Matrix Differential Operator Method of Finding a Particular Solution to a Nonhomogeneous Linear Ordinary Differential Equation with Constant Coefficients
aa r X i v : . [ m a t h . G M ] D ec MATRIX DIFFERENTIAL OPERATOR METHODOF FINDING A PARTICULAR SOLUTIONTO A NONHOMOGENEOUS LINEAR ORDINARYDIFFERENTIAL EQUATIONWITH CONSTANT COEFICIENTS
JOZEF FECENKO
Abstract.
The article presents a matrix differential operator and a pseu-doinverse matrix differential operator for finding a particular solution tononhomogeneous linear ordinary differential equations (ODE) with con-stant coeficients with special types of the right-hand side. Calculationrequires the determination of an inverse or pseudoinverse matrix. If thematrix is singular, the Moore-Penrose pseudoinverse matrix is used forthe calculation, which is simply calculated as the inverse submatrix of theconsidered matrix. It is shown that block matrices are effectively used tocalculate a particular solution. Introduction
The idea of representing the processes of calculus, differentiation and in-tegration, as operators has a long history that goes back to the prominentGerman polymath Leibniz, G.W. The French mathematician L. F. A. Arbo-gast was the first mathematician to separate the symbols of operation fromthose of quantity, introducing systematically the operator notation DF forthe derivative of the function. This approach was further developed by F. J.Servois who developed convenient notations. He was followed by a school ofBritish and Irish mathematicians. R. B. Carmichael and G. Boole describingthe application of operator methods to ordinary and partial differential equa-tions. Other prominent personalities who contributed to the development ofoperational calculus were, for example, physicist O. Heaviside, mathematiciansT.J. Bromwich, J.R. Carson, J. Mikusi´nski, N. Wiener, and others [13].
Mathematics Subject Classification.
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 owethanks to J´an Buˇsa and Graham Luffrum for their valuable comments and advice thathelped improve the text in many places.
Although the problem of solving ordinary nonhomogeneous linear differen-tial equations with constant coefficients is generally known, we will deal withthe method of finding a particular solution of differential equation using amatrix differential operator method. This method, like the method of unde-termined coefficients, can be used only in special cases, if the right-hand side ofthe differential equation is typical, i.e. it is a constant, a polynomial function,exponential function e αx , sine or cosine functions sin βx or cos βx , or finitesums and products of these functions ( α , β constants). In some cases, themethod can be used as a support method for determining the particular solu-tion of a differential equation using the method of undetermined coefficientsor 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 , D y , D y, · · · , D n y to denote dydx , d ydx , d ydx , · · · , d n ydx n . The symbols Dy , D y, . . . are called differ-ential operators [1] and have properties analogous to those of algebraic quan-tities [3].Table 1: Operator Techniques [1], [10] A. φ ( D ) n P k =0 c k R k ( x ) n P k =0 c k φ ( D ) R k ( x ) B. 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. D n e αx α n e αx , n is non-negative integer E. D n [ e ax R ( x )] e ax ( D + α ) n R ( x ) n is non-negative integer F. D n sin βx (cid:0) − β (cid:1) n sin βxD n cos βx (cid:0) − β (cid:1) n cos βx Using the operator notation, we shall agree to write the differential equation a n y ( n ) + a n − y ( n − + . . . + a y ′ + a y = f ( x ) , a n = 0 , a i ∈ R (1.1)as (cid:0) a n D n + a n − D n − + · · · + a D + a (cid:1) y = f ( x ) (1.2) ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 3 or briefly φ ( D ) y = f ( x ) , (1.3)where φ ( D ) = a n D n + a n − D n − + · · · + a D + a (1.4)is called an operator polynomial in D. If we want to emphasize the degree ofthe polynomial operator (1.4), we shall write it in the form φ n ( D ).The use of the operator calculus for solving linear differential equations iswell dealt with in several publications, e.g. [3], [1], [10].2. Matrix differential operator
Definition 1. If S = { v , v , . . . , v n } is a set of vectors in a vector space V, then the set of all linear combination of v , v , . . . , v n is called the span of v , v , . . . , v n and is denoted by span( v , v , . . . , v n ) or span(S).Let G be a vector space of all differentiable function. Consider the subspace V ⊂ G given by V = span ( f ( x ) , f ( x ) , · · · , f n ( x )) , (2.1)where we assume that the functions f ( x ) , f ( x ) , · · · , f n ( x ) are linearly inde-pendent. Since the set B = { f ( x ) , f ( x ) , · · · , f n ( x ) } is linearly independent,it is a basis for V.The functions f i ( x ) , i = 1 , , . . . , n expressed in basis B using base vectorcoordinates are usually written[ f ( x )] B = , [ f ( x )] B = , · · · , [ f n ( x )] B = The vector [ f i ( x )] B has in the i- th row 1 and 0 otherwise.Further, assume that the differential operator D maps V into itselfLet D ( f i ( x )) = n X j =1 c ij f j ( x ) , i = 1 , , · · · , n, where c ij ∈ R , i, j = 1 , , . . . , n are constants. Then[ D ( f i ( x ))] B = c i c i ... c in , i, j = 1 , , . . . , n JOZEF FECENKO and (see [9])[ D ] B = [[ D ( f ( x ))] B .... [ D ( f ( x ))] B .... · · · .... [ D ( f n ( x ))] B ] = c c · · · c n c c · · · c n ... ... . . . ... c n c n · · · c nn (2.2)If f ( x ) = n X i =1 α i f i ( x ) , α i ∈ R , i = 1 , , · · · , n, f ( x ) ∈ V then [ f ( x )] B = α α ... α n We express what is the derivative of the function f ( x ) Df ( x ) = n X i =1 α i Df i ( x ) = n X i =1 α i n X j =1 c ij f j ( x ) = n X j =1 n X i =1 α i c ij f j ( x )respectively[ D ( f ( x ))] B = n P i =1 c i α in P i =1 c i α i ... n P i =1 c in α i = c c · · · c n c c · · · c n ... ... . . . ... c n c n · · · c nn α α ... α n Let us next simply denote [ D ] B as D B .The matrix D B we will called a matrix differential operator correspondingto a vector space V with the considered basis B .Denote [ D ( f ( x )] B = f ′ B = β β ... β n and [ f ( x )] B = f B then f ′ B = D B f B ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 5
Note that the matrix transformation D B : V → V defined by D B f B = f ′ B is a linear transformation.As mentioned in (2.2), i- th, ( i = 1 , , · · · , n ) column of the matrix D ex-presses the derivative of the function f i ( x ) . The following properties of the matrix differential operator are given with-out proof.
