Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 54, pp. 1–22. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.54 SOLVING LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS GEORGE L. KARAKOSTAS Abstract. We find expressions for solutions to an initial value problem of a linear n-dimensional differential equation. This equation has time deviations of mixed type (advanced and delayed) that have fixed points. The key assumption is that these deviations are odd functions. 1. Introduction Berezansky, Braverman and Pinelas (2023) [3] studied differential equations of the form ẋ(t) + m∑ k=1 bk(t)x(hk(t)) = 0, t ≥ t0, where the functions bk and t → t− hk(t) are Lebesgue measurable and essentially bounded on an interval of the form [t0,+∞) and satisfy −σk ≤ t−hk(t) ≤ τk, with σk, τk ≥ 0. They provided conditions for the existence and uniqueness of solutions, with initial value ϕ(t), t ∈ (t0 − maxk τk, t0) → R, as well as, they studied the asymptotic behaviour. As an application they considered the differential equation ẋ(t) + b(t)x(t− sin t) = 0, t ≥ 0 (1.1) and proved the existence of an exponentially decaying solution in L∞(t0,+∞) [3, Example 4.3]. This is done under the hypothesis that b(t) ≥ b0 > 0 and supt≥0 b(t) < 1. To illustrate their results, in case of b(t) = αe−α sin t, the authors suggested as a solution the function x(t) = e−αt, for a suitable value of the pa- rameter α. Obviously, any function of the form x(t) = ξe−αt, is a solution, for the same values of α and, as we can see, this function is a special solution which does not depend on a certain given initial function at a point t0 > 0, but on the value ξ at 0. Also, in a discussion of the results (no. (2)) at the end of the paper, the authors, among others, posed the following open problem: Is it possible to find conditions such that there exists a unique solution (not necessarily bounded) of an initial value problem for some types of equations with mixed arguments? 2020 Mathematics Subject Classification. 34K06, 34K99. Key words and phrases. Differential equations with mixed arguments; nonsingular matrices. ©2024. This work is licensed under a CC BY 4.0 license. Submitted May 16, 2024. Published September 17, 2024. 1 2 G. L. KARAKOSTAS EJDE-2024/54 In this article we present a type of differential equations and give some answers to these questions. To be more precise we formulate and solve a initial value problem concerning the n-dimensional equation (DEA,B,f ): ẋ(t) = A(t)x(t) +B(t)x(t− τ(t)) + f(t), t ∈ (−R,R), (1.2) for some R ∈ R ∪ {+∞}, when the initial values are points in Rn, as the point ξ above. The function f and the n× n-matrix-valued functions A,B are continuous and defined on (−R,R). The argument τ is a real valued odd continuous function and no boundedness restriction on t − τ(t) (as in [3]) is posed. Thus, if in a set J , the function τ is positive, we have delay argument, while in the symmetric set −J it is negative and so we have advanced argument. Under these characteristics, we give the expression of the (unique) full solution (i.e. it is defined on the whole interval (−R,R)) in the form of a series with terms containing iterated integrals. It is well known that for a differential equation of the form ẋ(t) = H(t, x(t), x(g1(t)), x(g2(t)), . . . , x(gm(t))), t ∈ [a, b], we distinguish the following four cases: (1) Delay DE: All arguments gj are delays. (2) Advanced DE: All arguments gj are advance. (3) Deviated DE: Some of the arguments are delays and the rest of them are advance. (4) Mixed DE: There is a certain argument gj and subsets J1, J2 of the interval [a, b], with non-empty interior, such that gj(t) < t for t ∈ J1, and gj(t) > t for t ∈ J2. Recall that an argument g is a delay, if it satisfies g(t) ≤ t for all t ∈ [a, b] and g(t) < t for some t, and it is an advance argument, if g(t) ≥ t for all t ∈ [a, b] and g(t) > t for some t. The most investigated cases are the first three of them, but for the last one, a few results can be found in the literature. Differential equations with pure delay arguments have studied in many works originated from Volterra (1928) [26] and continued by Minorsky (1942) [22], Myschkis (1949) [23], Hayes (1950) [17], Cunningham (1954) [7], Bellman-Danskin (1954) [5], Razumikhin (1956) [25], Sverkin et al. (1959) [28], to name some of the contribu- tions before 1960. But in the last sixty years a great number of papers and books have been published on the subject. Among them we refer to the classical relative books such as [1, 6, 8, 10, 13, 14, 15, 20]. The problem with the advanced differential equations is, also, too old, originated, probably, by Myschkis (1955) [24]. Contributions to this kind of linear equations, perhaps, start in the work by Kato and McLeod (1971) [19], where an equation of the form y′(t) = ay(λt) + by(t), 0 ≤ t < +∞ (1.3) is investigated for various values of the parameter λ > 0. In [19], for values 0 < λ < 1, equation (1.3) is associated with the boundary condition y(0) = 1, (1.4) but for λ > 1, the authors claim that no such boundary condition seems to arise. Indeed, they prove that the boundary-value problem (1.3)-(1.4) is well-posed, if 0 < λ < 1, but not, if λ > 1. Actually, in [19] the authors seek for analytic solutions of the problem. However in [11] a similar approximation of the solutions of equations