CC BY
13
УДК 517.955, 517.55
Boundary Problems for Helmholtz Equation and the Cauchy Problem for Dirac Operators
Alexander A. Shlapunov*
Institute of Mathematics, Siberian Federal University, Svobodny, 79, Krasnoyarsk, 660041
Russia
Received 01.12.2010, received in revised form 01.12.2010, accepted 15.01.2011
Studying an operator equation Au = f in Hilbert spaces one usually needs the adjoint operator A* for A. Solving the ill-posed Cauchy problem for Dirac type systems in the Lebesgue spaces by an iteration method we propose to construct the corresponding adjoint operator with the use of normally solvable mixed problem for Helmholtz Equation. This leads to the description of necessary and sufficient solvability conditions for the Cauchy Problem and formulae for its exact and approximate solutions.
Keywords: mixed problems, Helmholtz equation, Dirac operators, ill-posed Cauchy problem.
Many problems in applications of mathematics can be formulated on the language of Functional Analysis as the study of operator equations of the first type in Hilbert spaces (cf. [1,2]). Namely, let Hilbert spaces H1 and H2 and a (continuous linear) map A : H1 ^ H2 be fixed. Then the problem is the following: given f G H2 find u G H1 with Au = f. Though the problem is usually ill-posed (see [1,2]), there are many approaches to it, for instance, fixed points method, iteration methods, methods of bases with double orthogonality property, method of small parameter perturbation, etc. Regarding the Cauchy problem for elliptic systems we refer, for instance, to [3-7]. Usually one needs the adjoint A* for A in the sense of Hilbert spaces in order to write down proper solvability conditions. Then there are many possibilities to construct regularization of the problem (i.e. a family of approximate solutions depending on a parameter and converging to an exact solution if and only if a solution exists).
Dirac type operators naturally appear in many applications (see, for instance, [8,9]). These include Cauchy-Riemann operator, gradient operator, operator Moisil-Teodorescu, stationary Maxwell operator and so on. We propose to consider the ill-posed Cauchy problem for Dirac type operators, see [7,22,23] in Lebesgue spaces. The Dirichlet problem for Helmholtz Equation plays an essential role in the formulation of the Cauchy problem as an operator equation of the first type in Hilbert spaces. Then we construct the corresponding adjoint operator with the use of a normally solvable mixed problem for Helmholtz Equation (cf. [6] for elliptic operators in Sobolev spaces where the Dirichlet Crack Problem for the Laplace equation was used instead).
1. Preliminaries
Let Rn be n-dimensional Eucledian space and Cn be n-dimensional complex space with points being n-vectors z = (z1,zn), where Zj = Xj + \J—Txj+n, j = 1, ...,n, x = (x1, ...,x2n) G R2n
i__ n d
and 1 being imaginary unit. Let A = J2 aj j,— be a Dirac operator in Rn, i.e., such a
j — i j
*ashlapunov@sfu-kras.ru © Siberian Federal University. All rights reserved
homogeneous first order matrix differential operator with constant coefficients that
nn
]T a*am£j£m = |£|2 Ik for all £ G Rn; (1)
j—1 m—1
here aj are (I x k) -matrices of complex numbers, a* = aj are their adjoint matrices and Ik
n
is the identity (k x k) -matrix. In particular, the symbol a(A)(£) = ^ aj £j is injective as
j—1
the map from Ck to C1 for all £ G Rn \ {0}, i.e. I > k. We say that A is elliptic if I = k
and overdetermined elliptic if I > k. Denote A* and AT the formally adjoint and transposed
n d n d
operators for A respectively: A* = - a* t;— and AT = - aj t;—. Then A*A = -AIk
j—1 dxj j—1 dxj
n d2
where A = V ——;2 is the Laplace operator in Rn. j—0 dxj
Typical elliptic Dirac operators are the derivative operator — in R and the (doubled) Cauchy-d d .— d
Riemann operator 2 — = —-----------+ V — 1—— in C. Typical overdetermined Dirac operators are
dz 0x1 dxj
( d d \T
the gradient operator V = — ——,... —— in Rn, the (doubled) multi-dimensional Cauchy-
\ 0x1 dxnJ
— ( d d \T
Riemann system 2d = — 2 —-,... - in Cn and the stationary Maxwell system M =
\oz1 dzn)
— (rot, div)T in R3. For Dirac operators on manifolds and in the Quantum Physics see, for instance, [8,9]. Our approach is also fit on a manifold but this lead to a mixed problem for more complicated second order elliptic operator.