Theorem 1.
Let D B be a matrix differential operator of vector space V witha basis B . Then • D B ( f B + g B ) = D B f B + D B g B • D B ( c f B ) = c D B ( f B ) , c ∈ R • f ′′ B = D B ( D B f B ) = D B f B • f ( n ) B = D B f ( n − B , f (0) B = f B • f ( n ) B = D nB f B • i- th ( i = 1 , , · · · , n ) column of the matrix D nB expresses n-th derivativeof function f i ( x ) . Example . Let us consider V = span( xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x ) (2.3)The Wronskian W ( xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x ) = 324 · e x =0 . It follows that the functions in (2.3) are linearly independent. Then the set B = { xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x } is a basis for V.Applying differential operator D to a general element of V , we see that D (cid:0) axe x sin 3 x + bxe x cos 3 x + ce x sin 3 x + de x cos 3 x (cid:1) = (2 a − b ) xe x sin 3 x + (3 a + 2 b ) xe x cos 3 x + ( a + 2 c − d ) e x sin 3 x + ( b + 3 c + 2 d ) e x cos 3 x, which is again in V. We found the corresponding matrix differential operator D B in the vectorspace V with the basis B . Here the construction of D B is shown schematically JOZEF FECENKO D (cid:0) x e x s i n x (cid:1) D (cid:0) x e x c o s x (cid:1) D (cid:0) e x s i n x (cid:1) D (cid:0) e x c o s x (cid:1) D B = xe x sin 3 xxe x cos 3 xe x sin 3 xe x cos 3 x − −
30 1 3 2 (2.4)Let us calculate the first and second derivative of the function f ( x ) = xe x sin 3 x + 2 xe x cos 3 x − e x cos 3 x using the matrix differential operator (2.4). We have[ D f ( x )] B = D B [ f ( x )] B = − −
30 1 3 2 − = − − So that Df ( x ) = − xe x sin 3 x + 7 xe x cos 3 x + 7 e x sin 3 x − e x cos 3 x The second derivative f ′′ B = D B f ′ B = − −
30 1 3 2 − − = − D f ( x ) = − xe x sin 3 x + 2 xe x cos 3 x + 16 e x sin 3 x + 24 e x cos 3 x For calculating higher powers of the matrix D B it is sometimes advanta-geous to express it as a block matrix (see example 11) or use Jordan matrixdecomposition.The point of a matrix differential operator is not that this method is easierthan a direct differentiation. Indeed, once the matrix D B has been established,it is easy to find the differentials with little to do. What is significant is thatmatrix methods can be used at all in what appears, on the surface, to bea calculus problem. ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 7 Some properties of block matrices and pseudoinverse matrix
Theorem 2. [5] Let A be a matrix partitioned into four × blocks A = (cid:20) P QR S (cid:21) where R is a regular matrix of the order r . Then the rank of matrix A is equalto r if and only if Q = P R − S . Proof.
To the first block row of the matrix A we add the second block rowmultiplied from the left by the matrix − P R − . This adjustment adds thelinear combination of matrix rows ( R S ) to the other matrix rows. Afterthis adjustment, we get the matrix (cid:20) Q − P R − SR S (cid:21) . (3.1)For matrix A and matrix (3.1) to have the same rank, the matrix Q − P R − S must be equals . It follows that Q = P R − S . (cid:3) Definition 2.
Let A be a m × n matrix. The matrix X for which AXA = A , is called the pseudoinverse matrix of matrix A. The pseudoinverse matrix ofa matrix A is denoted A − . Theorem 3.
Let A be a m × n matrix partitioned into four × blocks A = (cid:20) P QR S (cid:21) and let the rank of the matrix A be r . If the matrix R is regular of order r ,then the n × m matrix A − = (cid:20) R − (cid:21) is the pseudoinverse of the matrix A . Proof.
Compute AA − A = (cid:20) P QR S (cid:21) (cid:20) R − (cid:21) (cid:20) P QR S (cid:21) = (cid:20) P P R − SR S (cid:21) = (cid:20) P QR S (cid:21) = A We have used the consequence of the Theorem 2, i.e. Q = P R − S . (cid:3) Definition 3.
The pseudoinverse matrix A − of the matrix A satisfying all ofthe following four conditions(i) AA − A = A , ( A − is the pseudoinverse matrix of the matrix A ); JOZEF FECENKO (ii) A − AA − = A − , ( A is the pseudoinverse matrix of the matrix A − );(iii) (cid:0) AA − (cid:1) T = AA − , ( AA − is a symmetric matrix, ( · ) T means a trans-posed matrix);(iv) (cid:0) A − A (cid:1) T = A − A , ( A − A is a symmetric matrix)is called the Moore-Penrose pseudoinverse of the matrix A , denoted by A + . Theorem 4.
Let A be a m × n matrix partitioned into four × blocks A = (cid:20) R (cid:21) and let the rank of matrix A be r. If matrix R is regular of order r , then A − = (cid:20) R − (cid:21) n × m is the Moore-Penrose pseudoinverse matrix of the matrix A , thus A − = A + . Proof.
