On a Class of Functional Differential Equations with Symmetries

symmetry S S Article On a Class of Functional Differential Equations with Symmetries Nataliya Dilna 1, Michal Feˇckan 1,2 and András Rontó 3,4,* 1Mathematical Institute of Slovak Academy of Sciences, Štefánikova 49, 814 73 Bratislava, Slovakia; [email protected] (N.D.); [email protected] (M.F.) 2Department of Mathematical Analysis and Numerical Mathematics, Comenius University, Mlynská dolina, 842 48 Bratislava, Slovakia 3Institute of Mathematics, Czech Academy of Sciences, Brno branch, Žižkova 22, 616 62 Brno, Czech Republic 4Brno University of Technology, FBM, Kolejní 4, 612 00 Brno, Czech Republic *Correspondence: r[email protected] Received: 17 October 2019; Accepted: 22 November 2019; Published: 27 November 2019   Abstract: It is shown that a class of symmetric solutions of scalar non-linear functional differential equations can be investigated by using the theory of boundary value problems. We reduce the question to a two-point boundary value problem on a bounded interval and present several conditions ensuring the existence of a unique symmetric solution. Keywords: functional differential equation; argument deviation; periodic; antiperiodic; symmetry; two-point problem; unique solvability MSC: 34K05; 34K10 1. Introduction and Problem Formulation The purpose of this note is to present several conditions ensuring that a functional differential equation possessing a certain symmetry-invariance property has a symmetric solution. The motivation comes partly from previous studies on antiperiodic and more general classes of solutions (e.g., [ 1 – 4 ] and references therein). On the other hand, it is interesting to obtain meaningful solvability conditions in cases where the techniques specific for ordinary differential equations either cannot be easily applied or are not relevant at all. Equations with functional perturbations are interesting from many points of view (see, e.g., [5–9] and references therein). The class of equations we deal with includes, in particular, equations with variable argument deviations and integro-differential equations. More precisely, we consider the general functional differential equation u0(t) = (f u)(t),t∈J, (1) where J is the closure of an unbounded interval and f:ˇ C(J)→Lloc(J) is a mapping (generally speaking, non-linear). We choose ˇ C(J) to be either L∞ loc(J) or PCσ(J) with a specific σ (see (13) in the next section); the reasoning to follow is common for both versions. In the sequel, the following notation is used: PCσ(J) is the Fréchet space of piece-wise continuous functions J→R with possible jump discontinuities at points of a given countable set σ={tk:k∈Z} Symmetry 2019,11, 1456; doi:10.3390/sym11121456 www.mdpi.com/journal/symmetry Symmetry 2019,11, 1456 2 of 13 endowed with the system of seminorms νk(u):=max−k≤j≤ksupt∈(tj−1,tj)|u(t)| , u∈PCσ(J) , k≥ 1; C(J) is the Fréchet space of continuous functions on J with the sequence of seminorms |u|k=maxt∈[−k,k]∩J|u(t)| , k≥ 1, u∈C(J) ; Lloc(J) is the Fréchet space of functions that are Lebesgue integrable over every bounded interval contained in J , with the seminorms kukk=R[−k,k]∩J|u(t)|dt , k≥ 1, u∈Lloc(J) ; L∞ loc(J) is the Fréchet space of