Let D be a bounded domain (i.e. open connected set) in Rn, and let D be its closure. We always assume that the boundary dD of D is of class CTO. As usual we denote by D(D) the space of all the smooth functions with compact supports in D and by V'(D) the space of distributions over D. Besides, let CX(D) stand for the set of smooth functions in D with any derivative extending continuously to D and C^°(D) stand for the set of functions vanishing on a closed subset a c D.
Let E = Rn x Ck be the trivial k-vector bundle. The set of all k-vector functions over domain
D with the components from a functional space S(D) will be denoted by S(D,E).
It is known (see, for instance, [12]) that A induces a differential compatibility complex:
0 —► C™(Eo) —^ CTO(E1) —^ C^(Ej) —^ ... A—^ C™(En) —► 0 (2)
where Aj are differential operators with constsant coefficients, A0 = A and Ai+1 o Aj = 0; here Ei = Rn x Cki are trivial vector bundles of ranks ki (of course, k0 = k, k1 = l). From now on we assume that complex (2) is elliptic and orders of all the operators equal to one, i.e.
Ai = aji) Ti--------+ a0j) where ajj) are (ki+1 x ki) -matrices of complex numbers. It is true in
j—1 j dxj j
many cases, though it is not known to be true in general.
For instance, if A = V then the corresponding sequence is de Rham complex with Ei = Ai
(0 ^ i ^ n) being the set of all the exterior differential forms of degree i in Rn and Ai = di
being exterior derivatives for the forms. Similarly, if A = 2d then the corresponding sequence is
Dolbeault complex with Ei = A( 0,i) (0 ^ i ^ n) being the set of all the complex exterior forms
of bi-degree (0, i) in Cn and Ai = di be the (graduated) Cauchy-Riemann operator extended to
the differential forms. Obviously, if I = k then E0 = E1 and Ai = 0 for all i ^ 1.
We write L2(D) for the Hilbert space of all the measurable functions in D with a finite norm (u,v)L2(D) = JD u(x)v(x) dx. Then the Hermitian form
(u,v)L2(D,Ei) = / v*(x)u(x) dx D
defines the Hilbert structure on Lj(D,Ei). We also denote Hs(D) the Sobolev space of all the distributions over D, whose weak derivatives up to the order s G N belong to Lj(D). The closures of V(D) and C?Dd(D) in Hs(D) will be denoted by HS(D) and HsdD(D) respectively; for s = 1 these spaces coincide.
For non-integer positive s G R+ we define Sobolev spaces Hs with the use of the standard interpolation procedure (see [14] or [5, §1.4.11]). It is known (see, for instance, [14]) that functions of Hs(D), s G N, have traces on dD of class Hs_1/2(dD) and the corresponding trace operator is continuous. We will use Sobolev spaces of fractional smoothness for boundary data only.
Sobolev spaces of negative smoothness may be defined in many different ways (see [15],
[16]). We follow [16] and consider Sobolev spaces H_s(D) and H(D, | • |_s), s G R+, being the completions of CTO (D) with respect to the norms
,, ,, |(u,0)L2(D)| , , |(u,0)L2 (D)|
IMIh-s(D) = SUp_ —r--------------, |u|_s,D = SUp [LAN-----------
^eC^(D) \\V\\hs(D) ‘PtC-jD (D) \\V\\hs(D)
Easily, H_s(D,E) c H(D, | • |_s,E). These spaces are the strong duals for Hs(D,E) and HsdD(D,E) correspondingly with respect to Lj(D,E)-pairing (•, )d . Namely,
(u, 4>)d = lim (uv, <p)L2(D E), u G H_s(D, E), ^ G Hs(D, E),
V—— ^ ’
where {uV} c C'X’(D,E) is a sequence approximating u in H_s(D,E) (see [15] or [5, Theorem 1.4.28]). Easily, the pairing does not depend on the choice of the approximating sequence {uV} and
|(u,^)d| < ||u||h-={d,eM\hs{d,e) for all u G H_s(D,E),$ G Hs(D,E). (3)
Of course, we can do the similar procedure for the pair H(D, | • |_s, E) and HgD(D, E). Clearly, as d(dD) = 0 then we have H(dD, | • |_s,E) = H_s(dD, E).