We will verify all four conditions of definition 2.(i) AA − A = (cid:20) R (cid:21) (cid:20) R − (cid:21) (cid:20) R (cid:21) = (cid:20) R (cid:21) = A The condition i is met.(ii) Pseudoinverse matrix of the matrix A − = (cid:20) R − (cid:21) is the matrix A = (cid:20) R (cid:21) , because A − AA − = (cid:20) R − (cid:21) (cid:20) R (cid:21) (cid:20) R − (cid:21) = (cid:20) R − (cid:21) = A − We have also proved the validity of condition ii .(iii) (cid:0) AA − (cid:1) T = (cid:20)(cid:20) R (cid:21) (cid:20) R − (cid:21)(cid:21) T = (cid:20) I (cid:21) T = (cid:20) I (cid:21) We have shown that condition iii is met.(iv) (cid:0) A − A (cid:1) T = (cid:18)(cid:20) R − (cid:21) (cid:20) R (cid:21)(cid:19) T = (cid:20) I
00 0 (cid:21) T = (cid:20) I
00 0 (cid:21)
We have shown that condition iv is also met. (cid:3) Corollary 1.
For the matrices A and A − in Theorem 4 imply that AA − and A − A are incomplete identity matrices. ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 9
Theorem 5.
Let Ax = b (3.2) be a solvable system of linear equations and A − be a pseudoinverse of thematrix A , then x = A − b (3.3) is a solution of (3.2) .Proof. Assuming that the system of linear equations (3.2) is solvable, then avector y exists, such that Ay = b We show that the vector in (3.3) is the solution of the system of linear equations(3.2). It is valid that Ax = A (cid:0) A − b (cid:1) = (cid:0) AA − A (cid:1) y = Ay = b (cid:3) Remark . If the system of linear equations (3.2) is solvable and its solution isexpressed in the form x = A − b and A − is Moore-Penrose pseudoinverse of thematrix A , then this solution minimizes the Euclidean norm k Ax − b k and ofall n -dimensional solutions x that minimize this norm it has the lowest norm.Such a vector solution of the system of linear equations is called the solutionof the system of linear equations with the least squares solution of minimumnorm or the best approximate solution of the system .Note that in our considerations, when solving a solvable system of linearequations (3.2) we will consider a solution (3.3) expressed in the form x = A + b The solution of (3.2) exists if and only if AA + b = b , [4].4. Pseudoinverse matrix differential operator
We shall introduce some properties of the differential and inverse differentialoperators.The definition of the inverse differential operator [10]: Let 1 φ ( D ) f ( x ) bedefined as a particular solution y , such that (1.3) then 1 φ ( D ) f ( x ) is called theinverse differential operator . (cid:3) We will prove relation I in the next Table 2.
Proof.
With respect to C in the Table 1 is φ ( D )( xf ( x )) = xφ ( D ) f ( x ) + φ ′ ( D ) f ( x ) (4.1)Let φ ( D ) f ( x ) = R ( x )Then 1 φ ( D ) φ ( D ) f ( x ) = 1 φ ( D ) R ( x ) f ( x ) = 1 φ ( D ) R ( x ) (4.2)Substituting (4.2) into relationship (4.1), we get φ ( D ) (cid:18) x φ ( D ) R ( x ) (cid:19) = xφ ( D ) 1 φ ( D ) R ( x ) + φ ′ ( D ) 1 φ ( D ) R ( x ) φ ( D ) (cid:18) x φ ( D ) R ( x ) (cid:19) = xR ( x ) + φ ′ ( D ) 1 φ ( D ) R ( x )Let’s multiply the previous equation from the left side with 1 φ ( D ) , then x φ ( D ) R ( x ) = 1 φ ( D ) xR ( x ) + 1 φ ( D ) φ ′ ( D ) 1 φ ( D ) R ( x )From which 1 φ ( D ) xR ( x ) = x φ ( D ) R ( x ) − φ ( D ) φ ′ ( D ) 1 φ ( D ) R ( x ) (cid:3) ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 11
Table 2: Inverse Operator Techniques [1], [10] A. φ ( D ) n P k =0 c k R k ( x ) n P k =0 c k φ ( D ) R k ( x ) B. D − m R ( x ) e mx R e − mx R ( x ) dx C. e m x R e − m x e m x R e − m x . . . D − m )( D − m ) ... ( D − m n ) e m n x R e − m n x R ( x ) dx n This can also be evaluated by expandingthe inverse operator into partial fractionsand then using B. D. e px φ ( p ) if φ ( p ) = 0 φ ( D ) e px x k e px φ ( k ) ( p ) if φ ( p ) = φ ′ ( p ) = . . . φ ( k − ( p ) = 0 but φ ( k ) ( p ) = 0 E. φ ( D ) cos px cos pxφ ( − p ) if φ ( − p ) = 0 φ ( D ) sin px sin pxφ ( − p ) F. φ ( D ) cos px Re n e ipx φ ( ip ) o φ ( D ) sin px Im n e ipx φ ( ip ) o If φ ( ip ) = 0, otherwise use D. G. φ ( D ) x p (cid:18) p P k =0 c k D k (cid:19) x p by expanding φ ( D ) in powers of D since D p + n x p = 0 for n > . H. φ ( D ) e px R ( x ) e px φ ( D + p ) R ( x ) called the ”operator shift theorem” . I. φ ( D ) xR ( x ) x φ ( D ) R ( x ) − φ ( D ) φ ′ ( D ) φ ( D ) R ( x )= (cid:16) x − φ ( D ) φ ′ ( D ) (cid:17) φ ( D ) R ( x ) J. using substitute φ ( D ) ( e αx P m ( x ) cos βx + cos βx = e iβx + e − iβx Q n ( x ) sin βx ) sin βx = e iβx − e − iβx i and next to use an operator shift theorem Proofs of many of the statements in Table 2 can be found for example in [1]or [10].Note that all analytical relations derived for the differential operator havean adequate expression for the matrix differential operator. (cid:3)