locally essentially bounded functions with the seminorms kuk∞;k=ess supt∈[−k,k]∩J|u(t)| , k≥ 1, u∈L∞ loc(J) ; C(J0) and L(J0) are, respectively, the Banach spaces of continuous and integrable functions on a bounded interval J0⊂Jendowed with the standard norms. A solution of (1) is an absolutely continuous function satisfying equality (1) almost everywhere. In what follows, we consider the case where the restriction f|PCσ(J) (or directly f if ˇ C(J) = PCσ(J) ) is continuous as a mapping from PCσ(J) to Lloc(˜ J) with some ˜ J strictly containing J (singularities on the boundary are excluded). We study the problem of the existence of solutions u:J→R of Equation (1) possessing the symmetry property µu(ψ(t)) = u(t),t∈J, (2) where µ∈R\ { 0 } is a given constant and ψ:J→J is a monotone increasing, absolutely continuous function. Due to the nature of property (2) , we impose throughout the paper the following symmetry condition on the operator f: µψ0(t)( f u)(ψ(t)) = ( f u)(t),t∈J, (3) for any function u∈ˇ C(J) possessing property (2) . We also assume that J is invariant with respect to the action of ψ, i.e., ψ(J)⊂J. (4) Typically, Jis either (−∞,∞)or one of the intervals (−∞,0],[0, ∞). In (1) , (3) , and all similar relations below, we assume that the corresponding relations between integrable functions are satisfied almost everywhere and do not always indicate this fact explicitly. Condition (2) describes a class of properties of solutions such as evenness, oddness, periodicity, and antiperiodicity. For example, ψ may have the form ψ(t) = et+T , t∈(−∞ , ∞) , where T is fixed and e∈ {− 1,1 } ; then condition (2) defines a Floquet-type solution with e= 1 (for many-dimensional systems, solutions with this and more general properties are investigated in [ 1 – 4 ]), while for e=− 1, (2) describes the solutions studied in [ 2 , 10 ]. For more complicated functions ψ , the “symmetric” character of property (2) is less obvious compared to, e.g., periodicity or antiperiodicity. For example, with ψ(t) = 2 t− 1 and µ=2−n, property (2) holds for the function u(t) = (t−1)n,t∈R. Note that condition (3) naturally arises in the context of the study of solutions with property (2) . For example, if (1) is an ordinary differential equation of the form u0(t) = h(t,u(t)),t∈(−∞,∞), (5) then assumption (3) is satisfied when the function h:R2→Ris such that µψ0(t)h(ψ(t),µ−1z) = h(t,z)(6) for a.e. t∈(−∞ , ∞) and all z∈R . Relation (6) is a particular case of the property considered in [ 3 ] (see also the references therein for more details) for weakly non-linear systems of ordinary differential equations; it ensures the invariance of Equation (5) under the transformation t→ψ(t) , u→µu . The proof of the existence of symmetric solutions in [3] involves a small parameter argument. In this note, we focus on a general functional differential Equation (1) and formulate several conditions guaranteeing the existence of solutions with property (1) . The method is different from that employed Symmetry 2019,11, 1456 3 of 13 in [ 3 ]; here, we use results from the theory of boundary value problems. Note also that (1) may involve various kinds of argument deviations, in contrast to the most frequently studied delay equations (see, e.g., [11]). 