

Theorem 7. Let n ≥ 3, τ₁, …, τ_n > 0, τ : = max _i=1,…,nτ_i, and φ ∈ C¹([−τ, 0], ℝ^N), and let B₁, …, B_n be N × N pairwise permutable matrices; that is, B_iB_j = B_jB_i for each i, j ∈ {1, …, n}, and let f : [0, ∞) → ℝ^N be a given function. Solution x(t) of (11) satisfying initial condition (8) has the form

where $$ and $$.

Proof. Obviously, x(t) satisfies the initial condition on [−τ, 0), and, from definition (20), x(0) = φ(0). For the derivative, it holds $$. Moreover, if 0 ≤ t < min _i=1,…,nτ_i, then

since for each i = 1, …, n. Thus and $$. Clearly,

We show that, although 𝒳(t) is not C² at τ₁, …,  τ_n, function x(t) is C² at these points and, therefore, in (0, ∞). At once, we prove that x(t) is a solution of (11).

Assume that 0 ≤ t < min _i=1,…,nτ_i. Then identities (71) and (73) are valid, and by differentiating (73) for such t we get

since x(t − τ_i) = φ(t − τ_i) for each i = 1, …, n.

Now, let ∅ ≠ M_1,2 ⊂ {1, …, n} be such that τ_i ≤ t < τ_j for each i ∈ M₁, j ∈ M₂. Then

whenever j ∈ M₂, and (70) becomes By the point (5) of Lemma 4, we get and for the second derivative it holds since 𝒳(0) = E. Now, we apply the property (4) of Lemma 4 together with to see that both 𝒳 and 𝒴 are solutions of Therefore, In fact, this is exactly formula (11) since x(t − τ_j) = φ(t − τ_j) for each j ∈ M₂.

Finally, if max _i=1,…,nτ_i ≤ t, we have

So, differentiating this formula twice and applying (4) of Lemma 4 result in (11). Hence, one can see that function x(t) given by (70) really solves (11) and satisfies initial condition (8) and, moreover, that x ∈ C²((0, ∞), ℝ^N). To see the last one, one has to put τ₁, …,  τ_n into the computed derivatives, for example, if τ_k : = min _i=1,…,nτ_i < τ_i for each i = 1, …, k − 1, k + 1, …, n, then by (75) and (82) we get where M₂ = {1, …, n}∖{k}.

Corollary 8. Let n ≥ 3, τ₁, …, τ_n > 0, τ : = max _i=1,…,nτ_i, φ ∈ C¹([−τ, 0], ℝ^N), and let B₁, …, B_n be N × N pairwise permutable matrices; that is, B_iB_j = B_jB_i for each i, j ∈ {1, …, n}, and let f : [0, ∞) → ℝ^N be a given function. Solution x(t) of (86) satisfying initial condition (8) has the form

where $$ and $$.

Proof. The corollary can be proved exactly in the same way as Theorem 7.