For a generalized vector function u G V(D, E) we always consider Au in the sense of distributions. Thus, Au G V'(D,E1) is a vector-distribution over D. However there are no reasons for it being an element of Lj(D, E1) if u G Lj(D, E). We denote by HA(D) the so called strong extension of the differential operator A, i.e. the closure of CX(D, E) with respect to the graph norm
Ma,d = (\M2l2{d,e) + \\Au\\2l2{d,e,))112.
According to [13] this space coincides with the weak extension of the differential operator A, i.e. with the set of vector functions from L2(D,E) with Au G L2(D,E1). Of course, it is a Hilbert space with the scalar product
(u, v)a,d = (u, v)l2(d,e) + (Au, Av)l2(d,e1).
Thus, the differential operator A induces a bounded linear operator A : HA(D) ^ L2(D,E1) with ||A|| < 1. Clearly, ||u\a,D ^ CA\\u\\hi(D,E) for all u G H1 (D,E) with a positive constant CA being independent on u. It is worth to mention that, due to ellipticity of A and Garding inequality (see, for instance, [16]), the closure of V(D,E) in HA(D) coincides with Hq(D,E). Of course, Hy(D) = H 1(D), however, H2q(D) = H 1(D) since the space H2q(D) contains all the holomorphic L2(D)-functions. In any case, HA(D) c Hjoc(D,E) because of the ellipticity of A.
As Ai+1 o Ai = 0 we easily see that the differential operator Ai induces a bounded linear operator Ai : HAi (D) ^ HAi+1 (D). Moreover, we easily obtain a complex of such operators
0 Ha0 (D,Eo) HAl (D, Ei) -^ Ha2 (D,E2) ■■■ ^ L2(D,En) 0.
However it is not true that HAi(D) c Hjoc(D, Ei) for i > 1.
2. Traces on the Boundary
In order to formulate properly the Cauchy problem for A we need first to clarify what is the space of traces on dD for elements of HA(D). With this aim we denote by p the defining function of the domain D, i.e. p G CTO, |Vp| = 0 on dD and D = {x G Rn : p(x) < 0}. Then the following Green formula holds true for u G CX(D, E), g G CX(D, Ei):
n ji) dp
(Aiu,g)L2(D,Ei+1) - (u,A*g)L2(D,Ei) = JdD g*(x)|Vp| (x) u(x) ds(x)'
Set
* a? dp _ " ap . ,
Ti(u) = 1. |vp| s7,u' ‘''-■<g> = A^-jVprdxg’ i > 1
Ti = Vi-1 o fi. Then To = Ik.
Lemma 1. For every u G Ha(D) there is a weak trace T0(u) G H-1/2(dD, E), i.e.
(Au, '^)l2(d,e1) - (u,A*^)l2(d,e) = {To(u),fo(^))dD for all ^ G C™(D,E.).
Proof. Follows from [7, Lemma 1.1]. □
According to [7, Lemma 1.2] the subspace of elements in HA (D) having zero traces coincides with H.(D, E). Therefore in [7] the space of traces on dD for elements from Ha(D) was defined
as the factor space tHa(D'L,. Of course, Lemma 1 allows to identify it within H-1/2(dD, E) Ho(D, e)
but we will do it more precisely. Indeed, as we have seen above, if A = V then HA(D) = H.(D), i.e. in this case t0(u) g H 1/2(dD). This observation, combined with Lemma 1, leads us to the following definition. For v G Cx(dD, E), set
\(v,V0(^))L2 (dD,E)\
IIV || A,dD = sup 1 1
C^(D,E1)3^=0 nWnA*,D
Lemma 2. The functional || • || A,dD defines a norm on CX(8D, E). Moreover, there are positive constants ci, c2 such that
c1IIvIIH-1/2(8D,E) ^ ||v|| A,dD ^ c2 II'vIIH1/2(8D,E) for al1 v G C^(dD,E).