like (1.3) is obtained. EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 3 The main property of the solutions discussed in the literature refers to the asymp- totic behaviour and especially to the oscillatory character of them. For instance, such a property of the linear advanced differential equation of the form ẋ(t) = a(t)x(t+ h(t)) + b(t)x(t+ r(t)) = 0, t ≥ 0, where h(t), r(t) ≥ 0, is investigated in [9]. The oscillatory behaviour of solutions of the nth-order nonlinear differential equations with an advanced argument of the form x(n)(t) + δq(t)xλ(g(t)) = 0, is investigated in [12], where n ≥ 2 is even, δ = ±1 and g(t) ≥ t, for all t ≥ t0. The same property is, also, discussed for the differential equation x′(t) + δ1a(t)x(g(t)) + δ2b(t)x(h(t)) = 0, t ≥ t0, in [4], with coefficients a(t) ≥ 0, b(t) ≥ 0, and arguments g delay and h advance. In [21] the oscillatory behaviour of solutions of the second order functional dif- ferential equation with a mixed neutral term of the form( r(t)[(y(t) + p1(t)y(ρ1(t)) + p2(t)y(ρ2(t))) ′]γ )′ + q(t)yγ(σ(t)) = 0, t ≥ t0 is investigated, where ρ1(t) ≤ t ≤ ρ2(t). Similar problems can be found in the references in [21]. On the other hand there are some works, where the main subject is the existence of solutions of differential equations with mixed type arguments. See, for instance, [3] as above. N. Hale et al. (2024), in [16], suggest an absolutely and uniformly convergent series as solutions of the linear equation u′(t) + u(t) = 2u(αt), t > 0, (1.5) where α > 1, to approximate solutions to the nonlinear functional equation of the form u′(t) + u(t) = u2(αt), t > 0. For equation (1.5) it is shown that except of the trivial solution u = 0, with initial value u(0) = 0, there is a nontrivial solution for some special values of the parameter α. In the sequel we shall give sufficient conditions to guarantee uniqueness of the solutions. A similar problem was discussed in [27] concerning the problem f ′(t) = af(λt), t ∈ R, f(0) = 0, (1.6) where nontrivial solutions are constructed for constants λ > 1 and a ̸= 0. Notice that this equation is of advanced type for t > 0 and delay type for t < 0. More facts about differential equations with mixed arguments can be found in a great number of works in the literature, starting from the work of Kato and MacLeon (1971) [19], mentioned above, as well as in [2, 9, 12]. See, also, the refer- ences therein. In all these equations the response function depends on arguments which are either delay, or advance. The problem is what happens when the ar- gument has a mixed type dependence, namely, when in the equation both these situations occur in the same argument. In this note we deal with a special type of linear n-dimensional functional differ- ential equations with mixed argument of the form (1.2). We intend to give results on the existence of full solutions of such an equation with point-wise initial values. 4 G. L. KARAKOSTAS EJDE-2024/54 To obtain our results we use the simple (but very important) fact that the roots of the argument τ are fixed points of the function t→ t− τ(t). Then the results of [18] play a very important role. 2. Preliminaries We furnish the space of n × n matrices with a norm ∥ · ∥ induced from the Euclidean norm | · | of the vector space Rn. We start with some facts borrowed from [18]. Let J := (a, b) be an interval of the real line R. Consider the linear n-dimensional delay differential equation ẋ(t) = A(t)x(t) +B(t)x(g(t)) + f(t), t ∈ J, (2.1) where (C1) g : J → J is a continuous function such that g(t) ≤ t, for all t ∈ J and for some ω with g(ω) = ω it holds g(t) > ω for t ∈ (ω, b). (C2) A,B are n × n-matrix valued continuous functions defined on J and inte- grable on (ω, b). (C3) f : J → Rn is a function continuous and integrable on (ω, b). From the standard theory of linear functional differential equations (see [14, p. 11]) it is well known, that under these conditions, for any σ ∈ I and any continuous function φ : [ḡ(σ), σ] → Rn, there is a unique solution x(t) := x(t; (σ, φ)) such that x(t) = φ(t), for all t ∈ [ḡ(σ), σ]. Here ḡ(σ) = inft≥σ g(t). Let ω be a fixed point of g on I; then [ḡ(ω), ω] = [ω] and therefore any initial function φ for the solution x(ω, φ) is actually a point of Rn. Also, if g does not have a fixed point ω in J , (as, for instance, happens in case of constant delay, namely, g(t) = t− τ,) but we know that limt→ω+ g(t) = ω ∈ R∪{−∞}, then we say that ω is a generalised fixed point of g. In this case the solution x, whose limit limt→ω+ x(t) =: x(ω) = ξ exists, is unique and it is x(t) = EA,B(t, ω)ξ + z(t), t ∈ (ω, b), where EA,B(t, ω) is the evolution matrix of the homogeneous equation and z is the solution of the non-homogeneous equation such that z(ω) = 0. Here the evolution EA,B is defined by the formula EA,B(t, ω) := +∞∑ m=0 [Zm ω (A,B)In×n](t), if t ∈ [ω, b), (2.2) where In×n is the n× n identity matrix and [Zm ω (A,B)T ](t) :=  T (t), if m = 0,∫ t ω [ A(u)[Zm−1 ω (A,B)T ](u) +B(u)[Zm−1 ω (A,B)T ](g(u)) ] du, if m ≥ 1, for all continuous n× n-matrix valued functions T defined at least on the interval (ω, b). Notice that the evolution satisfies the equation ∂ ∂t EA,B(t, ω) = A(t)EA,B(t, ω) +B(t)EA,B(g(t), ω), t ∈ (ω, b), with EA,B(ω, ω) = In×n. EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 5 The solution z of the non-homogeneous equation with initial value z(ω) = 0, is z(t) = +∞∑ m=0 [Zm ω (A,B)F ](t), where F (t) := ∫ t ω f(u)du for t ∈ (ω, b). As an example, we borrow from [18] the information that the evolution E(t,−∞) of equation ẋ(t) = eλtx(t− 1), t ∈ R, where λ ≥ 0, is E(t,−∞) = +∞∑ m=0 exp[λ(mt− 1 2m(m− 1))] m!