Analytic Solutions of a Second-Order Functional Differential Equation

The functional differential equation, where all iim0,0, provide a physical or biological system statistical model, in which the rate of shift in a system is not only dictated by its current condition, but also by its past (see [1, 2]). Many scholars have researched the nature and uniqueness of a number of these equations in recent years. In 1984, in the Banch fixed point theorem, Eder [3] classified solutions for operational differential equations. In 2001[7], Si and Wang studied the nature of a second-order functional differential equation analytical solution: In 2009, Liu and Li [8] studied the equation Observe that (3) can be reduced to (2) by setting c = 0. Next, the equation has been studied by Si and Wang [9]. In order to obtain analytic solutions of (4), they constructed a corresponding auxiliary equation with parameter . The presence of auxiliary equation solutions depends on the parameter state , such as  is in the circle of units and  The source of peace fulfilling the state of the Diophantine. In this paper we examine the nature of the secondary order differential equation analytical solutions with a state derivative formal delay. If (z) = az, then (5) reduces to (4) Please note that three cases of parameters  – bis - would be studied in the relevant auxiliary equation in this article. One is that TER  is the origin of unity that satisfies the requirement of Brjuno. To construct an Equation supplementary, we set Then Where Z0 is a continuous complex, In particular, we have Applying relations (6) and (8) to (5), we obtain We construct the corresponding equation by differentiating both sides of (9) with respect to z. These yields.

In this section, we let (z) be a polynomial. Then, The polynomial solution of our research (5) Have the secondary equation With the relation 1p(z)bx(z)g(g(z)), we obtain an analytic solution x(z). Furthermore, we characterize a polynomial solution when p(z) is a polynomial. Theorem 1 : For a polynomial (z), the equation

has a nontrivial polynomial solution if and only if p(z) = p0

with 0011p0orp(z)ppzwithp0

Necessity. Assume that nkk0k(z)xz is a nontrivial polynomial solution of (12). Le nkk0k(z)pz with mp0. Observe that x(z)0when n = 0. This implies (z)0. From now on, we let n1.

Case 1(m = 0). That is, 0(z)p0. Equation (12) becomes Where n1012nq(z)(pbx)2bxz....nbxz If n = 1, then (13) changes to 0 = x0 + xi (p0 + bx1). Next, we consider n > 2. Equation (13) is reduced to Where n2012n1q(z)(pbx)2bxz....(n1)bxz. Comparing coefficient of (n1)(n2)z in (14), we have n1x0. Then, repeating the above method, we obtain n22x.....x0 and also have 01010xxx(pbx). By choosing an arbitrary nonzero x1, say , both situations yield a nontrivial solution 0x(z)(pb)z. Case 2 (m = 1). That is, (z) = p0 + piz, where 1p0. Equation (12) becomes Where 2n101121n(z)(pbx)(p2bx)z3bxz....nbxz By comparing coefficient of constant term and z in (15), we obtain (z)0forn1. Next, we consider n > 2. Comparing coefficient of n(n1)z in (15), we have nnnx(nbx)0, which implies xn = 0. Therefore, (15) is reduced to Where 201123q(z)(pbx)(p2bx)z3bxz...(n1) n2n1bxz Comparing coefficient of (n1)(n2)z in (16), we have n1x0. Continuing this process, we obtain 220122xxx(q(z))x(q(z)), where 0112q(z)(pbx)(p2bx)z. By comparing coefficient of z2 together with 2x0, we obtain a nontrivial solution. 21010(z)(p/b)(pb)(p/2b)(pb)z 21(p/2b)z, where  is an arbitrary constant. Case 3 (m > 2). We consider 2 subcases. Subcase 3.1 (m < n – 1). Equation (12) becomes Where 01mm1q(z)p)bx)...(p(m1)bx) mm1n1m2nz(m2)bxz....nbxz. By using method of undetermined coefficient, we obtain nm,2x...x0. Consequently, (46) is reduced to Where m0im1q(z(pbx)...(pm(m1)bx)z Comparing coefficient of z(m)(m+1) in (18), We have xm+1 (pm + (m+1) bxm+1)m+1 = 0. If xm+1 = 0, then xm = …. = x0 = 0. That is (z)0 which is a contradiction, Therefore, Pm + (m + 1) xm+1 = 0. Substituting this relation in (17) and repeating this process, we get xm-k = –pm–(k+1) / (m – k) b for k = –1 …..m–2. Using this fact in (18) and then comparing the coefficient of zt (I = 1,…., m –1), we obtain x2 = ….= xm+1 = 0. This yield p = 0, which is a contradiction. Subcase 3.2 (m > n – 1), Equation (12) becomes Where n1nm01n1nq(z)(pbx)....(pnbx)zpnz....pmz Comparing coefficient of in (19) together with we have . Then, repeating the above method. We obtain which is a contradiction.

Thus, (12) has a nontrivial polynomial solution

with 001ip0orp(z)ppzwithp0.

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Research Scholar, Department of Mathematics, Magadh University, Bodhgaya