In particular, the norms || • ||v,dD and || • ||Hi/2(dD) are equivalent.
Proof. By the definition, the functional || • HA,dD is non-negative. It is easy to see that IIv. + v2Ha,sd ^ ||v.Ha,sd + ||v21A,dD for all v.,v2 G CX(D, E),
HavHA,8D = |a|||v||A,dD for all v G CX(D, E), a G C.
In particular, ||0||A,dD = 0. Let now ||v||A,dD = 0. Then
(v, Vo(^))l2(sd,e) = 0 for all ^ G C^(D, E.). (5)
Now, given u0, h, consider Dirichlet problem for the Helmholtz operator in D c Rn:
a2u — Au =
h in D,
u0 on dD,
(6)
where a G R. The problem (6) is uniquely solvable on the Sobolev scale Hs(D), s e Z+, and its solution is given by the Poisson type formula
u = VD,a,uo + G D,ah
for data u e Hs-./2(dD) and f G Hs-2(D) (s > 2) or f e H(D, | • |s-2) (s < 1) where GD,a is the Green function of the Dirichlet Problem and PD,a the corresponding Poisson type integral. The integral operators are bounded in the Sobolev spaces:
PD,a : Hs-./2(dD) ^ Hs(D), s G Z,
GD,a : Hs-2(D) ^ Hs(D) (s > 2), GD,a : H(D, | • |a-2) ^ Hs(D) (s < 1)
(see, for instance, [16], [17, Theorem 2.26]). In particular, PD,0 is the classical Poisson integral of the Dirichlet Problem for the Laplace operator.
Then for every v G C^(dD,E) the vector-function gv = PD,0 J2
n aj dp
|Vp| dxj
v belongs to
Cx(D,E.) and, due to (1), satisfies
vo(gvL = v on дD, ||gvIIhi(d,Ei) < C Mm/2(sd,e)
with a constant C > 0 being independent on v. Therefore (5) implies, with ^ = gv,
i.e. v =0 on dD. Further,
i.e. c. = (CA* C)
-l
v* (x)v(x)ds(x) = 0,
IdD
ii _ \Jsd 4>*(x)v(x)ds(x)\
vnH-1/2(8D,E) = sup nTTi
0=0 II0NH1/2 (dD,E)
fgD f0(g^)*(x)v(x)ds(x)\
sup
0=0
A*,D
IMI
A*,D
\H 1/2(dD,E)
^ II || llg0 IIH1 (D,Ei) . ^ -—r11 N
^ llv|A,dD CA* suP TTTTj-------------- ^ CA* C ||v|| A,dD,
0=0 \\9\\H1/2(dD,E)
On the other hand, by Green formula (4),
\(ApD,0(v)^)L2(D,Ei) - (P>D,0(v),A*^)L2(D,E)\
WvHAdD = _sup
C™(D,Ei)3tp=0
A*,D
< WPD,0(v)WA,D < WP D,0(v)Wh1(D,E) < c2 11v 11H 1/2(дD,E),
where positive constant c2 does not depend on v because the Dirichlet Problem (6) is normally solvable on the Sobolev scale.
u=
Finally, taking ^ = VPD,.(v) and using Green formula (4) we obtain:
\(VVD,l(v), VPd,.(v))l2(d,e1) + (Pd,.(v), APd,.(v))l2(d) \
||VPD,l(v)||v*,D
= W'Pd^^Wh1 (D) > c3 ||т0('PD,1(v))||H1/2(дD) = cзHvHнl/2(дD)
with a positive constant c3 being independent on v because the trace operator t0 : H.(D) ^ H./2(dD) is continuous. □
The completion of C^(dD, E) with respect to || • IIA,dD we denote BA(dD).
Theorem 1. The trace operator t0 continuously maps Ha(D) onto BA(dD). In particular, for each v G Ba(dD) the vector PD.(v) belongs to Ha(D), satisfies t0(Pd,.(v)) = v on dD and
IIPD,.(v)IIA,D = ||v||A,dD for all v G BA(dD).
