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Matrix Differential Operator Method of Finding a Particular Solution to A MATRIX DIFFERENTIAL OPERATOR METHOD OF FINDING A PARTICULAR SOLUTION TO A NONHOMOGENEOUS LINEAR ORDINARY DIFFERENTIAL EQUATION WITH CONSTANT COEFICIENTS JOZEF FECENKO Abstract. The article presents a matrix differential operator and a pseu- doinverse matrix differential operator for finding a particular solution to nonhomogeneous linear ordinary differential equations (ODE) with con- stant coeficients with special types of the right-hand side. Calculation requires the determination of an inverse or pseudoinverse matrix. If the matrix is singular, the Moore-Penrose pseudoinverse matrix is used for the calculation, which is simply calculated as the inverse submatrix of the considered matrix. It is shown that block matrices are effectively used to calculate a particular solution. 1. Introduction The idea of representing the processes of calculus, differentiation and in- tegration, as operators has a long history that goes back to the prominent German polymath Leibniz, G.W. The French mathematician L. F. A. Arbo- gast was the first mathematician to separate the symbols of operation from those of quantity, introducing systematically the operator notation DF for the derivative of the function. This approach was further developed by F. J. Servois who developed convenient notations. He was followed by a school of British and Irish mathematicians. R. B. Carmichael and G. Boole describing the application of operator methods to ordinary and partial differential equa- tions. Other prominent personalities who contributed to the development of operational calculus were, for example, physicist O. Heaviside, mathematicians arXiv:2101.02037v1 [math.GM] 31 Dec 2020 T.J. Bromwich, J.R. Carson, J. Mikusi´nski, N. Wiener, and others [13]. 2010 Mathematics Subject Classification. 34A30, 15A09. Key words and phrases. operator method of solving differential equations, solving differ- ential equations using a matrix differential operator, More-Penrose pseudoinverse matrix, block matrices. This paper was supported by the Slovak Grant Agency VEGA No. 1/0647/19. I owe thanks to J´an Buˇsa and Graham Luffrum for their valuable comments and advice that helped improve the text in many places. 1 2 JOZEF FECENKO Although the problem of solving ordinary nonhomogeneous linear differen- tial equations with constant coefficients is generally known, we will deal with the method of finding a particular solution of differential equation using a matrix differential operator method. This method, like the method of unde- termined coefficients, can be used only in special cases, if the right-hand side of the differential equation is typical, i.e. it is a constant, a polynomial function, exponential function eαx, sine or cosine functions sin βx or cos βx, or finite sums and products of these functions (α, β constants). In some cases, the method can be used as a support method for determining the particular solu- tion of a differential equation using the method of undetermined coefficients or the differential operator method or for evaluation of indefinite integrals. We introduce the differential operator notation. It is sometimes convenient to adopt the notation Dy, D2y, D3y, · · · , Dny dy d2y d3y dny to denote , , , · · · , . The symbols Dy, D2y,... are called differ- dx dx2 dx3 dxn ential operators [1] and have properties analogous to those of algebraic quan- tities [3]. Table 1: Operator Techniques [1], [10] A. n n φ(D) ckRk(x) ckφ(D)Rk(x) k=0 k=0 B. P P φ(D) · ψ(D) ψ(D) · φ(D) φ(D), ψ(D) polynomial operators in D C. φ(D)(xf(x)) xφ(D)f(x)+ φ′(D)f(x) D. Dneαx αneαx, n is non-negative integer E. Dn [eaxR(x)] eax (D + α)n R(x) n is non-negative integer F. n D2n sin βx −β2 sin βx n D2n cos βx −β2 cos βx Using the operator notation, we shall agree to write the differential equation (n) (n−1) ′ any + an−1y + . + a1y + a0y = f(x), an 6= 0, ai ∈ R (1.1) as n n−1 anD + an−1D + · · · + a1D + a0 y = f(x) (1.2) MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 3 or briefly φ(D)y = f(x), (1.3) where n n−1 φ(D)= anD + an−1D + · · · + a1D + a0 (1.4) is called an operator polynomial in D. If we want to emphasize the degree of the polynomial operator (1.4), we shall write it in the form φn(D). The use of the operator calculus for solving linear differential equations is well dealt with in several publications, e.g. [3], [1], [10]. 