Solvability of a class of hyperbolic-cosine-type difference equations

Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 https://doi.org/10.1186/s13662-020-03027-8 RESEARCH Open Access Solvability of a class of hyperbolic-cosine-type difference equations Stevo Stevi´ c1,2,3*, Bratislav Iriˇ canin4,5, Witold Kosmala6and Zdenˇ ek Šmarda3 *Correspondence: [email protected] 1Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia 2Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan, Republic of China Full list of author information is available at the end of the article Abstract We describe a method for constructing one of the basic classes of solvable hyperbolic-cosine-type difference equations, generalizing a known difference equation by Laplace in a natural way. MSC: Primary 39A20; secondary 39A06; 39A45 Keywords: Difference equation; Solvable equation; Closed-form formula; Hyperbolic-cosine-type difference equation 1 Introduction Let N,Z,R,Cbe the sets of natural, whole, real, and complex numbers, respectively, and N0=N∪{0}.Ifk,l∈Z,thenj=k,lstandsforthesetofallj∈Zsuch that k≤j≤l. Finding closed-form formulas for solutions to difference equations is one of the basic problems in the area. The equation yn+2 =ayn+1 +byn,n∈N0,(1) whenb=0anda2+4b=0,wassolvedbydeMoivrein[1](seealso[2]).Hefoundaformula forsolutiontoequation(1), which is called the de Moivre formula for solutions to linear homogeneous second-order difference equation with constant coefficients, which is one of the first nontrivial results on solvability of difference equations. Before it, some special cases of equation (1)hadbeensolvedin[3]. Theseresultsattractedsomeattention,andsoonafterthatBernoulliin[4]foundanother method for solving linear difference equations with constant coefficients. A presentation of some old results on solvability can be found in [5]. For some later results see [6,7], as wellas[8],wheremanyclasses of differenceequationsandsystemsweresolved.For some twentieth century presentations of the theory, see, for example, [9–13]. Some recent resultsonsolvabilityofdifferenceequationsandsystemshavebeenobtainedandguessedby computer packages for symbolic calculations. They can help in getting or guessing some closed-formformulasforsolutionstotheequationsandsystems,butusingonlysuchtools could also produce some issues (see some comments, e.g., in [14–17]). This has been one ofthereasonswhichmotivatedustoconductmoreseriousinvestigationsonsolvabilityof ©The Author(s) 2020. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 2 of 12 difference equations and systems. Recent investigations show that still a great majority of papers on solvability use some substitutions of various complexity, which transform difference equations and systems to known solvable ones (see, for example, [14,18–26]and thereferencestherein).Itshouldbementionedthatsolvabledifferenceequationsandsystemshavemanyapplications(see,e.g.,[2,4,7–10,12,27–29]).Forinvariantsfordifference equations and systems, and their applications in solvability, see, for example, [30–35]. Recently, solvability of the so-called hyperbolic-cotangent-type difference equations, as well as of the corresponding systems of difference equations, has been studied (see [22– 26]). The difference equations and systems therein resemble the hyperbolic-cotangent sum formula which has been a good hint for solvability of the equations and systems. Generally speaking, the difference equations and systems which resemble some trigonometric or hyperbolictrigonometricformulasare naturalcandidatestobe