Boundary value problems in complex analysis II

Transcript

1 Boletín de la Asociación Matemática Veneolana, Vol. XII, No. 2 ( Boundary value problems in complex analysis II Heinrich Begehr Abstract This is the continuation of an investigation of basic boundary value problems for first order complex model partial differential equations. Model second order equations are the Poisson the inhomogeneous Bitsade equations. Different kinds of boundary conditions are posed as combinations of the Schwar, the Dirichlet, the Neumann conditions. Solvability conditions the solutions are given in explicit form for the unit disc. Exemplarily the inhomogeneous polyanalytic equation is investigated as a model equation of arbitrary order. Boundary value problems for second order equations There are two basic second order differential operators, the Laplace operator the Bitsade operator 2. The third one, 2 is just the complex conjugate of the Bitsade operator all formulas results for this operator can be attained by the ones for the Bitsade operator through complex conjugation giving dual formulas results. For the Laplace the Poisson, i.e. the inhomogeneous Laplace equation, the Dirichlet the Neumann boundary value problems are well studied. Before investigating them the Schwar problem will be studied for both operators. Theorem The Schwar problem for the Poisson equation in the unit disc w = f in D, Re w = γ 0, Re w = γ on D, Im w(0 = c 0, Im w (0 = c is uniquely solvable for f L (D; C, γ 0, γ C( D; R, c 0, c R. The solution is w( = ic 0 + ic ( + γ 0 ( + d γ ([ log( 2 log( 2 + ] d

2 28 H. Begehr + {f([log 2 log( ] f( log( } ( + { [ log( ] [ log( ]} f( 2 + log f( 2 + log [ f( + f( ] 2. Proof This result follows from combining the solution w( = ic 0 + of the Schwar problem γ 0 ( + d 2 [ ω( + + ω( + ] with the solution ω( = ic + w = ω in D, Re w = γ 0 on D, Im w(0 = c 0 of the Schwar problem γ ( + d 2 [ f( + + f( + ] ω = f in D, Re ω = γ on D, Im ω(0 = c. Here the relations + + = 2 log log( log 2 +, = 2 log( + log 2, + + = 2 + log( +, + = 2 log(,

3 Boundary value problems in complex analysis II 29 2 [ + + ] = are needed. More simple than this is to verify that ( is the solution. The uniqueness of the solution can easily be seen. In case w w 2 are two solutions then ω = w w 2 would be a harmonic function with homogeneous data, ω = 0 in D, Re ω = 0, Re ω = 0 on D, Im ω(0 = 0, Im ω (0 = 0. As ω is analytic, say ϕ in D, then integrating the equation ω = ϕ means ω = ϕ + ψ where ψ is analytic in D. Then Re ω = 0 on D, Im ω (0 = 0 means Re ϕ = 0, Im ϕ (0 = 0. From [3], Theorem 6 then ϕ is seen to be identically ero, i.e. ϕ a constant, say a. Then from Re ω = 0 on D, Im ω(0 = 0 it follows Re ψ = Re a Im ψ(0 = Im a. Thus again [3], Theorem 6 shows ψ( = a identically in D. This means ω vanishes identically in D. There is a dual result to Theorem where the roles of are interchanged. This can be attained by setting W = w complex conjugating (. Theorem The Schwar problem for the Poisson equation in the unit disc w = f in D, Re w = γ 0, Re w = γ on D, Im w(0 = c 0, Im w (0 = c, for f L (D; C, γ 0, γ C( D; R, c 0, c R is uniquely solvable by w( = ic 0 + ic ( + + γ 0 ( + d + + γ ([ log( 2 log( 2 + ] d {f([log 2 log( ] f( log( } ( { [ log( ] [ log( ]} f( 2 + log f( 2 + log [ f( + f( ] 2.

4 220 H. Begehr Theorem 2 The Schwar problem for the inhomogeneous Bitsade equation in the unit disc w = f in D, Re w = γ 0, Re w = γ on D, Im w(0 = c 0, Im w (0 = c, for f L (D; C, γ 0, γ C( D; R, c 0, c R is uniquely solvable through w( = ic 0 + i( γ ( + γ 0 ( + d ( + d ( f( + + f( + ( +. (2 Proof Rewriting the problem as the system w = ω in D, Re w = γ 0 on D, Im w(0 = c 0, w = f in D, Re ω = γ on D, Im ω(0 = c, combining its solutions w( = ic 0 + γ 0 ( + d 2 ( ω( + + ω( +, ω( = ic + γ ( + d 2 ( f( + + f( +,

5 Boundary value problems in complex analysis II 22 formula (2 is obtained. Here the relations 2 2 ( =, + = + (, = +, + = + = + (, ( are used. The uniqueness of the solution follows from the unique solvability of the Schwar problem for analytic functions, Theorem 6 Theorem 9 in [3]. It is well known that the Dirichlet problem for the Poisson equation w = f in D, w = γ on D is well posed, i.e. it is solvable for any f L (D; C, γ C( D; C the solution is unique. That the solution is unique is easily seen. Lemma The Dirichlet problem for the Laplace equation w = 0 in D, w = 0 on D is only trivially solvable. Proof From the differential equation w is seen to be analytic. Integrating this quantity w = ϕ + ψ is seen where ϕ ψ are both analytic in D. Without loss of generality ψ(0 = 0 may be assumed. From the boundary condition ϕ = ψ on D follows. This Dirichlet problem is solvable if only if, see [3], Theorem 7, 0 = ψ( d = ψ( d ψ( d = ψ(. This also implies ϕ = 0 on D so that w = 0 in D. As Bitsade [6] has realied such a result is not true for the equation w = 0.