λm , t ∈ R. Also, the evolution E(t, 0) of equation ẋ(t) = αtβx(γt), t > 0, (2.3) where α, β ≥ 0 and γ ∈ (0, 1), is E(t, 0) = +∞∑ m=0 αmγ 1 2m(m−1)(β+1) m!(β + 1)m tm(β+1), t ≥ 0. (2.4) From the previous arguments, it seems that an initial value problem formed for the differential equation (2.1) is defined on an interval (ω, T ) and satisfies x(ω) = ξ. Hence, it behaves like the analogous one for linear ordinary differential equations. And this is true because the initial value is a vector and not a function defined on the “past”. Now we come back to equation (1.2). In this case the basic assumption on the argument τ is the following (C4) The function τ : (−R,R) → R is continuous and odd and such that in any compact interval there is a finite number of points where it changes its sign. By assumption (C4), the roots of τ are distinc. We have τ(0) = 0 =: r0 and let rk ∈ (0, R), k ∈ {1, . . . , N}, (N ∈ N∪{+∞}), be the (positive) moments where the argument τ changes sign. Notice that, due to oddity of τ , the points −rk ∈ (−R, 0), (k ∈ {1, 2, . . . , N}) are the negative zeros of τ . We shall denote by Jrk the interval [rk, rk+1], if k = 0, 1, . . . , N − 1 and JrN := [rN , R), if k = N < +∞. Also, we denote by J−rk := [−rk,−rk−1], for all k = 1, . . . , N and in case k = N < +∞, J−R := (−R,−rN ]. We let J be the family of all such intervals. These items are such that, if in some interval Jrk , we have τ(t) ≥ 0, then, in the next interval Jrk+1 , it holds τ(t) ≤ 0. Moreover, if it holds τ(t) ≥ 0 on the interval J ∈ J , then τ(t) ≤ 0 on the dual interval −J and inversely. In other words, in this case, the equation is a delay differential equation on the interval J ∈ J and an advanced differential equation on the interval −J ∈ J . In the sequel we shall denote by J+ (resp. J−) the collection of all intervals where it holds τ(t) ≥ 0, t ∈ J (resp. τ(t) ≤ 0, t ∈ J). Further, we assume the following condition (C5) The function g(t) := t− τ(t) maps any J ∈ J to itself. A prototype of such a function τ is the trigonometric function sin appeared in differential equation (1.1). By using the expression (2.2) we conclude that for any interval J ∈ J+, there exists a solution of the problem defined on the whole interval J . The following 6 G. L. KARAKOSTAS EJDE-2024/54 result can be found in [18], but for completeness of this paper, we shall repeat the proof here. Lemma 2.1. Consider equation (1.2) defined on an interval J := [α, β] ∈ J+ and assume that conditions (C2)–(C5) are satisfied on J . Then, for each ξ ∈ Rn, there exists a unique solution defined on J , with the initial value x(α) = ξ. (In case α = −∞, or β = +∞ the values are meant in the limit sense.) Proof. Let x1 and x2 be two solutions with the same initial value ξ. Then we have xi(t) = ξ + ∫ t α A(s)xi(s)ds+ ∫ t α B(s)xi(g(s))ds+ F (t), for i = 1, 2, where F (t) := ∫ t α f(s)ds. Define p(t) = |x1(t)− x2(t)| and assume that for some α′, α′′ ∈ J it holds p(t) = 0, t ∈ [α, α′] and p(t) > 0, t ∈ (α′, α′′]. Then, for any r ∈ (α′, β], we have that Pr := max t∈[α′,r] p(t) > 0. Choose a t1 ∈ J such that∫ t1 α′ (∥A(u)∥+ ∥B(u)∥)du < 1. (2.5) Then, there is a t̄ ∈ [α′, t1], such that Pt1 = p(t̄) = ∣∣∣ ∫ t̄ α′ A(s)(x1(s)− x2(s))ds+ ∫ t̄ α′ B(s)(x1(g(s))− x2(g(s)))ds ∣∣∣ ≤ ∫ t̄ α′ ∥A(s)∥|x1(s)− x2(s)|ds+ ∫ t̄ α′ ∥B(s)∥|x1(g(s))− x2(g(s))|ds ≤ ∫ t1 α′ (∥A(s)∥+ ∥B(s)∥)p(s)ds ≤ Pt1 ∫ t1 α′ (∥A(s)∥+ ∥B(s)∥)ds < Pt1 , because of (2.5). This is a contradiction and so Pt1 = 0. Now, step by step, we conclude the uniqueness result. □ As we stated above, in the delay situation, our initial value problem behaves like the analogous one for linear ordinary differential equations. Moreover, in the ODE case the corresponding evolution is produced by the transition matrix and it is well known that the evolution is a nonsingular matrix. In the present functional case, this is not true, as it is proved in the scalar case in [18]. The question is under what conditions the evolution is a regular matrix. It is evident that in the real-scalar case, if in (1.2) the coefficients A,B are measurable functions and (nontrivially) nonnegative a.e., then the evolution of the homogeneous part of the equation is positive. In the n-dimensional case we have the following result. Lemma 2.2. Consider equation (1.2) with f = 0 and an interval J := [α, β] in J+. Assume that A,B are n× n-matrix valued continuous functions defined on J EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 7 and satisfy the condition (CA,B)∫ β α (∥A(u)∥+ ∥B(u)∥)du < ln 2. (2.6) Then, for each t ∈ J , the evolution matrix EA,B(t, α) is non-singular. Proof. Since EA,B(α, α) = In×n and the matrix In×n is nonsingular, it is enough to discuss it for any fixed t ∈ (α, β]. First, by using induction, we shall prove that, for m ≥ 1, it holds ∥Zm α (A,B)In×n](t)∥ ≤ 1 m! (∫ t α (∥A(u)∥+ ∥B(u)∥)du )m , t ∈ J. (2.7) Indeed, for m = 1, we have ∥Zm α (A,B)In×n](t)∥ = ∥∥∫ t α [ A(u)In×n +B(u)In×n) ] du ∥∥ ≤ 1 1! (∫ t α (∥A(u)∥+ ∥B(u)∥)du )1 . Notice that g(u) := u− τ(u) ≤ u, for all u ∈ J . Next, assume