In order to define a pseudoinverse matrix differential operator, we first con-sider a simple form of the differential equation (1.1) i. e. y ′ = f ( x ) (4.3)Let us assume that the function f ( x ) is differentiable and every derivativeof the function f ( x ) can be expressed as a finite linear combination of linearindependent functions f ( x ) , f ( x ) , . . . , f n ( x ) . Let’s consider the vector space V = span ( f ( x ) , f ( x ) , . . . , f n ( x ))with the basis B = { f ( x ) , f ( x ) , . . . , f n ( x ) } .( A ) Let us assume that the particular solution of the differential equation(4.3) belongs to the vector space V with the basis B . Let D B be amatrix differential operator corresponding to the basis B . Then theequation D B y B = f B , where f B = [ f ( x )] B , is solvable. And due to the Theorem 5 y B = D + B f B (4.4)is the solution of the differential equation (4.3) expressed in the basis B where D + B is the Moore-Penrose pseudoinverse of the matrix D B . Thus D + B is the pseudoinverse matrix differential operator to the oper-ator D B . (The unique solution (4.4) to the differential equation (4.3)do not contain a kernel of matrix differential operator D + B . This alsoapplies in the following cases.)( B ) Let us assume that the particular solution of the differential equation(4.3) does not belong to the vector space V and let D B be a matrixdifferential operator with the considered basis B. Then the equation D B y B = f B is not solvable. Let’s create a new system of functions B = { f ( x ) , f ( x ) , . . . , f n ( x ) } ∪ { xf ( x ) , xf ( x ) , . . . , xf n ( x ) } = { f ( x ) , f ( x ) , . . . , f m ( x ) } where m > n . Let functions f ( x ) , f ( x ) , . . . , f m ( x ) be a linearindependent and let V = span( f ( x ) , f ( x ) , . . . , f m ( x )) ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 13 with the basis B . Let a particular solution of the differential equation(4.3) belong to the vector space V and let D B be a matrix differentialoperator with the considered basis B . Then the equation D B y B = f B (4.5)is solvable, where f B = [ f ( x )] B . Due to Theorem 5 y B = D + B f B is the solution of the differential equation (4.3) expressed in the ba-sis B where D + B is the Moore-Penrose pseudoinverse of the matrix D B . Thus D + B is the pseudoinverse matrix differential operator to theoperator D B .( C ) If a solution of the differential equation (4.3) does not belong to thevector space V , we will create a new system of vectors in a mannersimilar to B and analyse the solvability of the differential equation.It is difficult to prove in general that the solution lies in some vectorspace, the construction of which we have described. It will be explainedin the following examples.For completeness, we still have to consider a differential equation in theform (1.1). Also in this case, we assume that the function f ( x ) is differentiableand every derivative of the function f ( x ) can be expressed as a finite linearcombination of linear independent functions f ( x ) , f ( x ) , . . . , f n ( x ) . We areconsidering a vector space V = span( f ( x ) , f ( x ) , . . . , f n ( x ))with the basis B = { f ( x ) , f ( x ) , . . . , f n ( x ) } and and we discuss the existenceof a solution to the equation (cid:0) a n D nB + a n − D n − B + · · · + a D B + a I (cid:1) y B = f B (4.6)or briefly φ ( D B ) y B = f B , where I is the identity matrix. The discussion is analogous to previous cases. (cid:3) Now let us present some elementary examples. We will focus this issue inmore detail below. Let’s start with a very elementary example.
Example . Determine using a matrix differential operator the particular so-lution of the equation y ′ = x. (4.7) Solution. f ( x ) = x and each derivative of f ( x ) can be expressed as a linearcombination of { x, } . Let V = span( x, D B = (cid:20) (cid:21) . The equation D B y B = [ x ] B (cid:20) (cid:21) (cid:20) y y (cid:21) = (cid:20) (cid:21) has no solution. It follows that a particular solution of (4.7) does not belongto V . Let’s create a new system of function B = { x , x } ∪ { x, } = { x , x, } which is linear independent. Then V = span( x , x, , D B = . The matrix equation D B y B = [ x ] B y y y = has a solution and (Theorem 5) y B = y y y = D + B According to Theorem 4 D + B = Thus y B = = which, in turn, implies the particular solution y = 12 x Example . Determine the particular solutions of the equations(a) y ′′ + 3 y ′ − y = xe x ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 15 (b) y ′′ + 3 y ′ − y = 2 xe x − e x Solution.
Every derivative of the functions on the right-hand side of the equa-tions is a linear combination of the functions x · e x , e x which are linear in-dependent Let’s assume that a particular solution of the differential equationsbelongs to the vector space V = span( xe x , e x ) with the base B = { xe x , e x } .Differential operator D B = (cid:20) (cid:21) (a) (cid:0) D B + 3 D B − I (cid:1) y B = [ xe x ] B (cid:20) (cid:21) + 3 (cid:20) (cid:21) − (cid:20) (cid:21)! y B = (cid:20) (cid:21)(cid:20) − (cid:21) y B = (cid:20) (cid:21) y B = (cid:20) − (cid:21) − (cid:20) (cid:21) =
16 0524 − = We have found the particular solution of differential equation in (a) y = 16 xe x + 524 e x . (b) y B =
16 0524 − − = y = 13 xe x + 76 e x Example . Find the particular solution of the differential equation y IV + 2 y ′′ + y = 2 sin x − x (4.8) Solution.