2. Extension by Symmetry It is convenient to work with the restriction of (1) to suitable bounded intervals. Although such restrictions are, generally speaking, impossible for general equations (1) without specifying additional initial data, it turns out that, in our case, this can be done in a straightforward way due to the symmetry assumption (3) . Let us fix some value t0∈J with ψ(t0)6=t0 . For definiteness, assume that ψ(t0)>t0 . It is clear from the problem formulation that the restriction v=u|[t0,ψ(t0)] of every solution u of (1) , (2) to the interval [t0,ψ(t0)] satisfies the two-point boundary condition v(t0) = µv(ψ(t0)). (7) Moreover, a converse statement, in a sense, is true under condition (3) . To formulate it in a rigorous way, we introduce some notation and make the equation setting more specific. The conditions assumed on the function ψ imply that the inverse function ψ−1 is well defined and the sequence of numbers . . . <ψ−2(t0)<ψ−1(t0)<t0<ψ(t0)<ψ2(t0)<. . . is strictly increasing. At this point, it is natural to make more precise the choice of the interval J on which Equation (1) is studied under the symmetry condition (3). Namely, we assume that t0is chosen so that lim k→∞ ψ−k(t0) = α−, (8) lim k→∞ ψk(t0) = α+(9) with |α−|+|α+|=∞and take J:=hα−,α+i, (10) where “ h ” means, respectively, “ [ ” or “ ( ”, depending on whether the corresponding value at the bracket is finite or not. For example, if ψ:t7→ t/2 and t0=−1, then (10) gives J= (−∞, 0]. Thus, we consider Equation (1) on the unbounded interval J of form (10) . Due to the properties mentioned above, Jcan be represented as the union of the half-open intervals Ik:= [ψk(t0),ψk+1(t0)),k∈Z, (11) if α+=∞and Ik:= (ψk(t0),ψk+1(t0)],k∈Z, (12) if α−=−∞ . It is obvious that in both cases, these intervals are mutually disjoint, which justifies the following notation: for every t∈J, put j(t)to be equal to j, where jis such that t∈Ij. In case the domain ˇ C(J) for f in the problem formulation is chosen to be PCσ(J) , from now on, we put σ:={ψk(t0):k∈Z}(13) in the definition of PCσ(J)appearing in the problem formulation. Lemma 1. If a continuous function v :¯ I0→Rsatisfies the two-point boundary condition (7), then the function v∗(t):=µ−j(t)vψ−j(t)(t),t∈J, (14) Symmetry 2019,11, 1456 4 of 13 has property (2). Proof. Let v∗:J→Rbe defined by (14). Since ψis increasing, it is clear from (11) that ψ(Ik)⊂Ik+1(15) for any k, and hence, for t∈I1, (14) yields v∗(t) = µ−1v(ψ−1(t)) = µ−1v∗(ψ−1(t)). Arguing by induction, we easily obtain that v∗satisfies the equality v∗(t) = µ−kv∗ψ−k(t),t∈Ik, (16) for every integer k. It follows immediately from (16) that µv∗(ψ(t)) = µ1−kv∗ψ1−k(t)=v∗(t),t∈Ik, for all k. Since Sk∈ZIk=J, this means that v∗satisfies (2) on J. For any v∈C(¯ I0) , let v∗:J→R stand for the corresponding function (14) . Let C0(¯ I0) be the subspace of C(¯ I0) constituted by the functions satisfying condition (7) (here and below, ¯ Ik denotes the closure of Ik , k∈Z). The following statement is an immediate consequence of formula (14). Lemma 2. Let v∈C(¯ I0) . Then v∗ is continuous if and only if v∈C0(¯ I0) . For v∈C(¯ I0)\C0(¯ I0) , the function v∗∈ˇ C(J)has countably many discontinuities of the first kind at points of set (13). Lemma 