Operator Method

v (z)6Exp (C~) and aj ~ 0, j = 1 , . . . ,n

where the integration goes over the lines (--VaToo. V~oo). passing through the points _+Va) of the complex plane sj. From this we immediately find that

-r

--r

i.e., ~(z) is the regular functional defined by the function contours indicated.

exp(--z2/4a) and

the family of

Remark. Another form can be given to the general form of exponential functionals. C n mxpa(C2)>we have Namely, as follows from Example 3 within the duality framework < Exp,(~), the formula 6 (z) = (2 V~) -~ exp ( -- D 2) exp ( -- Z214), ( 6.1 ) where exp ( - z 2 / 4 ) i s t h e r e g u l a r e x p o n e n t i a l f u n c t i o n a l d e f i n e d by t h i s f u n c t i o n and t h e r e a l c o n t o u r s of i n t e g r a t i o n - ~ < xj < ~, j = 1 , . . . , n . Hence, any e x p o n e n t i a l f u n c t i o n a l can be represented in the form h (z)----[A (-- D) exp (--D2)]o(2~)-~ exp (-- z~/4),

i.e., in the form of the action of a p/d operator with analytic symbol on the regular functional (2~/~)-nexp(--z2/4). To conclude the section we consider the algebra of p/d operators dual to the algebra ~(~)). For this we note that since the space ExPu(C2) is invariant relative to p/d operators A(D) with symbols A(~)60(e), the space Exp~ (C~) is invariant relative to p/d operators A(-D), i.e., the mapping A (-- m):Exp~ (C" ~ ) -~ Exp~ (C~) is continuous.

We denote b y ~t(fl-)the algebra of p/d operators A(D) with symbols A (~)6t~(~-), where ~- = {~:,~Ee},and with domain Exp~(C~). From what has been said we obtain THEOREM 6.2.

7.

The following algebraic isomorphism holds:

Operator Method i.

P/D Equations in the Entire Space C~_. We study the equation A (D) g (z) = h (z),

zEC~,

( 7. i )

where A(D) i s a p / d o p e r a t o r w i t h symbol A(~). THEOREM 7 . 1 . Suppose in some Runge domain f~cC~ t h e f u n c t i o n s A(r and A - I ( ~ ) a r e a n a lytic. Then f o r any f u n c t i o n h(z)~Exp~ (C~ ~) t h e r e e x i s t s a u n i q u e s o l u t i o n of e q u a t i o n ( 7 . 1 ) u(z)CExp.q( z] and t h i s s o l u t i o n i s g i v e n by t h e f o r m u l a

(z) A dual result

A@D) h (z).

(7.2)

can be f o r m u l a t e d s i m i l a r l y .

THEOREM 7 . 2 . Under t h e c o n d i t i o n s o f t h e p r e c e d i n g t h e o r e m f o r any f u n c t i o n a l h ( z ) (CA t h e r e e x i s t s a unique s o l u t i o n g(z)CExp'~-(C~) o f Eq. ( 7 . 1 ) , and t h i s s o l u t i o n ~s g i v e n E x p e l '* by f o r m u l a ( 7 . 2 ) . , C ~ the The proof of both theorems is obvious, since in the spaces Expa(C~)and Exp,-(~) operator I/A(D) is the inverse operator to the operator A(D).

Example i.

2758

We consider the complex Helmholtz equation Au (z) + ~o~u(z) = h (z), zCeC",

(7.3)

where A is the Laplace operator in zl,...,Zn, and co6Ci is a parameter. The symbol of the operator A + ~21 is obviously the entire function A(~) - ~2 + a~2, and the function A-I(r is analytic in the domain Q 0 = C ~ \ { ~ : ~ 2 + ~ = O } . Hence, for any functlon h(z)CExpa(Cz) or h(z)C Exp~ (C01 where ~2c-Q0 is a Runge domain, equation (7.3) has a unique solution u(z) = [I/(A + .,~21)]h(z) in the same space. 9

-

"

"

n

2. The Cauch'f_Problem (Exponential T h e o r y . In the space C n+i of variables tEC:, z~C n we consider the Cauchy problem ( n ~ > i is an integer) m--1

u(~) + ~

A~ (t, D) u(k) = h (t, z),

(7.4)

k=O

u(:)(O,z)=~j(z), where A k ( t , D) a r e p / d o p e r a t o r s

j = O , 1. . . . . m - - l ,

whose symbols A k ( t ,

~) a r e a n a l y t i c

We d e n o t e by CY(C~; Exp~(C~)) t h e s p a c e of f u n c t i o n s to the space~Exp~(C~) in zCC ~ for each fixed t6C t.

u(t,

(7.5) functions

z) a n a l y t i c

i n , t~C 1 and

in tEC ~ and b e l o n g i n g

~ E x p . ( C .~) ) a n d q~:(z)~Exp=(C.) .~ THEOREM 7 . 3 . For any f u n c t i o n s h(t, z)6 ,( C ,; u n i q u e s o l u t i o n of p r o b l e m (7.4), (7.5)u(t, z)Ea(CJ; Expa(Czn)).

there exists

a

Proof. By Duhamel~s principle it may be assumed that h(t, z) - 0. We now formally set D +-+ ~ and denote by uj(t, ~) the solution of the Cauchy problem for the ordinary differential equation . . m--I

'