solvable.Thisis an observation known to mathematicians for a longtime. Now, as a motivation for the study, we present a known example along with the most important details related to solvability of the equation in the example. Example1 The following difference equation xn+1 =x2 n–2, n∈N0,(2) was already known to Laplace [8]. He noticed that equation (2)issolvable.Namely,if x0∈C, then there is a∈C\{0}such that x0=a+1 a(3) (see, e.g., [36]). By using (3)in(2), then repeating the procedure, he noticed that x1=a+1 a2–2=a2+1 a2, x2=a2+1 a22–2=a4+1 a4, x3=a4+1 a42–2=a8+1 a8, and concluded xn=a2n+1 a2n,n∈N0,(4) which is easily proved by induction. Laplace did not conduct further analysis of solutions to equation (2). In what follows we mention several simple folklore things related to solvability of the equationinthecasewhenx0isa realnumber.Ifx0≥2,thenx0canbewrittenintheform givenin(3)forsomea>0.Forx0=2,wehavex1= 2, and by the method of induction, constant solution xn=2foreveryn∈N0is easily obtained. In this case there is unique a Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 3 of 12 suchthat(3)holds,namelya=1.Ifx0>2,thentherearetwopositivevaluesof asuchthat (3) holds. They are the roots of the quadratic polynomial t2–x0t+1,thatis, a1,2 =x0±x2 0–4 2.(5) If one of the numbers a1,2 is denoted by a, by the Viète formulas, we see that the second oneis 1/a.So,sinceformula(4)isinvariantunderthetransformationa→1/a,whichever of these two numbers is used, the formula is always obtained. By combining (4)and(5), we see that the solution to equation (2)inthiscasecanbe written as follows: xn=x0+x2 0–4 22n +x0+x2 0–4 2–2n ,n∈N0. Ifx0≤–2,thenx1=x2 0–2≥2.Thismeansthatthiscaseisreducedtothepreviousone. Namely, if x2 0–2=x1=b+1 b(6) for some b>0,then xn=b2n–1 +1 b2n–1 ,n∈N.(7) From (6)wehaveb2–(x2 0–2)b+1=0,sothat b1,2 =x2 0–2±|x0|x2 0–4 2, and b1=1/b2.Usingthisin(7), we get xn=x2 0–2–x0x2 0–4 22n–1 +x2 0–2–x0x2 0–4 2–2n–1 ,n∈N.(8) If a>0,notethat(4)canbewritteninthefollowingform: xn=e2nlna+1 e2nlna=2cosh2nlna=2cosh2n a,n∈N0,(9) where a=lna.Thismeansthatifx0=2cosh a,thenxn=2cosh(2n a), n∈N0. Bearing in mind the form of formula (9), we see that equation (2) is closely related to the hyperbolic cosine function. This connection is not so strange at all. Namely, by using thechangeofvariablesxn=2˜ xn,n∈N0,inequation(2), it is transformed to the following one: ˜ xn+1 =2˜ x2 n–1, n∈N0, Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 4 of 12 which resembles the formula cosh2x=2cosh2x–1. Because of this, it is natural to say that equation (2)inthecase|x0|≥2 is an example of hyperbolic-cosine-type difference equations. If x0∈[–2,2], then x0can be written in the form given in (3)forsomea=eiθ,where θ∈[0,2π). In this case we have x0=2cosθ,whereasfrom(4)weget xn=2cos2nθ,n∈N0. Hence, in the case x0∈[–2,2], equation (2) is an example of cosine-type difference equations. A natural problem is to try to find related hyperbolic-cosine-type difference equations, whicharealsosolvable.Thisproblemseemsclassicalone,butwecouldnotfindacomplete solution to the problemintheliteraturesofar.Besidethis,itis goodtohaveallthethings, someofwhichseemscatteredintheliterature,inthesameplace.Hence,weconsiderhere theproblemin detail. Weshowthatthereis a natural sequenceof hyperbolic-cosine-type difference equationswhicharesolvableinclosed form anddescribe asimpleconstructive way for obtaining the sequence of equations. 2 A basic class of solvable hyperbolic-cosine-type difference equations In this section we explain how a natural class/sequence of hyperbolic-cosine-type difference equationsrelatedtoequation (2) is obtained, which are also solvable. 