On the Case of Complex Roots of the Characteristic Operator Polynomial of a Linear $$n $$ th-Order Homogeneous Different
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NARY DIFFERENTIAL EQUATIONS
On the Case of Complex Roots of the Characteristic Operator Polynomial of a Linear nth-Order Homogeneous Differential Equation in a Banach Space V. I. Fomin1∗ 1
Tambov State Technical University, Tambov, 392000 Russia e-mail: ∗ [email protected]
Received July 28, 2018; revised January 23, 2020; accepted May 14, 2020
Abstract—We consider a linear homogeneous nth-order differential equation, n ∈ N, with constant bounded operator coefficients in a Banach space. Under some conditions on the (real and complex) roots of the corresponding characteristic equation, we obtain a formula expressing the general solution of the differential equation via the operator functions given by the exponential, sine, and cosine of the roots. The case in which n is even and the characteristic equation has n/2 pairs of complex-conjugate pure imaginary roots is investigated in detail. The case of a second-order differential equation is considered separately. DOI: 10.1134/S0012266120080054
In a real Banach space E, consider the differential equation y (n) + H1 y (n−1) + . . . + Hn−1 y 0 + Hn y = f (t),
0 ≤ t < ∞,
(1)
of order n ∈ N, where Hi ∈ L(E), 1 ≤ i ≤ n; f (t) ∈ C([0, ∞); E); L(E) is the complete normed algebra of bounded linear operators from E to E; and C([0, ∞); E) is the normed space of continuous functions defined on [0, ∞) and ranging in E. It is well known [1] that the general solution of Eq. (1) has the form y = y0,0 + y∗ , where y0,0 is the general solution of the corresponding homogeneous equation y (n) + H1 y (n−1) + . . . + Hn−1 y 0 + Hn y = 0,
0 ≤ t < ∞,
(2)
and y∗ is a particular solution of the inhomogeneous equation (1). The problem of finding the particular solution y∗ has been solved: in the case where the right-hand side f (t) of Eq. (1) is of a general form, y∗ has been determined by variation of arbitrary constants in the paper [1]; in the case where f (t) has a special form, y∗ has been obtained by the method of undetermined coefficients in the paper [2]. The general solution y0,0 was found in [1] for the case in which the characteristic operator polynomial of Eq. (2) has n distinct roots belonging to the algebra L(E) and in the paper [3], for the case in which this polynomial has p roots in L(E) with respective multiplicities r1 , r2 , . . . , rp (r1 + r2 + . . . + rp = n). In this case, to construct y0,0 , the operator exponential with the operator A ∈ L(E) was used, At
e
=
∞ X tk Ak k=0
k!
,
for which, as is well known [4, p. 41], eAt
0
= AeAt .
(3)
Note that eAt |t=0 = I and (eAt )(m) = Am eAt for each m ∈ N. In the present paper, we study the case in which there are complex roots among the roots of the characteristic operator polynomial. Let us set forth necessary notions and facts that we will need in the sequel. 1021
1022
FOMIN
Consider the Cartesian square E 2 = E × E = {w = (x, y) : x, y ∈ E} of the Banach space E with linear operations (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ) (θ = (0, 0) is the zero element; (−x, −y) is the opposite element for (x
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