that for some m = r, the claim is true, namely, ∥Zr α(A,B)In×n](t)∥ ≤ 1 r! (∫ t α (∥A(u)∥+ ∥B(u)∥)du )r . Then, for m = r + 1, we have ∥Zr+1 α (A,B)In×n](t)∥ = ∥ ∫ t α [ A(u)[Zr α(A,B)In×n](u) +B(u)[Zr α(A,B)In×n](g(u)) ] du∥ ≤ 1 r! ∫ t α [ ∥A(u)∥+ ∥B(u)∥ ]( ∫ u α (∥A(v)∥+ ∥B(v)∥)dv )r du =: φ(t). We shall show that φ(t) = ψ(t) := 1 (r + 1)! (∫ t α (∥A(v)∥+ ∥B(v)∥)dv )r+1 . To do it we observe that φ(0) = 0 = ψ(0). Also, φ′(t) = 1 r! [ ∥A(t)∥+ ∥B(t)∥ ]( ∫ t α (∥A(v)∥+ ∥B(v)∥)dv )r = ψ′(t). Hence we have φ = ψ, which proves the inequality (2.7). Now, assume that the matrix EA,B(t, α) is singular. Then its kernel has di- mension greater than or equal to 1. Thus a n-vector w exists such that |w| ̸= 0 satisfying equation EA,B(t, α)w = 0. This fact and (2.7) imply that 0 = |EA,B(t, α)w| = |In×nw + +∞∑ m=1 [Zm ω (A,B)In×n](t)w| ≥ |w| − +∞∑ m=1 ∥[Zm ω (A,B)In×n](t)∥|w| 8 G. L. KARAKOSTAS EJDE-2024/54 ≥ |w| − +∞∑ m=1 1 m! (∫ t α (∥A(u)∥+ ∥B(u)∥)du )m |w| = [ 2− exp ∫ β α (∥A(u)∥+ ∥B(u)∥)du ] |w| > 0, a contradiction. This proves the result. □ In the sequel we use the transformations s = −t, x̂(s) := x(t), Â(s) = −A(−s), B̂(s) = −B(−s), f̂(s) := −f(−s). (2.8) Remark 2.3. It is clear that if [α, β] is an interval in J and the condition (1.2)with  and B̂ holds on [α, β], then, from the relation∫ −α −β (∥Â(s)∥+ ∥B̂(s)∥)ds = ∫ β α (∥A(t)∥+ ∥B(t)∥)dt < ln 2, it follows that with  and B̂ with  and B̂ holds on [−β,−α]. 3. Expression of solutions Our main result is given using the folloing assumption (C6) Relation with  and B̂ holds on all intervals of the form J−r2k for k = 1, 2, . . . , N , if J0 ∈ J+; and of the form J−r2k+1 for k = 0, 1, 2, . . . , N , if J0 ∈ J−. Theorem 3.1. Consider the differential equation (1.2) and assume that conditions (C2)–(C6) are satisfied on any interval J ∈ J . Then, for each ξ ∈ Rn, there exists a unique full solution x(t), t ∈ (−R,R) such that x(0) = ξ. Proof. Our plan is to apply the idea in [18] to each interval of the form Jrk and J−rk+1 and obtain successively expressions of the solutions on each such interval. The question is what happens when in an interval of the form (a, b) it holds t−τ(t) > t, namely the differential equation is advanced. To give answer to this question, we use the dual interval (−b,−a), where we have a delay argument, and so the solution can be easily expressed. This justifies our assumption on the oddity of τ . Finally, having found the expression of the solutions on each interval piece- wise we “join” them by using the initial values at the fixed points of the argument, namely at the terms of the sequence . . . ,−rj ,−rj−1, . . . ,−r1, 0, r1, . . . , rj−1, rj , . . . , for j = 1, 2, . . . , N . To be more precise, we solve the problem by discussing two cases: case I: J0 ∈ J+ and case II: J0 ∈ J−. In each case we apply induction on successive stages S(n), n = 1, 2, . . . , where the n-th stage refers to extending the solution on the intervals in the order Jr2n−2 , J−r2n−1 , J−r2n , Jr2n−1 , in case I and on the intervals in the order J−r2n−1 , Jr2n−2 , Jr2n−1 , J−r2n , in case II. Making the transformation (2.8) we obtain the equation, (DEÂ,B̂,f̂ ): ˙̂x(s) = Â(s)x̂(s) + B̂(s)x̂(g(s))) + f̂(s), (3.1) which is associated with (1.2) and has the following property: If (1.2) is an advanced type differential equation on an interval J ∈ J , because g(s) = s − τ(s) = −t − τ(−t) = −t + τ(t) = −(t − τ(t)) ≤ −t = s, equation (3.1) is a delay differential equation and inversely. We call these equations symmetric. EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 9 Case I: J0 ∈ J+. Stage S(1): r r r r r r r r r r r r−r5 −r4 −r3 −r2 −r1 0 r1 r2 r3 r4 r5 r6 − + − + − + − + − + − ? 6? 6? 6 ? 6? 6 The solution on the intervals where τ gets negative values are obtained from the solution on their dual intervals where we have a delay differential equation. Step (I.1)1 [Solution on J0]: First we solve the differential equation (1.2), on the interval J0. Since 0 is a fixed point of the argument g(t) := t − τ(t), according to [18], the solution of (1.2) with initial value ξ at 0 is uniquely determined by its initial value ξ at 0 and it is x(t) = EA,B(t, 0)ξ + +∞∑ m=0 [Zm 0 (A,B)F0](t), t ∈ J0, (3.2) where F0(t) := ∫ t 0 f(u)du, [Zm 0 (A,B)F0](t) :=  F0(t), if m = 0,∫ t 0 [ A(u)[Zm−1 0 (A,B)F0](u) +B(u)[Zm−1 0 (A,B)F0](g(u)) ] du, if m ≥ 1 and EA,B(t, 0) is the function given by (2.2) for ω = 0. Step (I.2)1 [Extension to J−r1 ]: On this interval τ(t) ≤ 0 and so g(t) := t−τ(t) ≥ t, i.e. equation (1.2) is advanced. Observe that for t ∈ J−r1 , we have s = −t ∈ J0. So, its symmetric differential equation is a delay differential equation on J0. Then, as in the previous step, we can solve it and obtain its solution x̂(s) = EÂ,B̂(s, 0)ξ + +∞∑ m=0 [Zm 0 (Â, B̂)F̂0](s), s ∈ J0, (3.3) where F̂0(t) := ∫ t 0 f̂(u)du. Notice that x̂(0) = x(0) = ξ. Now, let w(t) := x̂(−t), t ∈ J−r1 . Since EÂ,B̂ is the evolution of equation (1.2) with Â, B̂ and f̂ in the place of A, B and f , we can easily obtain that ẇ(t) = − ˙̂x(−t) = −Â(−t)x̂(−t)− B̂(−t)x̂(g(−t))− f̂(−t) = A(t)w(t) +B(t)w(g(t)) + f(t). Setting x = w on J−r1 we find that the unique solution x of the original problem on the interval J−r1 is x(t) = EÂ,B̂(−t, 0)ξ + +∞∑ m=0 [Zm 0 (Â, B̂)F̂0](−t), t ∈ J−r1 . (3.4) Step (I.3)1 [Extension to J−r2 ]: On the interval J−r2 equation (1.2) is a delay