It’s easy to prove that the particular solution y of differential equa-tion (4.8) y / ∈ V = span(sin x, cos x ) y / ∈ V = span( x sin x, x cos x, sin x, cos x ) Let’s assume that y ∈ V = span( x sin x, x cos x, x sin x, x cos x, sin x, cos x ).Since the set of functions B = { x sin x, x cos x, x sin x, x cos x, sin x, cos x } is linearly independent, it is a basis for V . The corresponding matrix differ-ential operator D B = − − −
10 0 0 1 1 0
Then (cid:0) D B + 2 · D B + I (cid:1) y B = [2 sin x − x ] B − − y B = − y B = − − + − y B = −
00 0 0 0 0 − − = − The Moore-Penrose pseudoinverse to the matrix (cid:0) D B + 2 · D B + I (cid:1) we foundusing Theorem 4. Then y = − x sin x + 12 x cos x ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 17 Matrix differential operator and the method ofundetermined coeficients
In order to describe the algorithm for the finding of a particular solutionof an ordinary nonhomogeneous linear differential equation with constant co-efficients (1.2) using matrix differential operator, we use knowledge of thealgorithm for finding a particular solution by an undetermined coefficientsmethod.Let us consider a differential equation (1.2) with a characteristic equation a n k n + a n − k n − + . . . + a k + a = 0 (5.1)First, we make two conventions to simplify expressing the root multiplicityof the characteristic equation and the value of the function f ( x ) = x at thediscontinuity point.1. If the number α is not the root of the characteristic equation (5.1), wewill also say that it is a 0-fold root of the characteristic equation;if the number α is a simple characteristic root of (5.1), we will also saythat it is 1-fold root of the characteristic equation, and so on.2. Given the removable discontinuity of the function f ( x ) = x at thepoint x = 0, we define the function at this point as f (0) = 1 . Let us consider two cases of the right-hand side of the differential equation(1.2). We will describe the algorithm for finding a particular solution using amatrix differential operator:a) with the right-hand side (1.2) f ( x ) = e αx P m ( x ) (5.2)where α ∈ R and P m ( x ) is a polynomial of degree m . The algorithmto determine the particular solution of that differential equation, bythe method of undetermined coefficients, says:If α is the k -fold root ( k = 0 , , , . . . , ) of the characteristic equation(5.1) of the differential equation (1.2) with the right-hand side (5.2),then the particular solution of this equation is of the form y = x k e αx Q m ( x ) , (5.3)where Q m ( x ) = A m x m + A m − x m − + . . . + A x + A is a polyno-mial of degree m with undetermined coefficients. The values of theseundetermined coefficients can be determined by substituting (5.3) for y into the equation (1.2) and then comparing the coefficients for thesame functions on the right-hand and left-hand sides of the equation.The algorithm for determining the particular solution of the differen-tial equation (1.2) with the right-hand side (5.3) by the method of undetermined coefficients implies that this particular solution will bein the vector space V = span( x k + m e αx , x k + m − e αx , . . . , xe αx , e αx )with the basis for VB = n x k + m e αx , x k + m − e αx , . . . , xe αx , e αx o (5.4)The relevant matrix differential operator D B in the vector space V with the basis B be a matrix of the type ( k + m + 1) × ( k + m + 1) . b) with the right-hand side f ( x ) = e αx ( P r ( x ) sin βx + Q s ( x ) cos βx ) , (5.5)where α, β are real numbers, P r ( x ) is a polynomial of degree r , Q s ( x )is a polynomial of degree s .If α + βi is a k -fold root ( k = 0 , , , . . . , ) of the characteristic equation(5.1) of the differential equation (1.2) with the right-hand side (5.5),then the particular solution of this equation has the form y = x k e αx ( U m ( x ) sin βx + V m ( x ) cos βx ) (5.6)where m = max { r, s } , U m ( x ) = A m x m + A m − x m − + . . . + A x + A , , V m ( x ) = B m x m + B m − x m − + . . . + B x + B are polynomials of degree m with undetermined coefficients. Thevalues of these undetermined coefficients can be determined by sub-stituting (5.6) for y into the equation (1.2) and then comparing thecoefficients for the same functions on the right-hand and left-hand sidesof the equation. The algorithm for determining the particular solutionof the differential equation (1.2) with the right-hand side (5.6) by themethod of undetermined coefficients implies that this particular solu-tion will be in the vector space V = span( x k + m e αx sin βx, x k + m e αx cos βx, x k + m − e αx sin βx,x k + m − e αx cos βx . . . , xe αx sin βx, xe αx cos βx, e αx sin βx,e αx cos βx )with the basis for VB = (cid:8) x k + m e αx sin βx, x k + m e αx cos βx, x k + m − e αx sin βx,x k + m − e αx cos βx . . . , xe αx sin βx, xe αx cos βx, e αx sin βx,e αx cos βx } (5.7)The relevant matrix differential operator D B in the vector space V with the basis B be a matrix of the type 2( k + m + 1) × k + m + 1) . Subsequently, in both cases a) and b), we create a matrix equation (cid:0) a n D nB + a n − D n − B + · · · + a D B + a I (cid:1) y B = f B (5.8) ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 19 where f B = [ f ( x )] B and y B = [ y ( x )] B , where y ( x ) is a particular solution ofthe differential equation (1.2).If the right-side of the differential equation (1.1) is the sum of several func-tions, then the principle of superposition can be used to solve it. The principleof superposition of solutions says that if y i ( i = 1 , , . . . m ) is a solution of thedifferential equation (cid:16) a n y ( n ) + a n − y ( n − + · · · + a y ′ + a (cid:17) y = f i ( x )( i = 1 , , . . . m ), then for any constants k , k , . . . , k m , the function y = k y + k y + · · · + k m y m is a solution to the differential equation (1.1) with f ( x ) = k f ( x ) + k f ( x ) + · · · + k m f m In special case, if f ( x ) , f ( x ) , . . . , f m ( x ) form the basis B of the vectorspace V = span( f ( x ) , f ( x ) , . . . , f m ( x )) then it is enough to solve only theequation (cid:0) a n D nB + a n − D n − B + · · · + a D B + a I (cid:1) y B = k k ... k m B In the next example, we want to show how the matrix differential operatorcan be used to support the finding of a particular solution of the differentialequation by the method of undetermined coefficients.