Proof. Indeed, for every u G CX(D, E) we obtain with the use of Green formula (4)
|| < Ml \(Au,^)l2(D,Ei) - (u,A*^)L2(D,E)\ n ,,
IIT0(u)IIA,dD = JuP J- -(7^- - < IIuIIA,D. (7)
C™(D,Ei)3^ = 0 llrlU*,D
Therefore for each u G Ha(D) we may define the trace as the limit t0(u) := limv^TO T0(uv) in the space BA(dD) where {uv} c Cx(D,E) is a sequence approximating u in Ha(D). Estimate
(7) implies that the operator t0 : Ha(D) ^ BA(dD), defined in this way, is linear and bounded. Further, for each v G Cx(dD, E) we have t0(Pd,.(v)) = v on dD and
n „ ^^D.^^'P’d^^L2^^) + (P D,.(v), APD,.(v))L2(D,E) \
IMkdD > J------------------------1, ' ' ----------- —- = WP’dA^Wad.
||APD,l(v)|ІA*,D
Combining with (7) we easily obtain that HP^^Ha^ = ||v||A,dD for all v G C^(dD,E). Finally, if v G BA(dD) and {vv} c Cx(dD, E) is a sequence approximating v in BA(dD) then the sequence {PD.(vv)} c Cx(D,E) is fundamental in Ha(D). Hence it converges to a vector function w G HA (D) and, by the very definition of the trace, t0 (w) = v. Clearly, we may interpret w as PD.(v) in the sense of the strong extension of the operator PD,.. □
Corollary 1. The Poisson type integral PD. induces an isomorphism BA(dD) = . A(——
H.(D,E)-
Besides, Theorem 1 implies that H./2(dD,E) c BA(dD) c H-x/2(c)D,E).
3. The Cauchy Problem
In order to study the Cauchy problem we need one more type of boundary spaces. Let r be an open (in the topology of dD) connected set of dD. Denote by HA,r(D) the closure of Cr°(D,E) in Ha(D). Then the differential operator A induces a bounded linear operator Ar : HAiT(D) ^ L2(D, E.) with ||Ar|| < 1.
According to [7, Lemma 1.2], HA,r(D) c Hfoc(D U r,E). Moreover, [7, Theorem 1.4] states that HA,r(D) coincides with the set of elements in Ha(D) having zero traces on r, i.e. satisfying
(Au,g)L2(d,Ei) - (u,A*g)L2(D,E) = 0 for all g G CTO(D, E.) with f (g) =0 on dD \ r.
Following [7], it is natural to define the space of traces on r for elements from HA(D) as the
factor space H'a(^1^s, . However, similar to the situation r = dD above, we want to characterize Ha,t(D)
this space rather in terms of boundary functions.
With this aim we define BA,r(dD) as the closure of C^(dD, E) in BA(dD). Then let BA(r) be the factor space of BA(dD) over BA,r(dD). It is well-known that BA(r) is a normed space. By the very definition every its element extends from r up to an element of BA(dD).
Remark 1. Theorem 1 imply that the norm || • ||A,dD satisfy the parallelogram identity and hence it is a Hilbert space (with the standard scalar product coherent with the norm). This means that Ba(r) is actually the orthogonal complement of BA,r(dD) within BA(dD). Then for every uo G BA(r) there is a canonical representative u0 £ BA(dD) satisfying ||uo|| A,dd = ||uo|Ia r
Corollary 2. The Poisson type integral Pd 1 induces an isomorphism BA(r) = A(, r). .
HA,r(D)
Proof. Indeed, it follows from properties of the Poisson type integral Pd , 1 that for every v G
— ' Pd,i ha r(D)
C^(dD,E) we have Pd , 1(v) G C^(D,E). Then Theorem 1 implies BA,r(dD) = ’ .
Ho(D,E)
Finally, the desired statement follows from Corollary 1. □
In particular, it follows from Corollary 1 that there is correctly defined continuous linear trace operator t0,r : HA(D) ^ BA(r) and we can easily formulate the Cauchy problem.
Problem 1. Given g G HAl (D) and uo G BA(r) find w G HA(D) with
( Aw = g in D,
\ to,r(v) = uo on r,
i.e.
(w, A* ^) = (g, ^) — {uo, Vo(^))dD for all ^ G CTO(D, E1) with Vo(^) =0 on dD \ r. (8)
Since the time of Hadamard [18] this problem is known to be ill-posed. If r = 0 then it has no more than one solution, see, for instance, [5, theorem 10.3.5]. It was actively studied for various types of Dirac operators in various functional spaces [3,19-24].