differential equation and −r2 is a fixed point of the delay. Hence, its solution with 10 G. L. KARAKOSTAS EJDE-2024/54 initial value x(−r2) is x(t) = EA,B(t,−r2)x(−r2) + +∞∑ m=0 [Zm −r2(A,B)F−r2 ](t), t ∈ J−r2 , (3.5) where F−r2(t) := ∫ t −r2 f(u)du and [Zm −r2(A,B)F−r2 ](t) :=  F−r2(t), if m = 0,∫ t −r2 [ A(u)[Zm−1 −r2 (A,B)F−r2 ] (u)du, +B(u)[Zm−1 −r2 (A,B)F−r2 ](g(u)) ] du, if m ≥ 1. (3.6) Our purpose is to express the value x(−r2) in terms of the initial value ξ. First we notice that the matrices A,B satisfy condition with  and B̂ on the interval −Jr1 = J−r2 ∈ J+. Hence, by Lemma 2.2, the matrix EA,B(−r1,−r2) is non- singular. Now, from (3.5) we have x(−r1) = EA,B(−r1,−r2)x(−r2) + +∞∑ m=0 [Zm −r2(A,B)F−r2 ](−r1), and from (3.4) x(−r1) = EÂ,B̂(r1, 0)ξ + +∞∑ m=0 [Zm 0 (Â, B̂)F0](r1). These two equations give x(−r2) = ( EA,B(−r1,−r2) )−1 [ EÂ,B̂(r1, 0)ξ + +∞∑ m=0 ( [Zm 0 (Â, B̂)F0](r1)− [Zm −r2(A,B)F−r2 ](−r1) )] . Therefore on the interval J−r2 the solution x has the expression x(t) = EA,B(t,−r2) ( EA,B(−r1,−r2) )−1 [ EÂ,B̂(r1, 0)ξ + +∞∑ m=0 ( [Zm 0 (Â, B̂)F0](r1)− [Zm −r2(A,B)F−r2 ](−r1) )] + +∞∑ m=0 [Zm −r2(A,B)F−r2 ](t), t ∈ J−r2 . (3.7) Step (I.4)1 [Extension to Jr1 ]: On this interval the argument g(t) = t − τ(t) is of advanced type and for t ∈ Jr1 we have s = −t ∈ J−r2 . So, making the transformation (2.8) we obtain the symmetric equation, which, on the interval J−r2 is a delay differential equation. Hence the equation has the solution expressed as x̂(s) = EÂ,B̂(s,−r2)x̂(−r2) + +∞∑ m=0 [Zm −r2(Â, B̂)F̂−r2 ](s), s ∈ J−r2 , (3.8) EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 11 where, as previously, F̂−r2(s) := ∫ s −r2 f̂(u)du and Zm −r2(Â, B̂) is given as in (3.6), but evaluated with respect to the functions Â, B̂. From here we obtain x̂(−r1) = EÂ,B̂(−r1,−r2)x̂(−r2) + +∞∑ m=0 [Zm −r2(Â, B̂)F̂−r2 ](−r1). On the other hand we know that x̂(−r1) = x(r1) = EA,B(r1, 0)ξ + +∞∑ m=0 [Zm 0 (A,B)F0](r1). From Remark 2.7 and the fact that the interval [−r2,−r1] belongs to J+, the matrix EÂ,B̂(−r1,−r2) is non-singular. Then from the system of the last two relations we obtain x̂(−r2) = [ EÂ,B̂(−r1,−r2) ]−1 [ EA,B(r1, 0)ξ + +∞∑ m=0 ( [Zm 0 (A,B)F0](r1)− Zm −r2(Â, B̂)F̂−r2 ](−r1) )] . Therefore, from (3.8) we obtain the expression of the solution x̂ as x̂(s) = EÂ,B̂(s,−r2) [ EÂ,B̂(−r1,−r2) ]−1 [ EA,B(r1, 0)ξ + +∞∑ m=0 ( [Zm 0 (A,B)F0](r1)− [Zm −r2(Â, B̂)F̂−r2 ](−r1) )] + +∞∑ m=0 [Zm −r2(Â, B̂)F̂−r2 ](s), s ∈ J−r2 . (3.9) Thus the solution of the original equation on the interval Jr1 is x(t) = x̂(−t) = EÂ,B̂(−t,−r2) [ EÂ,B̂(−r1,−r2) ]−1 [ EA,B(r1, 0)ξ + +∞∑ m=0 ( [Zm 0 (A,B)F0](r1)− [Zm −r2(Â, B̂)F̂−r2 ](−r1) )] + +∞∑ m=0 [Zm −r2(Â, B̂)F̂−r2 ](−t), t ∈ Jr1 . (3.10) At this first stage S(1) we obtained the solution x of the differential equation (1.2) on the interval [−r2, r2] and the solution x̂ of the differential equation (3.1) on the same interval. The latter happens, because x̂ is obtained to be defined by (3.3) on the interval J0, by (3.9) on the interval J−r2 and moreover it satisfies x̂(s) = x(t), for s = −t ∈ J−r1 ∪ Jr1 . Notice that the values of x on the intervals [−J−r1 ]∪[−Jr1 ] = J0∪J−r2 are already known. The conclusion is that the solutions x and x̂ have known values on the interval [−r2, r2] and these values are given in terms of the initial value ξ at 0. Now we assume that the process up to the k-stage S(k) gave the solution of (1.2) and of its symmetric equation (3.1) defined on the interval [−r2k, r2k], as functions depending on the initial value ξ at 0. Let X(k)(·; (0, ξ)) and X̂(k)(·; (0, ξ)), respec- tively, be these solutions. Hence, the values X(k)(r2k; (0, ξ)) and X (k)(−r2k; (0, ξ)) 12 G. L. KARAKOSTAS EJDE-2024/54 are known and they will be used in the S(k + 1) stage. Notice that it holds X̂(k)(s; (0, ξ)) = X(k)(−s; (0, ξ)), for all s ∈ [−r2k, r2k]. To proceed to the (k + 1)- stage, S(k + 1), namely, to extend the solution to the interval [−r2k+2, r2k+2], we work as follows: Stage S(k + 1): Step (I.1)k+1 [Extension to Jr2k ]: On the interval Jr2k the argument is a delay, so the solution with initial value x(r2k) at r2k is x(t) = EA,B(t, r2k)x(r2k) + +∞∑ m=0 [Zm r2k (A,B)Fr2k ](t), t ∈ J2k, (3.11) where Fr2k(t) := ∫ t r2k f(u)du, and [Zm r2k (A,B)Fr2k ](t) :=  Fr2k(t), if m = 0,∫ t r2n [ A(u)[Zm−1 r2k (A,B)Fr2k ](u) +B(u)[Zm−1 r2k (A,B)Fr2k ](g(u)) ] du, if m ≥ 1. The value x(r2k) can be found from the k-th stage for t = r2k, which is equal to X(k)(r2k; (0, ξ)). Hence the solution on Jr2k is x(t) = EA,B(t, r2k)X (k)(r2k; (0, ξ)) + +∞∑ m=0 [Zm r2k (A,B)Fr2k ](t), t ∈ Jr2k . (3.12) Step (I.2)k+1 [Extension to J−r2k+1 ]: On this interval the equation is of advanced type. Then, as above, making the transformation (2.8) we obtain the symmetric equation defined on Jr2k . Then, by the previous step, we obtain that its solution is x̂(s) = EÂ,B̂(s, r2k)X̂ (k)(r2k; (0, ξ)) + +∞∑ m=0 [Zm r2k (Â, B̂)F̂r2k ](s), s ∈ Jr2k , (3.13) where, we used that X̂(k)(r2k; (0, ξ)) = X(k)(−r2k; (0, ξ))). This implies that the solution x on the interval J−r2k+1 , with initial value x(0) = ξ, is x(t) = x̂(−t) = EÂ,B̂(−t, r2k)X (k)(−r2k; (0, ξ)) + +∞∑ m=0 [Zm r2k (Â, B̂)F̂r2k ](−t), t ∈ J−r2k+1 . (3.14) Step (I.3)k+1 [Extension to J−r2k+2 ]: On this interval equation (1.2) is a delay differential equation and −r2k+2 is a fixed point of the argument. So, its solution