2.1 Basic ideas and equations First, note that the main thing connected to solvability of equation (2) is the fact that the following relation holds: a2+1 a2=a+1 a2–2 (10) for every a∈C\{0}, which is a simple, but no doubt very useful, relation between the quantities Ik:=ak+1 ak for k=1andk=2. The consideration in Example 1suggests that if we can express the quantity I3in terms of I1in a similar way, then we can obtain another solvable difference equation. It is not difficult to see that such a relation exists. Namely, we have a+1 a3=a3+1 a3+3a+1 a(11) for every a∈C\{0}. Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 5 of 12 Let the sequence (xn)n∈N0be a solution to the following difference equation: xn+1 =x3 n–3xn,n∈N0, (12) and x0∈C. Write the initial value x0in the form in (3). Then we have x1=a+1 a3–3a+1 a=a3+1 a3, and by a simple inductive argument, we obtain xn=a3n+1 a3n,n∈N0. (13) The corresponding consideration in Example 1shows that equation (12)when|x0|≥ 2 is also an example of a hyperbolic-cosine-type difference equation, which is solvable. Moreover, we see that the following result holds. Proposition1 Consider equation (12). Then the following statements hold: (a) If x0∈Cis given by (3), then the solution to the equation is given by (13). (b) If x0≥2,then the solution to the equation is given by xn=x0+x2 0–4 23n +x0+x2 0–4 2–3n ,n∈N0. (c) If x0≤–2,then the solution to the equation is given by xn=x2 0–2–x0x2 0–4 23n–1 +x2 0–2–x0x2 0–4 2–3n–1 ,n∈N. (d) If |x0|≤2and x0=2cosθfor some θ∈[0,2π),then the solution to the equation is given by xn=2cos3nθ,n∈N0. Following the above idea, we can try to express thequantity I4in terms of I1ina similar way, and then use the relation in order to obtain another solvable difference equation. Namely, we have a+1 a4=a4+1 a4+4a2+1 a2+6=a4+1 a4+4a+1 a2–2. (14) Let the sequence (xn)n∈N0be a solution to the following difference equation: xn+1 =x4 n–4x2 n+2, n∈N0, (15) and x0∈C. Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 6 of 12 Write the initial value x0in the form in (3). Then we have x1=a+1 a4–4a+1 a2+2=a4+1 a4, and by a simple inductive argument, we obtain xn=a4n+1 a4n(16) for n∈N0. So, equation (15) is also an example of a hyperbolic-cosine-type difference equation, which is solvable. Moreover, we see that the following result holds. Proposition2 Consider equation (15). Then the following statements hold: (a) If x0∈Cis given by (3), then the solution to the equation is given by (16). (b) If x0≥2,general solution to the equation is given by xn=x0+x2 0–4 24n +x0+x2 0–4 2–4n ,n∈N0. (c) If x0≤–2,general solution to the equation is given by the following formula: xn=x2 0–2–x0x2 0–4 24n–1 +x2 0–2–x0x2 0–4 2–4n–1 ,n∈N. (d) If |x0|≤2and x0=2cosθfor some θ∈[0,2π),then the solution to the equation is given by xn=2cos4nθ,n∈N0. The corresponding relation between I5and I1is the following: a+1 a5=a5+1 a5+5a3+1 a3+10a+1 a =a5+1 a5+5a+1 a3–5a+1 a, (17) where in the last equality we have used relation (11). Let the sequence (xn)n∈N0be a solution to the following difference equation: xn+1 =x5 n–5x3 n+5xn,n∈N0, (18) and x0∈C. Write the initial value x0in the form in (3). Then we have x1=a+1 a5–5a+1 a3+5a+1 a=a5+1 a5, Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 7 of 12 and by a simple inductive argument, we obtain xn=a5n+1 a5n,n∈N0. (19) So, equation (18) is another example of a hyperbolic-cosine-type difference equation, which is solvable. Moreover, we see that the following