Example . Determine the particular solution of the differential equation( D − ( D + 4) y = 3 e x (5.9) Solution.
First we solve the equation( D + 4) y = 3 e x (5.10)Since α = 2 is not a solution of the characteristic equation ( k + 4) = 0, itfollows that the solution of (5.10) belongs to the vector space V = span( e x )with the basis B = { e x } . Then the matrix differential operator is D B = [2] and ( D B + 4 I ) y B = [3 e x ] B ([2] + 4[1]) y B = [3][6] y B = [3] y B = [6] − [3] y B = (cid:20) (cid:21) y = 112 e x Now we have to determine the particular solution of the differential equation( D − y = 112 e x (5.11)Because α = 2 is 2-fold root of the characteristic equation ( k − = 0, it fol-lows that the solution of (5.11) belongs to the vector space V = span( x e x , xe x , e x ) with the basis B = { x e x , xe x , e x } . Then thematrix differential operator is B = and ( D B − I ) y B = (cid:20) e x (cid:21) B y B = y B = + y B = = The particular solution of the differential equation (5.9) is y = 124 x e x ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 21
Example . Determine the particular solution of the differential equation y ′′ − y ′ + 13 y = 2 xe x cos 3 x (5.12)using a pseudoinverse matrix differential operator. Solution.
Because 2 + 3 i is 1-fold root of the characteristic equation k − k + 13 = 0, it follows that the particular solution of (5.12) will be(due to (5.5),(5.6)) in the form y = xe x (( Ax + B ) sin βx + ( Cx + D ) cos βx )In other words, the particular solution y will be in the vector space V = span( x e x sin 3 x, x e x cos 3 x, xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x )with the basis B = (cid:8) x e x sin 3 x, x e x cos 3 x, xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x (cid:9) The matrix differential operator D B = − − −
30 0 0 1 3 2 (cid:0) D B − D B + 13 I (cid:1) y B = [2 xe x cos 3 x ] B After editing the previous equation, we get −
12 0 0 0 012 0 0 0 0 02 0 0 − y B = y B = −
12 0 0 0 012 0 0 0 0 02 0 0 − + In accordance with Theorem 4 we determine the Moore-Penrose pseudoin-verse. Then y B = −
112 0 0 00 0 136 0 0 160 0 0 136 −
16 00 0 0 0 0 00 0 0 0 0 0 = So, the particular solution is y p = 16 x e x sin 3 x + 118 xe x cos 3 x (cid:3) Example . Find the integral
Z (cid:0) xe x sin 3 x − xe x cos 3 x + 5 e x sin 3 x − e x cos 3 x (cid:1) dx Solution . We want to solve the differential equation Dy = 13 xe x sin 3 x − xe x cos 3 x + 5 e x sin 3 x − e x cos 3 x The solution will be in the vector space V = span( xe x sin 3 x,xe x cos 3 x,e x sin 3 x,e x cos 3 x )with the basis B = (cid:8) xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x (cid:9) In (2.4) we calculated D B = − −
30 1 3 2
We are having to solve the matrix equation D B y B = − − ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 23
Then y B = D − B − − =
213 313 0 0 −
313 213 0 05169 − −
313 213 − − = − − − We have calculated that
Z (cid:0) xe x sin 3 x − xe x cos 3 x + 5 e x sin 3 x − e x cos 3 x (cid:1) dx = − xe x sin 3 x − xe x cos 3 x + 1513 e x sin 3 x − e x cos 3 x + C (cid:3) Sometimes it is useful to combine a differential operator with a matrixdifferential operator. Relationship I in Table 2 allows us to reduce a matrixdifferential operator by two rows and two columns. In the following we willgive an example of this.
Example . Find a particular solution of( D − D + 16) y = xe x sin 3 x (5.13) Solution.
Using I in the Table 2 we have y = 1 D − D + 16 xe x sin 3 x = (cid:18) x − D − D + 16 (2 D − (cid:19) D − D + 16 e x sin 3 x (5.14)All we need to do is to create a 2 × ×
4. The particular solution of (5.13) belongs to the vector space V = span (cid:0) xe x sin 3 x, xe x cos 3 x, e x sin 3 x, e x cos 3 x (cid:1) . It can be reduced us-ing (5.14) to the vector space V = span (cid:0) e x sin 3 x, e x cos 3 x (cid:1) with the basis B = { e x sin 3 x, e x cos 3 x } .Then D B = (cid:20) −
33 2 (cid:21) D B − D B + 16 I = (cid:20) − (cid:21) (cid:0) D B − D B + 16 I (cid:1) − = − (2 D B − I ) − = (cid:20) − − − (cid:21) where I is the 2 × (cid:16) x I − (cid:0) D B − D B + 16 I (cid:1) − (2 D B − I (cid:17) (cid:0) D B − D B + 16 I (cid:1) − (cid:20) (cid:21) = x − − − − − − = (cid:20) x x (cid:21) − − − − − = x x − − − = x + 725310 x + 2750 We can rewrite this using functions as1 D − D + 16 xe x sin 3 x = (cid:18) x + 725 (cid:19) e x sin 3 x + (cid:18) x + 2750 (cid:19) e x cos 3 x so the particular solution of the differential equation (5.13) is y = (cid:18) x + 725 (cid:19) e x sin 3 x + (cid:18) x + 2750 (cid:19) e x cos 3 x (cid:3) In the previous example, the number resulting from the right-hand side ofthe differential equation (5.13) was not the root of the corresponding charac-teristic equation of (5.13). In the next example it will be.
Example . Determine the particular solution of the equation (cid:0) D + 1 (cid:1) y = x cos x (5.15) ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 25
Solution.