;)=o, k=O

a ~ ) (0, ;) = 6:k (j, k = O, 1 . . . . .

m--

1),

where 6jk is the gronecker symbol and ~ is a complex parameter. Since Ak(t, ~) depends on ~6~2 analytically, the solutions uj(t, ~) also depend on ~ 2 analytically (see, for example, Kamke [17]). We assign to each such solution uj(t, ~) the p/d operator uj(t, D) which in correspondence with the results of Sec. 4 acts in the space Expn(~) continuously for each C~ t6CL Obviously, the formula

U (t, z) -~- ~/Zj (t, D) q~7(z)

( 7.6 )

defines the desired solution. To prove uniqueness of the solution of problem (7.4), (7.5) it suffices to note that the Cauchy data make it possible to uniquely determine the values of all derivatives uk(0, z), k = 0~ i,. .... The theorem is proved. Example 2.

The solution of the Cauchy problem ~t" ~ a D 2 u

can be written in the form the integral form

= O, u (0, z) = ~ (z)6Exp (C~),

u(t, z ) = e x p a D ~ o ( P ( z )

or, equivalently (see Example 3, See. 6), in

Vh)~,

1 (4 z) ~- 2 g'~at

:

f e - -4~,~ - ( z - - s) ds. - V~

Exile 3. We consider the Cauchy problem for the complex analogue of the SchrSdinger equation of a relativistically free particle (see BSrken, Drell [3]) i/z'----V I-

~u,

~ (0, z) = q~(Z).

( 7.7 )

The symbol A(5) = v/~ - ~: is not a single-valued function. To distinguish a unique branch on the complex plane s6C ~ we consider the function A(s) = ./i - s 2 and distinguish some unique branch for it, making a cut of the s plane along a smooth contour F joining the branch points s = • of this function. For all ~s = we have distinguished a single-valued branch of the function / f - ~ with domain'~q0={~:~F}. Applying Theorem 7.3, we find that for any function q(z)6Expe (~), where f~<~0 is a Runge domain there exists a unique solution of problem (7.7): 2759

u (t, z) = exp t V I - AoCp(z). The dual result can be established in a similar manner. THEOREM 7 94.

.

For any

h(t' z)CO(C~; Exp ' ~_(C~)), ~ ~k(z)ffExp'a_ (C ,n) t h e r e

it(t,z)@O(C~; Exp'n_(C~))of

tion

exists a unique soluproblem (7.4), (7.5) which is defined by formula (7.6).

The Case of an Arbitrary Domain

If Q c C ~ is an arbitrary open set, then the operator A(D), as we have seen in Sec. 5, is, generally speaking, multivalued. In this case in place of p/d equations we consider p/d inclusions. it is required to find a solution u(z)~ E xp~(Cz) of the inclusion

For example, suppose

A (D) tt (z)~h (z),

(8.1)

~ is a given function. where h(z)EExpa( Cz) I n t h i s c a s e t h e i n c l u s i o n u(z)e[I/A(D)]h(z)defines a s o l u t i o n o f t h e i n c l u s i o n ( 8 . 1 ) , and, s i n c e t h e o p e r a t o r I / A ( D ) a l s o may n o t be s i n g l e - v a l u e d , t h e r e may be s e v e r a l s o l u t i o n s . I n p r a c t i c e f i n d i n g s o l u t i o n s i n t h i s c a s e d o e s n o t d i f f e r a t a l l from t h e c a s e o f a p / d equation. Indeed,

let h (z) = ~ e~'% (z) %

be some representation of the function h(z)CExp~(Cnz}. Then the function u (z) = ~ where

e~

b~ (:~) D ~

b~(;~)-=-O~A-lO~)/a, i s o b v i o u s l y a s o l u t i o n

Thus, t o clusion (8.1) tion obviously a t o r I/A(D) is

(z) ,

of the inclusion

(8.1).

each representation of the right side h(z) there corresponds a solution of inhaving a representation o v e r t h e same c o l l e c t i o n o f p o i n t s ~E~. This situaa l s o a r i s e s i n t h e c a s e where t h e o p e r a t o r A(D) i s s i n g l e - v a l u e d b u t t h e o p e r not single-valued.

Example 1.

We c o n s i d e r

the differential

equation with constant

coefficients

A (D) tt (z) = h (z), zEC ~.

( 8.3 )

We set fl={~:A(~)sA0}. As a rule, this is not a Runge domain. (D)]h(z), h(z)fi Expa(C~n), are solutions of Eq. (8.3).

Hence, all functions

u(z)q[l/A

Example 2. The p/d inclusion [D/(Dq-f)]tt(z)~exp%z, zEC I, %~0. I, has at least two solutions (cf. the counterexample in Sec. 5) r~ +__Jl ~z. . . D + I / 1 u (z)=---b-- e ~ = ( ~ _~ _l e z 'e --~-. &

In conclusion we note that there is no difficulty in formulating analogous results on the Cauchy problem for the inclusions

~'~)-~- "~ A~ (t, D) uc~h (t, z), k=O

ur (0, z)---- % (z), Example 3.

k = O, 1 . . . . .

m--

1.

We consider the Cauchy problem u ' - - [ l / D ] u~O, u (0, z) = e ~ (~ ~ 0),

where [I/D] solutions 2760

is a p/d operator

i n t h e s p a c e Expn(Czn); ~ C 1 \ { 0 } .

The problem has at least two