result holds. Proposition3 Consider equation (18). Then the following statements hold: (a) If x0∈Cis given by (3), then the solution to the equation is given by (19). (b) If x0≥2,general solution to the equation is given by xn=x0+x2 0–4 25n +x0+x2 0–4 2–5n ,n∈N0. (c) If x0≤–2,general solution to the equation is given by the following formula: xn=x2 0–2–x0x2 0–4 25n–1 +x2 0–2–x0x2 0–4 2–5n–1 ,n∈N. (d) If |x0|≤2and x0=2cosθfor some θ∈[0,2π),then the solution to the equation is given by xn=2cos5nθ,n∈N0. The corresponding relation between I6and I1is the following: a+1 a6=a6+1 a6+6a4+1 a4+15a2+1 a2+20 =a6+1 a6+6a+1 a4–9a+1 a2+2, (20) where in the last equality we have used (10)and(14). Let the sequence (xn)n∈N0be a solution to the following difference equation: xn+1 =x6 n–6x4 n+9x2 n–2, n∈N0, (21) and x0∈C. Write the initial value x0in the form in (3). Then we have x1=a+1 a6–6a+1 a4+9a+1 a2–2=a6+1 a6, and by a simple inductive argument, we obtain xn=a6n+1 a6n,n∈N0. (22) So, equation (21) is another example of a hyperbolic-cosine-type difference equation, which is solvable on a domain. Moreover, we see that the following result holds. Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 8 of 12 Proposition4 Consider equation (21). Then the following statements hold: (a) If x0∈Cis given by (3), then the solution to the equation is given by (22). (b) If x0≥2,general solution to the equation is given by xn=x0+x2 0–4 26n +x0+x2 0–4 2–6n ,n∈N0. (c) If x0≤–2,general solution to the equation is given by the following formula: xn=x2 0–2–x0x2 0–4 26n–1 +x2 0–2–x0x2 0–4 2–6n–1 ,n∈N. (d) If |x0|≤2and x0=2cosθfor some θ∈[0,2π),then the solution to the equation is given by xn=2cos6nθ,n∈N0. Remark 1Relations(10), (11), (14), (17), and (20) are well known and are frequently used in various situations such as in solving the polynomial equations akxk+ak–1xk–1 +···+a1x+a0=0, in the case aj=ak–j,j=0,k(see, e.g., [37]). 2.2 Main equation ByusingtheprocedureprecedingPropositions1–4,othersolvablehyperbolic-cosine-type difference equations can be found. However, the corresponding relations become more and more complicated, so the method is not so effective. Note that equations (2), (12), (15), (18), and (21)canbewrittenintheform xn+1 =Pk(xn), n∈N0, (23) where P2(t)=t2–2, P3(t)=t3–3t, P4(t)=t4–4t2+2, P5(t)=t5–5t3+5t, P6(t)=t6–6t4+9t2–2. Hence, it is of some interest to find a polynomial class (Pk)k∈Ncontaining them. Todothis,itshouldbesaidthataveryusefulfactrelatedtothesequenceofpolynomials Pk(t), k∈N, is that they satisfy a linear recursive relation of second order. Namely, let t:=a+1 a, Stevi´ cetal.Advances in Difference Equations (2020) 2020:564 Page 9 of 12 then Pk(t)=ak+1 ak. (24) Since ak+1 aka+1 a=ak+1 +1 ak+1 +ak–1 +1 ak–1 , we have Pk+1(t)–P1(t)Pk(t)+Pk–1(t)=0,thatis, Pk+1(t)–tPk(t)+Pk–1(t)=0 (25) for k≥2, which is the desired recursive relation. From this, since initial values are P1(t)=tand P2(t)=t2–2, (26) allthepolynomialsPk(t)canbecalculatedrecursively.Moreover,sinceitisahomogeneous linear difference equation, it can be solved in a closed form. Indeed,thecharacteristicpolynomialassociatedwithequation(25)is P2(λ)=λ2–tλ+1, and its roots are λ1,2 =t±√t2–4 2. (27) Hence, general solution to equation (25) has the following form: Pk(t)=c1t+√t2–4 2k+c2t–√t2–4 2k,k∈N. (28) From (26)and(28), we have c1t+√t2–4 2+c2t–√t2–4 2=t, c1t+√t2–4 22+c2t–√t2–4 22=t2–2. (29) The determinant of system (29)is = t+√t2–4 2t–√t2–4 2 (t+√t2–4 2)2(t–√t2–4 2)2=–√t2–4. Hence, after some calculations, we have c1=1  tt–√t2–4 2 t2–2 (t–√t2–4 2)2= 1 (30)