The complex number i is a single root of the characteristic equa-tion k + 1 = 0, therefore the particular solution of the differential equation(5.15) will be in the vector space V = span (cid:0) x sin x, x cos x, x sin x, x cos x, sin x, cos x (cid:1) with the basis for VB = { x sin x, x cos x, x sin x, x cos x, sin x, cos x } The pseudoinverse matrix differential operator will be a 6 × D B = − − −
10 0 0 1 1 0 ( D B + I ) y B = [ x cos x ] B − − y B = Using a Moore-Penrose pseudoinverse matrix (Theorem 4 and 5) we have y B = ( D B + I ) + [ x cos x ] B y B = − −
00 0 0 0 0 00 0 0 0 0 0 = So the particular solution of (5.15) is y = 14 x sin x + 14 x cos x b) We can reduce the size of the matrix D B by using I in the Table 2.1 D + 1 x cos x = x D + 1 cos x − D + 1 2 D D + 1 cos x Let’s solve 1 D + 1 cos x which corresponds to the particular solution of thedifferential equation ( D + 1) y = cos x . This particular solution belongs tothe vector space V = span( x sin x, x cos x, sin x, cos x )with the basis B = { x sin x, x cos x, sin x, cos x } Then D B = − −
10 1 1 0 D B + I ) y B = [cos x ] B − y B = y B = − + y B = −
00 0 0 00 0 0 0 = We have calculated that 1 D + 1 cos x = x sin x (cid:18) D + 1 sin x (cid:19) B = −
00 0 0 00 0 0 0 = − So 1 D + 1 sin x = − x cos xx ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 27
Let’s continue to calculate1 D + 1 x cos x = x x sin x − D + 1 2 D D + 1 cos x D + 1 x cos x = x x sin x − D + 1 (sin x + x cos x )1 D + 1 x cos x = x x sin x x cos x − D + 1 x cos x From this, by expressing 1 D + 1 x cos x , we have1 D + 1 x cos x = x sin x x cos x y = 1 D + 1 x cos x = 1 D + 1 (cid:18) x e ix + e − ix (cid:19) = 12 (cid:18) e ix D + i ) + 1 x + e − ix D − i ) + 1 x (cid:19) = 12 (cid:18) e ix D ( D + 2 i ) x + e − ix D ( D − i ) x (cid:19) = 12 (cid:18) e ix D (cid:18) − i D (cid:19) x + e − ix D (cid:18) i D (cid:19) x (cid:19) = 12 (cid:18) e ix D (cid:18) − ix (cid:19) + e − ix D (cid:18) ix (cid:19)(cid:19) = 12 (cid:18) e ix (cid:18) − ix x (cid:19) + e − ix (cid:18) ix x (cid:19)(cid:19) = ix − e ix + e − ix x e ix + e − ix x e ix − e − ix i + x e ix + e − ix x sin x x cos x (cid:3) In the previous example we compared the solution method with other meth-ods. In the following example, we return to the differential operator for solvinga linear ODE with a right-hand polynomial function. In this case, it is pro-posed to expand the operator φ ( D ) in powers of D . For example, in [1] for1 φ ( D ) f ( x ) = 1 a n D n + a n − D n − + . . . + a D + a f ( x ) is the expand for a = 0 y = 1 φ ( D ) f ( x ) = ∞ X k =0 ( − k a (cid:18) a n a D n + . . . + a a D (cid:19) k f ( x ) (5.16)In the case a = a = . . . = a k − = 0 (1 ≤ n ≤ n )) and ( a k = 0, then y = 1 φ ( D ) f ( x ) = D − k a n D n − k + a n − D n − k − + . . . + a k f ( x ) , (5.17)where we would expand an inverse operator as a Maclaurin series in the senseof (5.16).The disadvantage (5.16) of expanding as a Maclaurin series is the calculation (cid:18) a n a D n + . . . + a a D (cid:19) k although it is not necessary to count all members ofthe power. We will show a different approach to solving this problem. Theorem 6.
Let a = 0 , then the Maclaurin expansion a n D n + a n − D n − + · · · + a D + a = c + c D + c D + . . . where c k = 0 for k < , c = 1 a , c k = q · ( c k − n , c k − n +1 , . . . , c k − ) , for k = 1 , , . . . is a dot product of vectors q = 1 a ( − a n , − a n − , . . . , − a ) , ( c k − n , c k − n +1 , . . . , c k − ) . Proof.
The statement follows from the identity1 = (cid:0) a n D n + a n − D n − + · · · + a D + a (cid:1) (cid:0) c + c D + c D + . . . (cid:1) (5.18)after expanding the right-hand side of (5.18) and comparing coefficients of thesame powers of D we get c = 1 a , The coefficient of D k , k = 1 , , , . . . we get from the equation c k a + c k − a + c k − a + . . . + c k − n +1 a n − + c k − n a n = 0 , where c k = 0 for k < . From this, we immediately have the proof of theTheorem 6. (cid:3)
Example . Find a particular solution of the equation( D − D + 2 D + 1) y = x + 2 x + 3 x (5.19)using the Maclaurin expansion of the inverse differential operator. ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 29
Solution.