1is a natural generalization of the corresponding well-known statements for ordinary differential equations (in particular, on the extension of a periodic solution of an equation with the right-hand-side periodic in time). The function v∗ extends v:¯ I0→R to J by symmetry and, by Lemma 2, v∗ is always continuous if v satisfies condition (7) . For example, if µ= 1 / 2, ψ:t7→ 2 t , J= [ 0, ∞) , and t0= 1, v∗ is continuous for v(t) = t3− 2 t2+ 2, t∈[ 1,2 ) (Figure 1a) and has a jump at each of the points {2k:k=0, ±1, . . . }for v(t) = t3−2t2+1, t∈[1,2)(Figure 1b). (a) (b) Figure 1. An example of a symmetric extension. Symmetry 2019,11, 1456 5 of 13 3. Equation on a Bounded Interval Define the operator ˜ f:C(¯ I0)→L(¯ I0)by putting (˜ f v)(t):= (f v∗)(t),t∈¯ I0, (17) for any v∈C(¯ I0) . Note that since f is well defined on ˇ C(J) , the expression in the right-hand side of (17) makes sense not only for v∈C0(¯ I0), for which v∗is continuous, but for any v∈C(¯ I0). Lemma 3. If a function v :¯ I0→Ris a solution of the equation v0(t) = ( ˜ f v)(t),t∈¯ I0, (18) satisfying condition (7), then the function v∗:J→Ris a solution of (1)possessing property (2). Proof. Let v satisfy (7) and (18) . By Lemma 1, the function v∗ is the extension of v to J with the preservation of symmetry, i.e., v∗(t) = µv∗(ψ(t)),t∈J. (19) Let t∈I−1. Then formula (14) yields v∗(t) = µv(ψ(t)), and therefore, v0 ∗(t) = µψ0(t)v0(ψ(t)). (20) By (15), ψ(t)∈I0for tfrom I−1, and it follows from (18) that v0(ψ(t)) = ( f v∗)(ψ(t)), (21) whence, by (20), v0 ∗(t) = µψ0(t)(f v∗)(ψ(t)),t∈I−1. (22) Since v∗has property (19), using assumption (3) we obtain from (22) that v0 ∗(t) = (f v∗)(t)(23) for t∈I−1. Now let t∈I1. Then, by (14), we have v∗(t) = µ−1v(ψ−1(t)) and v0 ∗(t) = v0(ψ−1(t)) µψ0(ψ−1(t)),t∈I1. (24) The denominator in (24) is non-zero almost everywhere because ψis increasing. Using (15), (18), we find that ψ−1(t)∈I0and v0(ψ−1(t)) = ( f v∗)(ψ−1(t)), whence, by (24), µψ0(ψ−1(t)) v0 ∗(t) = (f v∗)(ψ−1(t)),t∈I1. (25) On the other hand, in view of assumption (3), we have (f v∗)(ψ−1(t)) = µψ0(ψ−1(t)) (f v∗)(t),t∈I1. (26) Symmetry 2019,11, 1456 6 of 13 Combining (25) with (26) , we conclude that (23) holds for t∈I1 . In a similar manner, by induction with respect to negative and positive values of k , we show that equality (23) is true for t∈Ik for any k , i.e., v∗ is a solution of (1). It is worth pointing out that the formulation of Equation (18) on the bounded interval ¯ I0 is correct and no initial functions are needed: all the operations with values of u that may appear in the right-hand side of (18) are well defined since the function is extended by symmetry. For example, if µ= 1 / 2, ψ:t7→ 2 t , J= [0, ∞), the operator fappearing in (1) has the form (f u)(t) = u(τ(t)),t∈J, (27) with a certain τ:J→J , and one needs to compute the values {(˜ f v)(t):t∈[ 1,2 ]} on the function v(t) = t2+ 2, t∈[ 1,2 ] , then we substitute into (27) the corresponding function v∗ the graph of which is presented on Figure 2. We see, in particular, that v∗ is continuous because v satisfies (7) . An easy computation shows that in this case, (˜ f v)(t) = ((τ(t))2+2 if τ(t)∈I0, 2−3k(τ(t))2+21−kif τ(t)∈Ik,k=±1, ±2, . . . . Figure 2. The graph of v∗for vfrom the