We want to reduce Problem 1 to an operator equation of first type in Hilbert spaces. However Theorem 1 implies that Problem 1 can be easily reduced to the following one.
Problem 2. Given f G HAl (D) find u G HA,r(D) with Aru = f.
Obviously, the reduction of Problem 1 to Problem 2 can be made via
f = h — APd , i(uo), u = w — Pd , i(uo),
where uo G BA(dD) is the canonical representative from the class uo G BA(r).
In order to obtain solvabilty conditions for Problem 2 we denote by A£ the adjoint for Ar in the sense of Hilbert spaces. Let also Hr(D) be the closed subspace in HAl (D) consisting of elements h satisfying A*h = 0 in D, iso(h) =0 on dD \ r, i.e
(h,A0)L2(D,E1) = 0 for all 4 G C^(D,E),
and A1h = 0 in D, T1(h) =0 on r, i.e.
(h,A*i^)L2(D,E1) = 0 for all ^ G C%D\r(D, E2).
Recall that the adjoint operator for A is a bounded linear map A£ : HAl (D) ^ HA,r(D) satisfying
(Au, g)L2 (D, El) = (Au, AATg)L2(D, El) + (u, Ar g)L2(D, E) for aU u G HA, Г(D), g G HAi (D). (9)
Theorem 2. Problem 2 is solvable if and only if 1) A1f = 0 in D, T1(f) = 0 on r; 2) (f,h)L2(D,e1) = 0 for all h G Hr(D); 3) the series u(f) = ( I — ArAr)vArf converges
v=0
in the space Ha(D). Moreover, under conditions 1)-3), the series u(f) is the unique solution to Problem 2.
Proof. According to [7, Lemma 2.4] the space Hr(D) coincides with kernel of the operator A£. Then the desired statement follows from [26, Corollary 2.10] because ||A£|| = ||Ar|| < 1. □ Thus, solving Problem 2 we need to identify the operator Af.
4. Mixed Problems for Helmholtz Equation
In order to identify Arg we note that (since Ctt(D,E) c HA(D)) equation (9) can be interpreted as a mixed problem for Helmholtz type equation:
Ar g — AIk (Af g) = A*g in D,
To(Ar g) =0 on r, _ (10)
Vo(A(Arg) — g) = 0 on dD \ r.
We consider a little different type of problems with a parameter a G R.
Problem 3. Given triple (h,uo,u1) of vector-distributions find a vector-distribution w such that in a proper sense
a2w — AIk w = h in D,
To(w) = uo on r,
Vo (Aw) = u1 on dD \ r.
This type of problems are usually called Zaremba Problems (cf. [27], [28] for A = V; in this
dw
case Vo(Aw) = —— is the normal derivative with respect to dD). However, Problem 3 could be dv _
ill-posed for a = 0 (cf. [7] regarding the case A = 2d). An Existence and Uniqueness Theorem in the space HA(D) n H2oc(D U r) was obtained for such type of problems in [7, Theorem 3.2] for natural but rather disappointing class of regular data h G L2(D,E), uo G H3/2(r, E), u1 G H 1/2(dD \ T,E) if a = 0.
Working with a Dirac operator A = 2d one easily can see that in this case Vo o A coincides with the complex normal derivative dv and the trace Vo(Aw) exists on dD for w G H 1(D) if and only if Aw G H-1(D), while, in general, Aw G H(D, | • |-1) (see, for instance, [24,25]). We seek for a solution to Problem 3 even in a worse than H1 (D,E) class Ha(D). However Theorem 1 allows us to indicate the right classes for the data-triples and for the solution in the formulation of Problem 3. With this aim we will use more negative norms. Namely, set
l(u,4)L2 (D, E)1 "u|H-'WD> = «?(PD.E)^ik^
The completion of Ctt(D, E) with respect to the norm ||u||H-l (D) will be denoted by H— T(D).
