12 1. PROBLEMS ON THE HALF-AXIS

DEFINITION

1.1.1. Assume that the Green formula (1.1.6) is valid. Then the

problem

(1.1.15) L+(Dt)v(t) = f{t) for£0 ,

(1.1.16) P(Dt) v(t)\t=o + Q*v = g, C*v = h

is said to be formally adjoint to (1.1.2), (1.1.3).

By the representation of the elements q^j of the matrix (J, the boundary con-

ditions (1.1.16) of the formally adjoint problem have the following form

771+J

Pj(Dt)v(t)\t=o+ ^2 bk,j-iVk= g3 , j = l, . . . , 2 m ,

M/cj- i

m-\-J

^ Ck,jVk = h3; , j = 1, . . . , J .

fe=l

The formally adjoint problem has the same structure as the starting problem. How-

ever, the number of the boundary conditions and of the unknowns is greater than

in (1.1.2), (1.1.3).

1.1.3. Boundary operators of higher order. Now we consider the bound-

ary value problem (1.1.2), (1.1.3) without the restriction ^ 2m on the orders of

the differential operators £?*.. Let K be an integer number such that

K

2m,

K

m a x ^ for k — 1,... ,m + J,

and let V^ be the column vector with the components 1, Dt,... ,

D^~l.

Then the

vector B(Dt) can be written in the form

(1.1.17) B(Dt) =

QlK)-V(K\

where Q^ is a (m + J) x n matrix of complex numbers. Furthermore, according

to (1.1.11), we have

oo oo

(1.1.18) / Lu-vdx= fu -IT^vdx + ({V^u)(0), (PMv){0))£K ,

0 0

where P^ is the vector with the components Pi(Dt),.. • , P2m{Dt),0,... ,0. We

introduce the (K — 2m) x K, matrix

RM

[

a0 ai •••

a2m

0 ••• 0 ^

0 CLQ • • • Q2m-1

a

2m ' ' ' 0

\ 0 0 • • • a0 ai • • • a2m /

Obviously, the vector p(^-

2 m

) L(Dt) has the representation

(1.1.19) £(*-2™) L ( A ) = ii(/c) • VM