6 222 H. Begehr Lemma 2 The Dirichlet problem for the Bitsade equation w = 0 in D, w = 0 on D has infinitely many linearly independent solutions. Proof Here w is an analytic function in D. Integrating gives w( = ϕ( + ψ( with some analytic functions in D. On the boundary ϕ( + ψ( = 0. As this is an analytic function this relations hold in D too, see [3], Theorem 7. Hence, w( = ( 2 ψ( for arbitrary analytic ψ. In particular w k ( = ( 2 k is a solution of the Dirichlet problem for any k N 0 these solutions are linearly independent over C. Because of this result the Dirichlet problem as formulated above is ill-posed for the inhomogeneous Bitsade equation. With regard to the Dirichlet problem for the Poisson equation the representation formula [3], (5 is improper as is also [3], (5. The middle terms are improper. They can easily be eliminated by applying the Gauss Theorem, see [3]. For the respective term in [3], (5 in the case D = D w ( log 2 d = w ( log 2 d = = [w ( log 2 ] w ( log 2 + w ( follows. Applying the Gauss Theorem again shows w ( = ] [w( = w( d = w( d. Thus inserting these in [3], (5 leads to the representation w( = ( w( + d w ( log 2. (3 The kernel function in the boundary integral is the Poisson kernel, the one in the area integral is called Green function for the unit disc with respect to the Laplace operator.

7 Boundary value problems in complex analysis II 223 Definition The function G(, = (/2G (, with G (, = log 2,, D,, (4 is called Green function of the Laplace operator for the unit disc. Remark The Green function has the following properties. For any fixed D as a function of ( G(, is harmonic in D \ {}, (2 G(, + log is harmonic in D, (3 lim t G(, = 0 for all t D, (4 G(, = G(, for, D,. They can be checked by direct calculations. Green functions exist for other domains than just the unit disc. The existence is related to the solvability of the Dirichlet problem for harmonic functions in the domain. The Riemann mapping theorem can be used to find it e.g. for regular simply connected domains. Having the Green function [3], (5 [3], (5 can be altered as above leading to the Green representation formula, see e.g. []. Green functions exist also in higher dimensional spaces for other strongly elliptic differential operators. For the unit disc the following result is shown. Theorem 3 Any w C 2 (D; C C ( D; C can be represented as w( = ( w( + d w (G (,, (3 where G (, is defined in (4. Formulas [3], (5 [3], (5 are both unsymmetric. Adding both gives some symmetric formula which is for the unit disc w( = 4i 4i + ( w( + d (w ( + w ( log 2 d w ( log 2. (5

8 224 H. Begehr Motivated by the procedure before, the Gauss Theorems are applied in a symmetric way to w ( log 2 = 2 { [w ( log 2 ] + [w ( log 2 ] ] [ ]} + [w( + w( = log 2 [w ( + w 4i (] d + 4i [ w( = 4i + 4i + ] d log 2 [w ( + w (] d [ w( + ] d. Here are two possibilities. At first the second term in (5 can be eliminated giving w( = ( w( + d w ( log 2, i.e. (3. Next the first term in (5 is simplified so that w( = w( d (w ( + w ( log 2 d + w ( log ( ( 2. (6 Here the normal derivative appears in the second term while a new kernel function arises in the area integral. Definition 2 The function N(, = (/2N (, with N (, = log ( ( 2,, D,, (7

9 Boundary value problems in complex analysis II 225 is called Neumann function of the Laplace operator for the unit disc. Remark The Neumann function, sometimes [7] also called Green function of second kind or second Green function, has the properties ( N(, is harmonic in D \ {}, (2 N(, + log is harmonic in D for any D, (3 ν N(, = for D, D, (4 N(, = N(, for, D,. (5 N(, d 2 = 0. They can be checked by direct calculations. The last result may therefore be formulated as follows. Theorem 4 Any w C 2 (D; C C (D; C can be represented as w( = + w( d ν w( log 2 d w (N (,. (6 This formula can also be written as w( = [w( ν N (, ν w(n (, ] d 4i + w (N (,. (6 Theorem 3 immediately provides the solution to the Dirichlet problem. Theorem 5 The Dirichlet problem for the Poisson equation in the unit disc w = f in D, w = γ on D, for f L (D; C γ C( D; C is uniquely given by w( = ( γ( + d f(g (,. (8 This is at once clear from the properties of the Poisson kernel the Green function. As the Dirichlet problem formulated as for the Poisson equation is not uniquely solvable for the Bitsade equation another kind Dirichlet problem is considered which is motivated from decomposing this Bitsade equation in a first order system.