Again, similar pairing {•, -)D may be defined for pairs (u,v) G H— 1r(D)®HA,r(D). Then H— r(D) is naturally embedded to the strong dual (HA,r(D))' for HA,r(D) keeping in the mind the corresponding L2(D, E)-pairing (cf. (3)). As the norm | • !hi(D,E) is not weaker than the norm
| • |a, D, there is a positive constant C with |u|H(D,|.|_l, E) < C|u|H-l (D) for all u G Ctt(D, E). Then H— 1r(D) c H(D, | • |_1, E).
Similarly, for v G Cx(dD, E), we set
\(V $)L2(dD,E)\
IMIbtVSDX r) = SUp
C^(D,E)3<p=0 I'?IA,D
It follows from Theorem 1 that in fact ||v||B-i/dD\n = sup ^ i\ ^t\L (9D,e) . For this
A ; C^(8D,E)3^ = 0 WnBAidD)
reason the completion of Cx(dD, E) with respect to II • IB—i(dD\r) will be denoted by B_^dD\r).
Again, B—1(dD \ r) is naturally embedded to the strong dual (HA,r(D))' for HA,r(D) keeping in the mind the corresponding L2(dD, E)-pairing (cf. (3)).
Now we are ready to introduce the right spaces for solving Problem 3. We denote HA,r ,a(D)
and HA,voA(D) the completions of Cx(D,E) with respect to the norms
IMlA,r,A = ^IMIA.D + I(a2 - Ik A)wII2H-rr{D), IMU,r,voA = ^IMlA,D + I^o(Aw)Il_i{aD\r)
correspondingly.
Theorem 3. The linear spaces HA,r,A(D) and HA,r,voA(D) coincide and their norms are equivalent. Besides, an element w G Ha(D) belongs to the space HA,r,A(D) if and only if (a2 — IkA)w G HaT(D).
Proof. The first part of the statement immediately follows from the Green formula
(Aw, Au)l2(d,e1) + a2(w, u)l2(d,e) = ((a2 — IkA)w, u)l2(d,e) — (Vo(Aw), u)l2(sd,e) (11)
being true for all w G Cx(D, E), u g Cx(D, E).
Further, by the definition, (a2 — IkA)w G H_r(D) if w G HA,r,A(D). The proof of the
converse statement is similar to that of [24, Corollary 2]. Namely, if w G Ha(D) and the
vector h = (a2 — IkA)w belongs to H—1r(D) c H(D, | • \_1,E) then according to [16] (cf.
also [17, Theorem 2.26] and the proof of Theorem 1) we have w = Ga,Dh + Pa,Dw, where
Pa,Dw G HA,r,A(D) and Ga,Dh G Hl(D, E).
Fix now a sequence {hv} c Cx(D, E), approximating h in the space H_1r(D). Then, Green formula (11) imply, for u G Cx(D,E),
(Vo(AGa,D hv ),u)l2(8D,E) = (hv, u) l2(D,E) (AGa,D hv ,Au)l2(D,E1 ) + a2 (Ga,D hv ,u)l2(D,E)- (12) As the operator Ga,D : H(D, | • |_1, E) ^ Hq(D) is bounded we see that
hm || G a,D hv —Ga,D h|U,D = 0- (13)
v^0
Therefore (12), (13) yield that the sequence {v0(AGa,Dhv} c Cx(dD,E) is fundamental in the space B)_1(3D \ r). Hence the sequence {Ga,Dhv} c Cx(D,E) converges in the space HA,r,voA(D) which coincides with the space HA,r,A(D) according to the already proved part of the theorem. □
Using Green formulae, it is easy to see that, for a triple (h,uo,u1) G H—1(D) ® BA(r) © B_1(dD \ r), Problem 3 in the space Ha,
t,a(D) can be interpreted in the following weak sense:
(Aw,'^)l2(d,Ei) — (w,A*^)l2(d,e) = {uo,i>oW)dD for all ^ G C™(D,E), vq(^) = 0 on dD \ r,
_ (14)
(Aw, A4>)l2(d,e1) + a2(w, $)l2(d,e) = (h, $)d — (u1 ,4’)sd for all $ G Cjr°(D, E). (15)
Corollary 3. Let a = 0. Then for every triple (h,u0,u1) G H-1 (D) © BA(r) © B-1(dD \ r) there is a unique solution w G H-,r,A(D) to Problem 3 in the weak sense of (14) and (15). Moreover, there are positive constants c1, c2, c3 such that
llwl|A,r,A ^ c1IIu0 IIbaCT) + c2 H u 1 H B—1(dD\r) + C3llhII.HA 1r(D) (16)
Proof. Indeed, fix the canonical representative U0 G BA(dD) of the datum u0 G BA(r). Then the potential PD,a(U0) belongs to HA,dD,A(D) c HA,r,A(D) and it coincides with u0 on dD (see Theorem 1). Hence, because of Theorem 3, there is a trace i0(APD,a(U0)) = v1 belonging to B—1(dD \ r). Therefore mixed problem (14) and (15) is equivalent to finding u G HA r a(D) satisfying
(AuU, A$)l2(D,E1) + $)L2(D,E) = (h $) D — (u1 — v1 ,$)dD for all $ G CY (D, E). (17)
Of course w = u + PD,a(U0).