2. Matrix differential operator Definition 1. If S = {v1, v2,..., vn} is a set of vectors in a vector space V, then the set of all linear combination of v1, v2,..., vn is called the span of v1, v2,..., vn and is denoted by span(v1, v2,..., vn) or span(S). Let G be a vector space of all differentiable function. Consider the subspace V ⊂ G given by V = span (f1(x),f2(x), · · · ,fn(x)) , (2.1) where we assume that the functions f1(x),f2(x), · · · ,fn(x) are linearly inde- pendent. Since the set B = {f1(x),f2(x), · · · ,fn(x)} is linearly independent, it is a basis for V. The functions fi(x), i = 1, 2,...,n expressed in basis B using base vector coordinates are usually written 1 0 0 0 1 0 [f1(x)]B = . , [f2(x)]B = . , · · · , [fn(x)]B = . 0 0 1 The vector [fi(x)]B has in the i-th row 1 and 0 otherwise. Further, assume that the differential operator D maps V into itself Let n D(fi(x)) = cijfj(x), i = 1, 2, · · · , n, j X=1 where cij ∈ R, i, j = 1, 2,...,n are constants. Then ci1 c i2 [D (fi(x))]B = . , i, j = 1, 2,...,n . cin 4 JOZEF FECENKO and (see [9]) c11 c21 · · · cn1 c c · · · c . 12 22 n2 [D]B = [[D (f1(x))]B . [D (f2(x))]B . · · · . [D (fn(x))]B]= . .. c1n c2n · · · cnn (2.2) If n f(x)= αifi(x), αi ∈ R, i = 1, 2, · · · ,n, f(x) ∈ V i X=1 then α1 α 2 [f(x)]B = . αn We express what is the derivative of the function f(x) n n n n n Df(x)= αiDfi(x)= αi cijfj(x)= αicijfj(x) i i j j i X=1 X=1 X=1 X=1 X=1 respectively n ci1αi i=1 n c11 c21 · · · cn1 α1 P ci2αi c12 c22 · · · cn2 α2 [D(f(x))] = i=1 = . . B . . .. P . . . n c1n c2n · · · cnn αn cinαi i=1 P Let us next simply denote [D]B as DB. The matrix DB we will called a matrix differential operator corresponding to a vector space V with the considered basis B. Denote β1 β ′ 2 [D(f(x)]B = f B = . and [f(x)]B = f B . βn then ′ f B = DBf B MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 5 Note that the matrix transformation DB : V → V defined by ′ DBf B = f B is a linear transformation. As mentioned in (2.2), i-th, (i = 1, 2, · · · ,n) column of the matrix D ex- presses the derivative of the function fi(x). The following properties of the matrix differential operator are given with- out proof. Theorem 1. Let DB be a matrix differential operator of vector space V with a basis B. Then • DB (f B + gB)= DBf B + DBgB R • DB (cf B)= cDB (f B) , c ∈ ′′ 2 • f B = DB (DBf B)= DBf B (n) (n−1) (0) • f B = DBf B , f B = f B (n) n • f B = DBf B n • i-th (i = 1, 2, · · · ,n) column of the matrix DB expresses n-th derivative of function fi(x). Example 1. Let us consider V = span(xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x) (2.3) The Wronskian W (xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x) = 324 · e8x 6= 0. It follows that the functions in (2.3) are linearly independent. Then the set B = {xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x} is a basis for V. Applying differential operator D to a general element of V, we see that D axe2x sin 3x + bxe2x cos 3x + ce2x sin 3x + de2x cos 3x 2x 2x 2x = (2a − 3b)xe sin 3x + (3a + 2b)xe cos 3x + (a + 2c − 3d)e sin 3x + (b + 3c + 2d)e2x cos 3x, which is again in V. We found the corresponding matrix differential operator DB in the vector space V with the basis B. Here the construction of DB is shown schematically 6 JOZEF FECENKO x x x x cos3 sin 3 cos3 sin 3 x x 2 2 x x 2 2 xe e e xe D D D D xe2x sin 3x 2 −3 0 0 xe2x cos 3x 3 2 0 0 D = (2.4) B e2x sin 3x 1 0 2 −3 e2x cos 3x 0 1 3 2 Let us calculate the first and second derivative of the functi on f(x)= xe2x sin 3x + 2xe2x cos 3x − 2e2x cos 3x using the matrix differential operator (2.4). We have 2 −3 0 0 1 −4 3 2 0 0 2 7 [Df(x)] = D [f(x)] = = B B B 1 0 2 −3 0 7 0 1 3 2 −2 −2 So that Df(x)= −4xe2x sin 3x + 7xe2x cos 3x + 7e2x sin 3x − 2e2x cos 3x The second derivative 2 −3 0 0 −4 −29 ′′ ′ 3 2 0 0 7 2 f = D f = = B B B 1 0 2 −3 7 16 0 1 3 2 −2 24 D2f(x)= −29xe2x sin 3x + 2xe2x cos 3x + 16e2x sin 3x + 24e2x cos 3x For calculating higher powers of the matrix DB it is sometimes advanta- geous to express it as a block matrix (see example 11) or use Jordan matrix decomposition.
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