example. Lemma 4. If f|PCσ(J) is continuous as a mapping from PCσ(J) to Lloc(˜ J) , ˜ f is continuous as a mapping from C(¯ I0)to L(¯ I0). Proof. Let ˇ C(J) = PCσ(J) with σ of form (13) . Let {vm:m≥ 1 } ⊂ C(¯ I0) and vm→v , m→∞ , in C(¯ I0) . Construct the functions v∗:J→R and vm∗:J→R , m≥ 1, according to (14) . If t∈I−1 , then by (15) , ψ(t)∈I0and (14) yields |vm∗(t)−v∗(t)|=|µ||vm(ψ(t)) −vm(ψ(t))|≤|µ|max s∈¯ I0 |vm(s)−v(s)| for all t∈I−1and m≥1. Similarly, for t∈I1, we have ψ−1(t)∈I0, and by (14), |vm∗(t)−v∗(t)|=|µ|−1|vm(ψ−1(t)) −vm(ψ−1(t))|≤|µ|−1max s∈¯ I0 |vm(s)−v(s)| Symmetry 2019,11, 1456 7 of 13 for all t∈I1,m≥1. Continuing by analogy, we find that sup t∈Ij |vm∗(t)−v∗(t)|≤|µ|jmax t∈¯ I0 |vm(t)−v(t)|(28) for all j∈Zand m≥1. Therefore, max −k≤j≤ksup t∈(ψ(t0)j,ψ(t0))j+1 |vm∗(t)−v∗(t)| ≤ max −k≤j≤k|µ|jmax t∈¯ I0 |vm(t)−v(t)| =|µ|ksign(|µ|−1)max t∈¯ I0 |vm(t)−v(t)| → 0 (29) as m→∞ for every k≥ 1, i.e., limm→∞vm∗=v∗ in PCσ(J) . The continuity of f:PCσ(J)→Lloc(˜ J) and definition (17) imply that ˜ f(vm) = f(vm∗)→f(v∗) = ˜ f(v) , m→∞ , in Lloc(˜ J) . Since ˜ J is an open neighborhood of the interval J , it follows that ˜ f(v) has no non-integrable singularities on ¯ I0 , and hence ˜ f(v)∈C(¯ I0). If ˇ C(J) = L∞ loc(J) and {v , v1 , v2 , . . . } ⊂ C(¯ I0) are as above, then ess supt∈Ik|vm∗(t)−v∗(t)| can be estimated using (29), and the same argument can be applied. Lemmata 1and 3allow us to replace the problem of finding a solution u:J→R of (1) with property (2) by the two-point problem (7) and (18) on the bounded interval ¯ I0 . Lemma 4ensures that we are under standard assumptions concerning boundary value problems for first-order functional differential equations. The above-mentioned facts are true in particular for the operators involving inner superpositions, which may have the form (f u)(t) = h(t,u(τ1(t)), . . . , u(τm(t))),t∈J, (30) where h:J→R is a Carathéodory function and τi:J→J , i= 1,2, . . . , m , are measurable. In this case, (1) is an equation with argument deviations, and the procedure of restriction of Equation (1) to the bounded interval ¯ I0 , in fact, corresponds to the well-known techniques from [ 5 ]. The symmetry condition (3) for operator (30) can be verified, e.g., using the following simple lemma. Lemma 5. Let there exist integers k1, k2,. . . , kmsuch that τi◦ψ=ψki◦τi,i=1, 2, . . . , m, (31) and ψ0(t)h(ψ(t),µ−k1z1,µ−k2z2, . . . , µ−kmzm) = µ−1h(t,z1,z2, . . . , zm)(32) for all real z1 , . . . , zm and almost every t∈J . Then the operator f:ˇ C(J)→Lloc(J) given by (30) satisfies condition (3). Proof. The proof is based on assumption (31) . Indeed, let u∈C(J) be such that (2) holds. Then u(ψ(t)) = µ−1u(t),t∈J, and (31) yields u(ψki(τi(t))) = µ−1u(ψki−1(τi(t))) = · · · =µ−kiu(τi(t)),t∈J,i=1, 2, . . . , m. Symmetry 2019,11, 1456 8 of 13 By virtue of (32), µ ψ0(t)h(ψ(t),µ−k1u(τ1(t)), . . . , µ−kmu(τm(t))) = h(t,u(τ1(t)), . . . , u(τm(t))), which in view of the arbitrariness of uwith property (2), proves (3). Lemma 5allows us to check condition (3) for a class of equations with argument deviations and carry out the transition from (1) and (2) to (7) and (18). For example, consider the problem u0(t) = b(t) + (u(τ1(t)))2ν+1 a(t) + (u(τ2(t)))2ν, (33) u(t) = µu(t+µ),t∈(−∞,∞), (34) where