x with initial value x(−r2k+2) is x(t) = EA,B(t,−r2k+2)x(−r2k+2) + +∞∑ m=0 [Zm −r2k+2 (A,B)F−r2k+2 ](t),+ (3.15) for t ∈ J−r2k+2 , where F−r2k+2 (t) := ∫ t −r2k+2 f(u)du and [Zm −r2k+2 (A,B)F−r2k+2 ](t) :=  F−r2k+2 (t), if m = 0,∫ t −r2k+2 [ A(u)[Zm−1 −r2k+2 (A,B)F−r2k+2 ] (u)du, +B(u)[Zm−1 −r2k+2 (A,B)F−r2k+2 ](g(u)) ] du, if m ≥ 1. (3.16) EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 13 We shall express the value x(−r2k+2) in terms of the initial value ξ. First we notice that the matrices A,B satisfy condition (2.6) with  and B̂ on the interval J−r2k+2 ∈ J+. To proceed we observe that from (3.5) we have x(−r2k+1) = EA,B(−r2k+1,−r2k+2)x(−r2k+2) + +∞∑ m=0 [Zm −r2k+2 (A,B)F−r2k+2 ](−r2k+1), and from (3.14), x(−r2k+1) = EÂ,B̂(r2k+1, r2k)X (k)(−r2k; (0, ξ)) + +∞∑ m=0 [Zm r2k (Â, B̂)F̂r2k ](r2k+1), for t ∈ J−r2k+1 . These two equations give x(−r2k+2) = ( EA,B(−r2k+1,−r2k+2) )−1 [ EÂ,B̂(r2k+1, r2n)X (k)(−r2k; (0, ξ)) + +∞∑ m=0 [ [Zm r2k (Â, B̂)F̂2k](r2k+1)− [Zm −r2k+2 (A,B)F−r2k+2 ](−r2n+1) ]] . because, by Lemma 2.2, the matrix EA,B(−r2k+1,−r2k+2) is non-singular. There- fore on the interval J−r2k+2 the solution x has the expression x(t) = EA,B(t,−r2k+2) ( EA,B(−r2k+1,−r2k+2) )−1 × [ EÂ,B̂(r2k+1, r2k)X (k)(−r2k; (0, ξ)) + +∞∑ m=0 [ Zm r2k (Â, B̂)F̂2k](r2k+1)− [Zm −r2k+2 (A,B)F−r2k+2 ](−r2k+1) )] + +∞∑ m=0 [Zm −r2k+2 (A,B)F−r2k+2 ](t), t ∈ J−r2k+2 . (3.17) Step (I.4)k+1 [Extension to Jr2k+1 ]: On this interval the argument g(t) = t− τ(t) is of advanced type and, for t ∈ Jr2k+1 , we have s = −t ∈ J−r2n+2 . So, making the transformation (2.8) we obtain its symmetric equation which is a delay differential equation. This equation has the solution x̂(s) = EÂ,B̂(s,−r2k+2)x̂(−r2k+2) + +∞∑ m=0 [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](s), (3.18) for s ∈ J−r2k+2 , where, as before, F̂−r2k+2 (s) := ∫ s −r2k+2 f̂(u)du and Zm −r2k+2 (Â, B̂) is given as in (3.6), but evaluated with respect to the functions Â, B̂, f̂ defined on the interval J−r2k+2 . From here we obtain x̂(−r2k+1) = EÂ,B̂(−r2k+1,−r2k+2)x̂(−r2k+2) + +∞∑ m=0 [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](−r2k+1). On the other hand, from (3.12), we know that x̂(−r2k+1) = x(r2k+1) 14 G. L. KARAKOSTAS EJDE-2024/54 = EA,B(r2k+1, r2k)X (n)(r2k; (0, ξ)) + +∞∑ m=0 [Zm r2k (A,B)Fr2k ](r2k+1). Since the functions Â, B̂ satisfy condition (2.6) with  and B̂ on the interval [−r2k+2,−r2k+1], from Remark 2.7, it follows that the matrix EÂ,B̂(−r2k+1,−r2k+2) is nonsingular. The system of these two relations give the value x̂(−r2k+2) = [ EÂ,B̂(−r2k+1,−r2k+2) ]−1 { EA,B(r2k+1, r2k)X (n)(r2k; (0, ξ)) + +∞∑ m=0 ( [Zm r2n(A,B)Fr2k ](r2k+1)− [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](−r2k+1) )} . Thus, from (3.18) we obtain the expression of the solution x̂ as x̂(s) = EÂ,B̂(s,−r2k+2) [ EÂ,B̂(−r2k+1,−r2k+2) ]−1 × { EA,B(r2k+1, r2k)X (k)(r2k; (0, ξ)) + +∞∑ m=0 ( [Zm r2k (A,B)Fr2k ](r2k+1) − [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](−r2k+1) )} + +∞∑ m=0 [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](s), s ∈ J−r2k+2 and so the solution of the original equation on the interval Jr2k+1 is x(t) = x̂(−t) = EÂ,B̂(−t,−r2n+2) [ EÂ,B̂(−r2k+1,−r2k+2) ]−1 × { EA,B(r2k+1, r2k)X (k)(r2k; (0, ξ)) + +∞∑ m=0 ( [Zm r2k (A,B)Fr2k ](r2k+1)− [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](−r2k+1) )} + +∞∑ m=0 [Zm −r2k+2 (Â, B̂)F̂−r2k+2 ](−t), t ∈ Jr2k+1 . (3.19) So far we have arrived to stage S(k+1) and we extended the solution x of (1.2) on the interval Jr2k ∪ J−r2k+1 ∪ J−r2k+2 ∪ Jr2k+1 , by the types (3.12), (3.14), (3.17), (3.19). At the same time we obtained the solution of its symmetric on the same intervals. Thus we obtained the expression of the solution x on the entire interval [−r2k+2, r2k+2], for all positive integers k. Case II: J0 ∈ J−. Stage S(1): r r r r r r r r r r r r−r5 −r4 −r3 −r2 −r1 0 r1 r2 r3 r4 r5 r6+ − + − + − + − + − + 6 ? 6 ?6 ? 6 ?6 ? EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 15 Step (II.1)1 [Solution on J−r1 ]: Consider the delay differential equation (1.2) defined on the interval J−r1 . Since −r1 is a fixed point of the argument g(t) := t − τ(t), according to [18], it admits a unique solution x with initial value x(−r1) at −r1, which is uniquely determined by its initial value and it is x(t) = EA,B(t,−r1)x(−r1) + +∞∑ m=0 [Zm −r1(A,B)F−r1 ](t), t ∈ J−r1 , (3.20) where F−r1(t) := ∫ t −r1 f(u)du and, for t ∈ [−r1, 0], [Zm −r1(A,B)F−r1 ](t) :=  F−r1(t), if m = 0,∫ t −r1 [ A(u)[Zm−1 −r1 (A,B)F−r1 ](u) +B(u)[Zm−1 −r1 (A,B)F−r1 ](g(u)) ] du, if m ≥ 1. Keeping in mind that x(0) = ξ, we have x(−r1) = [EA,B(0,−r1)]−1 [ ξ − +∞∑ m=0 [Zm −r1(A,B)F−r1 ](0) ] , since, by (C6), condition (2.6) with  and B̂ is satisfied on the interval J−r1 = −J0 ∈ I+. And so, by Lemma 2.2, the matrix EA,B(0,−r1) is nonsingular. Hence the solution x on the interval J−r1 is x(t) = EA,B(t,−r1)[EA,B(0,−r1)]−1 [ ξ − +∞∑ m=0 [Zm −r1(A,B)F−r1 ](0) ] + +∞∑ m=0 [Zm −r1(A,B)F−r1 ](t), t ∈ J−r1 . Step (II.2)1 [Extension to J0]: On this interval it happens τ(t) ≤ 0, and so its symmetric is a delay differential equation on J−r1 . Then, as in Step (II.1)1, we can solve it and obtain its solution x̂(s) = EÂ,B̂(s,−r1)[EÂ,B̂(0,−r1)] −1 [ ξ − +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](0) ] + +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](s), s ∈ J−r1 , where F̂−r1(t) := ∫ t 0 f̂(u)du. Notice that x̂(0) = ξ. Notice that the matrix EÂ,B̂(0,−r1) is nonsingular, due to Remark 2.7 and condition (2.6) with  and B̂ is satisfied on the interval J−r1 ∈ J+. Now, let w(t) := x̂(−t), t ∈ J0. Since EÂ,B̂ is the evolution of equation (1.2) with Â, B̂ and f̂ in the place of A, B and f , we obtain ẇ(t) = − ˙̂x(−t) = −Â(−t)x̂(−t)− B̂(−t)x̂(g(−t))− f̂(−t) = A(t)w(t) +B(t)w(g(t)) + f(t). 