Compute the coefficients of the Maclaurin expansion c = 1 c = ( − , , − · ( c − , c − , c ) = ( − , , − · (0 , ,
1) = − c = ( − , , − · ( c − , c , c ) = ( − , , − · (0 , , −
2) = 5 c = ( − , , − · ( c , c , c ) = ( − , , − · (1 , − ,
5) = − c = ( − , , − · ( c , c , c ) = ( − , , − · ( − , , −
13) = 33...Then we have 1 D − D + 2 D + 1 ( x + 2 x + 3 x )= (1 − D + 5 D − D + 33 D + · · · )( x + 2 x + 3 x )= (1 − D + 5 D − D ) x + (1 − D + 5 D )2 x + (1 − D )3 x = ( x − x + 30 x −
78) + (2 x − x + 20) + (3 x − x − x + 25 x − y = x − x + 25 x − V = span( x , x , x, B = { x , x , x, } . Then D B = , D B = , D B = D nB = , for n > y B = ( I − D B + 5 D B − D B )[ x + 2 x + 3 x ] B = − − −
78 10 − = − − The particular solution y = x − x + 25 x − (cid:3) Example . Using block matrices determine a particular solution of the dif-ferential equation( D + 6 D + 13) y = 2 xe − x sin 2 x − xe − x cos 2 x (5.20) Solution . Since the roots of the characteristic equation are − i, − − i compared to the right-hand side of the differential equation, it follows thatthe particular solution will be in a vector space V =span (cid:0) x e − x sin 2 x, x e − x cos 3 x, xe − x sin 2 x, xe − x cos 3 x,e − x sin 2 x, e − x cos 3 x (cid:1) with basis B of V B = (cid:8) x e − x sin 2 x, x e − x cos 3 x, xe − x sin 2 x, xe − x cos 3 x,e − x sin 2 x, e − x cos 3 x (cid:9) D B = − − − − − − − −
20 0 0 1 2 − We express this matrix as a block matrix D B = C I C I C where C = (cid:20) − − − (cid:21) , I = (cid:20) (cid:21) , = (cid:20) (cid:21) Using mathematical induction it can be proved that D nB = C n n C n − C n n ( n − C n − n C n − C n , n = 0 , , , . . . (5.21)It is easy verify that equation (5.21) is true also for n = − , − , . . . , i.e. D − nB = C − n − n C − n − C − n n ( n + 1) C − n − − n C − n − C − n n = 1 , , , . . . ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 31
Then D B + 6 D B + 13 I I
00 0 I = C C C I C C + 6 C I C I C + 13 I I
00 0 I = C + 6 C + 13 I C + 12 I C + 6 C + 13 I I C + 6 I C + 6 C + 13 I = C + 12 I I C + 6 I because C + 6 C + 13 I = (cid:20) −
12 5 (cid:21) + 6 (cid:20) − − − (cid:21) + 13 (cid:20) (cid:21) = (cid:20) (cid:21) = Next 2 C + 6 I = (cid:20) −
44 0 (cid:21) C + 12 I = 2 (2 C + 6 I ) = (cid:20) −
88 0 (cid:21) I = (cid:20) (cid:21) Dividing the matrix into 2 × C + 12 I I C + 6 I From Theorem 4, we have y B = C + 12 I I C + 6 I + b b b = (2 C + 6 I ) − − (2 C + 6 I ) − (2 C + 6 I ) − b b b = (2 C + 6 I ) − b − (2 C + 6 I ) − b b ∗ (5.22)where b = (cid:20) (cid:21) , b = (cid:20) − (cid:21) , b = (cid:20) (cid:21) , b ∗ = (cid:20) (cid:21) . Let us verify that the product of the following block matrices is the identitymatrix (cid:20) C + 12 I I C + 6 I (cid:21) "
12 (2 C + 6 I ) − − (2 C + 6 I ) − (2 C + 6 I ) − = (4 C + 12 I ) 12 (2 C + 6 I ) − I
12 (2 C + 6 I ) − − (2 C + 6 I (2 C + 6 I ) − (2 C + 6 I )(2 C + 6 I ) − = (cid:20) I (2 C + 6 I ) − − (2 C + 6 I ) − I (cid:21) = (cid:20) I I (cid:21) The same is true in reverse order of multiplying the matrices. Let us computethe elements of the last matrix in (5.22)12 (2 C + 6 I ) − b = 12 (cid:20) −
44 0 (cid:21) − (cid:20) − (cid:21) = 12 −
14 0 (cid:20) − (cid:21) = − − − (2 C + 6 I ) − b = − (cid:20) −
44 0 (cid:21) − (cid:20) − (cid:21) = − −
14 0 (cid:20) − (cid:21) = − −
116 00 − (cid:20) − (cid:21) = − b + b + b = (cid:20) (cid:21) From this we can find the particular solution
ATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 33 y B =
12 (2 C + 6 I ) − b − (2 C + 6 I ) − b b ∗ = − − − = − − − or y = − x e − x sin 2 x − x e − x cos 2 x + 18 xe − x sin 2 x − xe − x cos 2 x Although the original matrix D B was 6 ×
6, during the calculation of partic-ular solution of the differential equation (5.20) we calculated the 2 × (cid:3) Conclusion
The operational method is a fast and universal mathematical tool for ob-taining solutions of differential equations. Determining particular solutionsof ordinary nonhomogeneous linear differential equations with constant coeffi-cients using the undetermined coefficients method and the differential operatormethod are generally known.In particular, distinct from the differential opera-tor method introduced in the literature, we propose and highlight utilizing thedefinition of the pseudoinverse matrix differential operator to determine a par-ticular solution of differential equations. This method is simple to understandand to apply, compared to some cases of the differential operator method.However, it requires the determination of an inverse or pseudoinverse matrix,which generally has a special type. If the matrix is singular, then we will take apseudoinverse matrix - Moore Penrose pseudoinverse matrix - instead of the in-verse matrix. The paper shows that for its determination, we need to calculateonly the inverse submatrix of the considered matrix. Finally, the technique ofcalculating a particular solution by expressing a matrix differential operatoras a block matrix is illustrated. Combination of the operational method, themethod of undetermined coefficients and application of the matrix differentialmethod provides a powerful instrument to determine particular solutions ofdifferential equations.
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Operational calculus. History .https://en.wikipedia.org/wiki/Operational calculus. Online; accessed 30-December-2020
University of Economics, Faculty of Economic Informatics, Department ofMathematics and Actuarial Science, Dolnozemsk´a St., 832 04 Bratislava, Slo-vakia
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