ν> 0 and µ> 0 are constants, a , b are functions integrable on every bounded interval, and such that ais positive, a(t+µ) = µ−2νa(t), (35) b(t+µ) = µ−1b(t)(36) for all t∈(−∞,∞), and τi(t):=nit+βi,t∈(−∞,∞),i=1, 2, (37) with {n1,n2} ⊂ Zand {β1,β2} ⊂ R. Define f by (30) with m= 2, h:(t , z1 , z2)7→ b(t) + z2ν+1 1/(a(t) + z2ν 2) . Then problem (33) and (34) is a particular case of (1) and (2) with ψ:t→t+µ on J= (−∞ , ∞) (it is obvious that α−=−∞ and α+=∞ in (10) for this case). It is easy to see that functions (37) satisfy condition (31), which means here that τi(t+µ) = τi(t) + niµ,t∈(−∞,∞), where n1,n2are integers. Furthermore, by (35) and (36), µh(t+µ,µ−1z1,µ−1z2) = µb(t+µ) + µz2ν+1 1µ−2ν−1 a(t+µ) + z2ν 2µ−2ν =b(t) + µz2ν+1 1µ−2ν−1 µ−2νa(t) + z2ν 2µ−2ν=h(t,z1,z2) for any t and z1 , z2 , i.e., h satisfies condition (32) with the given ψ and k1= 1, k2= 1. Consequently, the problem of finding solutions u:R→R of (33) possessing property (34) can be replaced by the corresponding two-point problem (7) and (18) on a bounded interval of length µ. 4. Existence of a Unique Symmetric Solution We formulate conditions in terms of the “restriction” operator ˜ f given by (17) . Consider the case where the operator ˜ fadmits the estimate |(˜ f u)(t)−(˜ f v)(t) + g1(u−v)(t)| ≤ g0(u−v)(t),t∈¯ I0, (38) for all u , v from C(¯ I0) , where gi:C(¯ I0)→L(¯ I0) , i= 0,1, are certain positive linear operators. By a positive operator, we mean an operator p:C(¯ I0)→L(¯ I0) such that essinft∈¯ I0(pu)(t)≥ 0 for all u∈C(¯ I0) such that mint∈¯ I0u(t)≥0. Symmetry 2019,11, 1456 9 of 13 Theorem 1. Let there exist positive linear operators gi:C(¯ I0)→L(¯ I0) , i= 0,1, such that (38) holds for all u , v from C(¯ I0). Let |µ| ≤ 1and 1− |µ| |µ|<θkg1(1)k ≤ |µ|(39) with a certain θ∈(0, 1). In addition, assume that kg0(1)k+ (1−2θ)kg1(1)k<1− |µ|(40) if 0<θ≤1/2 or kg0(1)k<1, kg0(1)k 1− kg0(1)k−1− |µ| |µ|<(2θ−1)kg1(1)k ≤ |µ|(41) if 1/2 <θ<1. Then Equation (1)has a unique solution possessing property (2). In (39) – (41) and similar relations below, gi( 1 ) stands for the value of gi on the constant function equal to 1, and k·kis the norm in L(¯ I0). Theorem 2. Let there exist positive linear operators gi:C(¯ I0)→L(¯ I0) , i= 0,1, such that (38) holds for all u , v from C(¯ I0). Let |µ| ≥ 1and |µ| − 1<θkg1(1)k ≤ 1 (42) with some θ∈(0, 1). In addition, assume that kg0(1) + (1−2θ)g1(1)k<1 |µ|(43) if 0<θ≤1/2 or kg0(1)k<1 |µ|,|µ| 1− |µ|kg0(1)k−1<(2θ−1)kg1(1)k ≤ 1(44) if 1/2 <θ<1. Then Equation (1)has a unique solution possessing property (2). Condition (38) is satisfied, in particular, if (1) is a linear equation of the form u0(t) = (p0u)(t)−(p1u)(t) + r(t),t∈J, (45) where pi:ˇ C(J)→Lloc(˜ J) , i= 0,1, are positive linear operators such that their restrictions to C(J) are continuous mappings from C(J) to Lloc(˜ J) with some ˜ J strictly containing J . In this case, the symmetry condition holds if r∈Lloc(˜ J)is such that µψ0(t)r(ψ(t)) = r(t),t∈J, (46) and µψ0(t)(piu)(ψ(t)) = (piu)(t),t∈J,i=0,1, (47) for any absolutely continuous function u possessing property (2) , and Theorems 1and 2can be applied (in fact, p0=˜ g0 and p1=˜ g1 in this case). Other conditions for the existence of symmetric solutions of (45) are given by the next two statements.