If a = 0 then the Hermitian form (Au,Av)L2(D,El) + a2(u,v)L2(DE) induces a scalar product on the space HA,r(D) and the corresponding norm is equivalent to the original one.
Further, by the very construction, the right hand side of (17) defines a bounded linear functional on HAr(D). Now Riesz Theorem on the general form of a continuous linear functional on Hilbert spaces guarantees the existence of a unique element u G HA,r(D) satisfying (17). In particular, (17) implies (a2 — IkA)u = h G HA,r,A (D) and, using Theorem 3, we conclude that u actually belongs to HA,r ,a(D).
Finally, estimate (16) follows from Banach Theorem on the inverse operator. □
Remark 2. Note that the right hand side in (15) can be replaced by F(v) with an arbitrary continuous linear functional F G (H-(D))'. Then again Riesz Theorem guarantees a unique solution w G Ha(D) to Problem 3 in the weak sense of (14) and (15) if a = 0. However the element w may have no boundary values of i0(Aw) in the sense of distributions over dD \ r (cf.
(10) where we do not claim that the trace v0(A(Arg)) on dD \ r exists).
Remark 3. We also note that, though Problem 3 is Fredholm (according to Corollary 3), it is not an Elliptic Boundary Problem in the sense of Lopatinski because the Dirichlet boundary system (t0, i0 o A) is not coercive in general. Of course, it is coercive if A = V but it is not the case for A = 2d.
The advantage of using Mixed Problem 3 is the following. One may follow the classic approach to Fredholm Boundary Problems and construct the Green function of the problem, say, $r. As usual §r(x,y) = §(x,y) — y(x,y), where &(x,y) is the bilateral fundamental solution of the Helmholtz operator a2 — A and the rest Y(x,y) is the solution to the mixed problem
{a2 y(x, •) — AIkY(x, •) = 0 in D,
T0(y(x, •)) = $(x, •) on r, _
v0(Ay(x, •)) = i0(A&(x, •)) on dD \ r.
with the smooth data dependent on the parameter x G D (see [7] for solving). It is only left to say that the fundamental solution §(x,y) may be taken as b(\x — y|) where b(r) is the solution to the famous ordinary differential Bessel type equation:
((r^) +(n — 2){rdr)— aV) b(r) = 0
which is unbounded at the origin if n > 2 (see [7] and [29, Ch. 7, §2]); in particular b(r) =
e^V_1ar i
--------- for n = 3. Obviously, b(r) = e_v_1ar for n = 1.
4nr
The research was supported by grants Federal Agency of Education "The Development of Science Potential of High School" 2.1.1/4620 and NSH-7347.2010.1.
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Граничные задачи для уравнения Гельмгольца и задача Коши для операторов Дирака
А.А. Шлапунов
При изучении операторного уравнения Au = f в пространствах Гильберта обычно требуется знать сопряженный A* оператор для A. Решая некорректную задачу Коши для операторов типа Дирака в пространствах Лебега одним итерационным методом, мы предлагаем построить соответствующий сопряженный оператор при помощи нормально 'разрешимой смешанной задачи для уравнения Гельмгольца. Это ведет к описанию условий разрешимости задачи Коши и к построению ее точного и приближенных решений.
Ключевые слова: смешанная задача, уравнение Гельмгольца, операторы Дирака, некорректная задача Коши.