16 G. L. KARAKOSTAS EJDE-2024/54 Put x = w on J0 and so, the unique solution x of the original problem on the interval J0 is x(t) = x̂(−t) = EÂ,B̂(−t,−r1) [ EÂ,B̂(0,−r1) ]−1[ ξ − +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](0) ] + +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](−t), t ∈ J0. (3.21) Step (II.3)1 [Extension to Jr1 ]: On the interval Jr1 equation (1.2) is a delay differential equation and r1 is a fixed point of the delay. Hence, its solution with initial value x(r1) is x(t) = EA,B(t, r1)x(r1) + +∞∑ m=0 [Zm r1 (A,B)Fr1 ](t), t ∈ Jr1 , (3.22) where Fr1(t) := ∫ t r1 f(u)du and [Zm r1 (A,B)Fr1 ](t) :=  Fr1(t), if m = 0,∫ t r1 [ A(u)[Zm−1 r1 (A,B)Fr1 ] (u)du +B(u)[Zm−1 r1 (A,B)Fr1 ](g(u)) ] du, if m ≥ 1. (3.23) We shall express the value x(r1) in terms of the initial value ξ. From relation (3.21) we have x(r1) = [EÂ,B̂(0,−r1)] −1 [ ξ− +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](0) ] + +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](−r1), because EÂ,B̂(−r1,−r1) = In×n. Hence on the interval Jr1 the solution x has the expression x(t) = EA,B(t, r1) ( [EÂ,B̂(0,−r1)] −1 [ ξ − +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](0) ] + +∞∑ m=0 [Zm −r1(Â, B̂)F̂−r1 ](−r1) ) + +∞∑ m=0 [Zm r1 (A,B)Fr1 ](t). Step (II.4)1 [Extension to J−r2 ]: On this interval the argument g(t) = t−τ(t) is of advanced type and for t ∈ J−r2 , we have s = −t ∈ Jr1 . Making the transformation (2.8) we obtain its symmetric which is a delay differential equation on the interval Jr1 . Hence, by the previous step, this equation has the solution expressed as x̂(s) = EÂ,B̂(s, r1) ( [EA,B(0,−r1)]−1 [ ξ − +∞∑ m=0 [Zm −r1(A,B)F−r1 ](0) ] + +∞∑ m=0 [Zm −r1(A,B)F−r1 ](−r1) ) + +∞∑ m=0 [Zm r1 (Â, B̂)F̂r1 ](s), where, recall that, F̂r1(s) := ∫ s r1 f̂(u)du and Zm r1 (Â, B̂) is given as in (3.23), but with Â, B̂ in the place of A,B, respectively.. Thus, the expression of the solution EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 17 x̂ is x(t) = x̂(−t) = EÂ,B̂(−t, r1) ( [EA,B(0,−r1)]−1 [ ξ − +∞∑ m=0 [Zm −r1(A,B)F̂−r1 ](0) ] + +∞∑ m=0 [Zm −r1(A,B)F−r1 ](−r1) ) + +∞∑ m=0 [Zm r1 (Â, B̂)F̂r1 ](−t), t ∈ J−r2 . Now, assume that we have applied the stages S(1), S(2), . . . , S(k) and we found the solutions X(k)(·, (0, ξ)) and X̂(k)(·, (0, ξ)) on the interval [−r2k, r2k]. Recall that it holds X(k)(t, (0, ξ)) = X̂(k)(−t, (0, ξ)), t ∈ [−r2k, r2k]. We shall extend it to [−r2k+2, r2k+2] by proving stage S(k + 1). Stage S(k + 1): Step (II.1)k+1 [Extension to J−r2k+1 ]: On the interval J−r2k+1 the argument is a delay, so the solution with initial value x(−r2k+1) at −r2k+1 is x(t) = EA,B(t,−r2k+1)x(−r2k+1) + +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](t), (3.24) for t ∈ J−r2k+1 , where F−r2k+1 (t) := ∫ t −r2k+1 f(u)du, and [Zm −r2k+1 (A,B)F−r2k+1 ](t) :=  F−r2k+1 (t), if m = 0,∫ t −r2k+1 [ A(u)[Zm−1 −r2k+1 (A,B)F−r2k+1 ](u) +B(u)[Zm−1 −r2k+1 (A,B)F−r2k+1 ](g(u)) ] du, if m ≥ 1. The value x(−r2k+1) can be found from the fact that X(k)(−r2k; (0, ξ)) is already known. Then we have X(k)(−r2k; (0, ξ)) = x(−r2k) = EA,B(−r2k,−r2k+1)x(−r2k+1) + +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](−r2k) and so x(−r2k+1) = [EA,B(−r2k,−r2k+1)] −1 [ X(k)(−r2k; (0, ξ)) − +∞∑ m=0 [Zm −r2n+1 (A,B)F−r2k+1 ](−r2k) ] . Notice that the matrix EA,B(−r2k,−r2k+1) is nonsingular, since, by our assump- tion, condition CA,B is satisfied by the matrices A,B on the interval J−r2k+1 ∈ J+. Now substitute the value x(−r2k+1) into (3.24) and obtain the expression of x on 18 G. L. KARAKOSTAS EJDE-2024/54 the desired interval as follows. x(t) = EA,B(t,−r2k+1)[EA,B(−r2k,−r2k+1)] −1 [ X(k)(−r2k; (0, ξ)) − +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](t), t ∈ J−r2k+1 . (3.25) Step (II.2)k+1 [Extension to Jr2k ]: On this interval it happens τ(t) ≤ 0, and so we have s = −t ∈ J−r2k+1 . So, making the transformation (2.8), it is enough to seek for solutions of (3.1) defined on J−r2k+1 , which is a delay differential equation on it. Then, as in Step (II.1)k+1, we can solve it and obtain its solution x̂(s) = EÂ,B̂(s,−r2k+1)[EÂ,B̂(−r2k,−r2k+1)] −1 × [ X̂(k)(−r2k; (0, ξ))− +∞∑ m=0 [Zm −r2k+1 (Â, B̂)F−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (Â, B̂)F̂−r2k+1 ](s), s ∈ J−r2k+1 , where F̂−r2k+1 (s) := ∫ s −r2k+1 f̂(u)du. The matrix EÂ,B̂(−r2k,−r2k+1) is nonsingu- lar, due to Remark 2.7 and condition (CÂ,B̂) is satisfied on the interval [−r2k+1,−r2k]. Now, let w(t) := x̂(−t), t ∈ Jr2k . Since EÂ,B̂ is the evolution of equation (3.1), we have ẇ(t) = − ˙̂x(−t) = −Â(−t)x̂(−t)− B̂(−t)x̂(g(−t)))− f̂(−t) = A(t)w(t) +B(t)w(g(t))) + f(t), t ∈ Jr2k , where, notice that x̂(−r2k) = X(k)(−r2k; (0, ξ)). Now we set x = w on Ir2k and so we see that the unique solution x of the original problem on the interval Jr2k is x(t) = x̂(−t) = EÂ,B̂(−t,−r2k+1)[EÂ,B̂(−r2k,−r2k+1)] −1 × [ X̂(k)(−r2k; (0, ξ))− +∞∑ m=0 [Zm −r2n+1 (Â, B̂)F−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (Â, B̂)F̂−r2k+1 ](−t), t ∈ Jr2k . (3.26) Step (II.3)k+1 [Extension to Jr2k+1 ]: On the interval Jr2k+1 equation (1.2) is a delay differential equation and r2k+1 is a fixed point of the delay. Hence, its solution with initial value x(r2k+1) is x(t) = EA,B(t, r2k+1)x(r2k+1) + +∞∑ m=0 [Zm r2k+1 (A,B)Fr2k+1 ](t), t ∈ Ir2k+1 , (3.27) EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 19 where Fr2k+1 (t) := ∫ t r2k+1 f(u)du and [Zm r2k+1 (A,B)Fr2k+1 ](t) :=  Fr2k+1 (t), if m = 0,∫ t r2k+1 [ A(u)[Zm−1 r2k+1 (A,B)Fr2k+1 ] (u)du +B(u)[Zm−1 r2k+1 (A,B)Fr2k+1 ](g(u)) ] du, if m ≥ 1. We shall express the value x(r2k+1) in terms of the initial value ξ. From (3.26) we have x(r2k+1) = [EÂ,B̂(−r2k,−r2k+1)] −1 [ X̂(k)(−r2n; (0, ξ)) − +∞∑ m=0 [Zm −r2k+1 (Â, B̂)F̂−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (Â, B̂)F̂−r2k+1 ](−r2k+1), t ∈ Jr2k . because EÂ,B̂(−r2k+1,−r2k+1) = In×n. Hence on the interval Jr2k+1 the solution x has the expression x(t) = EA,B(t, r2n+1) { [EÂ,B̂(−r2k,−r2k+1)] −1 × [ X̂(k)(−r2k; (0, ξ))− +∞∑ m=0 [Zm −r2n+1 (Â, B̂)F̂−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (Â, B̂)F̂−r2k+1 ](−r2k+1) } + +∞∑ m=0 [Zm r2k+1 (A,B)Fr2k+1 ](t), t ∈ Jr2k+1 . (3.28) Step (II.4)k+1 Extension to J−r2k+2 : On this interval the argument g(t) = t−τ(t) is of advanced type and for t ∈ J−r2k+2 , we have s = −t ∈ Jr2k+1 . So, by the transformation (2.8), we obtain its symmetric on the interval Jr2k+1 , which is a delay differential equation. By the previous step, this equation has the solution x̂ expressed as x̂(s) = EÂ,B̂(s, r2k+1) { [EA,B(−r2k,−r2k+1)] −1 × [ X(k)(−r2k; (0.ξ))− +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](−r2k+1) } + +∞∑ m=0 [Zm r2k+1 (Â, B̂)F̂r2k+1 ](s). 20 G. L. KARAKOSTAS EJDE-2024/54 where, recall that, F̂r2k+1 (s) := ∫ s r2k+1 f̂(u)du and Zm r2k+1 (Â, B̂) is given as in (3.23). Thus, the expression of the solution x is x(t) = x̂(−t) = EÂ,B̂(−t, r2k+1) { [EA,B(−r2k,−r2k+1)] −1 × [ X(k)(−r2k; (0, ξ))− +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](−r2k) ] + +∞∑ m=0 [Zm −r2k+1 (A,B)F−r2k+1 ](−r2k+1) } + +∞∑ m=0 [Zm r2k+1 (Â, B̂)F̂r2k+1 ](−t), t ∈ J−r2k+2 . (3.29) By arriving to stage S(k+1) we extended the solution x of equation (1.2) with Â, B̂ and f̂ in the place of A, B and f , on the interval J−r2k+1 ∪Jr2k ∪Jr2k+1 ∪J−r2k+2 , by the types (3.25), (3.26), (3.28) and (3.29) and its symmetric on the same interval [−r2k+2, r2k+2], for all positive integers k. The proof is complete. □ 4. Applications (1) Consider the differential equation ẋ(t) + b(t)x(t− sin t) = f(t), t ≥ 0 (4.1) with mixed argument, where the function f : R → R is continuous. Write it in the form (??) as ẋ(t) = −b(t)x(t− sin t) + f(t), t ∈ R. The points rk = kπ, k = 0, 1, . . . are the fixed points of the argument t − sin t. Moreover all conditions (C1)–(C4) are satisfied. Also, condition (C5) is satisfied, since, for all integers j and t ∈ [jπ, (j + 1)π] it holds jπ ≤ t− sin t ≤ (j + 1)π. Notice that the function t→ t− sin t is increasing. Now, if the function b satisfies∫ +∞ −∞ |b(u)|du < +∞, ∫ (2j+2)π (2j+1)π |b(u)|du < ln 2, ∫ (2j+2)π (2j+1)π |b(−u)|du < ln 2, for all integers j = 1, 2, . . . , then condition (2.6) with A = 0 and B = −b is satisfied and so, by Theorem 3.1, it follows that, for any ξ ∈ R, there is a full solution x of equation (4.1) such that x(0) = ξ. (2) Consider equation (1.6) for which we assume that the coefficient a is not constant, but it depends continuously on time and it satisfies∫ 0 −∞ |a(s)|ds < ln 2 and ∫ +∞ 0 |a(s)|ds < ln 2. Then it is easy to see that all conditions of Theorem 3.1 are satisfied. The collection J consists of two intervals J−∞ := (−∞, 0] ∈ J+ and J0 := [0,+∞) ∈ J−. Notice that λ > 1. Thus Theorem 3.1 is applied and the solutions exist and are defined by x(t; (0, ξ)) = { E0,a(t,−∞)ξ + ∑+∞ m=0[Z m −∞(0, a)](t)ξ, t ≤ 0, E0,â(−t,−∞)ξ + ∑+∞ m=0[Z m −∞(0, â)](−t)ξ, t > 0, EJDE-2024/54 LINEAR DIFFERENTIAL EQUATIONS WITH MIXED ARGUMENTS 21 where [Zm −∞(0, a)](t) = { 1, if m = 0,∫ t −∞ a(u)[Zm−1 −∞ (0, a)](λu)du, if m ≥ 1, for t ≤ 0 and â(s) := −a(−s). (3) Consider equation ẋ(t) = a(t)x(t) + b(t)x(t2k+1), t ∈ R, where a, b are continuous real valued functions that satisfy∫ 1 −1 (|a(s)|+ |b(s)|)ds < +∞, ∫ −1 −∞ (|a(s)|+ |b(s)|)ds < ln 2,∫ +∞ 1 (|a(s)|+ |b(s)|)ds < ln 2. In this case the collection J consists of the intervals J−∞ := (−∞,−1] ∈ J+, J−1 := [−1, 0] ∈ J−, J0 := [0, 1] ∈ J+ and J1 := [1,+∞) ∈ J−. Then Theorem 3.1 is applied and the existence of a unique full solution with initial value ξ ∈ R at 0 is guaranteed. References [1] Bellman, R.; Cooke, K. 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[26] Volterra, V.; Sur la théory mathematique des phénoménes héréditaires, J. Math. Pures Appl., 7 (1928), 249-298. [27] Yoneda, T.; On the functional-differential equation of advanced type f ′(x) = af(λx), λ > 1, with f(0) = 0, J. Math. Anal. Appl., 332 (2007), 487-496. [28] Zverkin, A. M.; Kamenskii, G. A.; Norkin, S. B.; E‘l’sgol’ts, L. E.; Differential equations with a perturbed argument, Russian Mathematical Surveys, Volume 17 (1962), Issue 2, 61-146. DOI: 10.1070/RM1962v017n02ABEH001128. George L. Karakostas Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece (Hellas) Email address: gkarako@uoi.gr, gkarako@hotmail.com 1. Introduction 2. Preliminaries 3. Expression of solutions 4. Applications References