10 226 H. Begehr Theorem 6 The Dirichlet problem for the inhomogeneous Bitsade equation in the unit disc w = f in D, w = γ 0, w = γ on D, for f L (D; C, γ 0, γ C( D; C is solvable if only if for < ( γ0 ( γ ( d + γ ( d The solution then is w( = γ 0 ( d f( = 0 (9 f( = 0. (0 γ ( d + Proof Decomposing the problem into the system composing its solutions w( = ω( = w = ω in D, w = γ 0 on D, ω = f in D, ω = γ on D, γ 0 ( d γ ( d the solvability conditions γ 0 ( d = γ ( d = ω(, f(, ω(, f(, f(. (

11 Boundary value problems in complex analysis II 227 proves ( together with (9 (0. Here ( ( = d = ( ( = ( = are used. This problem can also be considered for the Poisson equation. Theorem 7 The boundary value problem for the Poisson equation in the unit disc w = f in D, w = γ 0, w = γ on D, for f L (D; C, γ 0, γ C( D; C is uniquely solvable if only if = γ 0 ( d + γ ( log( d f( log( (2 d γ ( = f(. (3 The solution then is w( = + γ 0 ( d γ ( log( d f((log 2 log(. (4 Proof The system w = ω, ω = f in D, w = γ 0, ω = γ on D

12 228 H. Begehr is uniquely solvable according to Theorem 0 if only if The solution then is w( = ω( = d γ 0 ( = d γ ( = γ 0 ( d γ ( d ω(, f(. Inserting ω into the first condition gives (2 for ω( = γ ( = < f( ω(, f(. ( ( d ( ( d ξd η with ( ( = log( = log( + = log(. log( d log( d Combining the two integral representations for w ω leads to (3 as ω( = γ ( ( ( d + = < f( ( ( d ξd η

13 Boundary value problems in complex analysis II 229 where ( ( = log 2 = log 2 + Remark In a similar way the problem log 2 d log( d d log( ( = log 2 log(. w = f in D, w = γ 0, w = γ on D with f L (D; C, γ 0, γ C( D; C can be solved. That integral representations may not always be used to solve related boundary value problems as was done in the case of the Dirichlet problem with formula (3, can be seen from (6. If w is a solution to the Poisson equation w = f in D satisfying ν w = γ on D being normalied by w( d = c for proper f γ then on the basis of Theorem 6 it may be presented as w( = c γ( log 2 d + f( log ( ( 2. (5 But this formula although providing always a solution to w = f does not for all γ satisfy the respective boundary behaviour. Such a behaviour is also known from the Cauchy integral. Theorem 8 The Neumann problem for the Poisson equation in the unit disc w = f in D, ν w = γ on D, w( d = c, for f L (D; C, γ C( D; C, c C is solvable if only if γ( d = 2 f(. (6

14 230 H. Begehr The unique solution is then given by (5. Proof As the Neumann function is a fundamental solution to the Laplace operator the boundary integral is a harmonic function, (5 provides a solution to the Poisson equation. For checking the boundary behaviour the first order derivatives have to be considered. They are w ( = d γ( ( ( f( +, so that w ( = ν w( = = + d γ( ( [ f( + ( f( +, ( γ( + d + ( γ( + d 2 [ f( 2 + ] ]. For = this is using the property of the Poisson kernel ν w( = γ( γ( d + 2 f(. Therefore ν w = γ on = if only if condition (6 holds. At last the normaliation condition has to be verified. It follows from = for = = log 2 d = = log( d = log( d = 0. Theorem 9 The Dirichlet-Neumann problem for the inhomogeneous Bitsade equation in the unit disc w = f in D, w = γ 0, ν w = γ on D, w (0 = c,

15 Boundary value problems in complex analysis II 23 for f L (D; C C( D; C, γ 0, γ C( D; C, c C is solvable if only if for D c d γ 0 ( + 2 f( = 0 (7 ( d (γ ( f( ( + f( = 0. (8 ( 2 The solution then is w( = c γ 0 ( d 2 (γ ( f( f( 2 2 ( Proof The problem is equivalent to the system w = ω in D, w = γ 0 on D, ω = f in D, ν ω = γ on D, ω(0 = c. The solvability conditions are d γ 0 ( = the unique solutions d (γ ( f( + w( = log( d. (9 γ 0 ( d ω( f( ( 2 = 0 ω(

16 232 H. Begehr ω( = c (γ( f( log( d according to [3], Theorems 0. From =, log( = f( ( log( d ( = 0, = 2 ( condition (7 follows. Similarly (9 follows from =, 2 log( = log( ( = 2 2. ( Theorem 0 The boundary value problem for the inhomogeneous Bitsade equation in the unit disc w = f in D, w = γ 0, w = γ on D, w (0 = c, is solvable for f L (D; C, γ 0, γ C( D; C, c C if only if for D condition (7 together with d γ ( ( + f( ( 2 = 0 (20 holds. The solution then is uniquely given by w( = c + γ 0 ( d + + f( 2 2 ( 2 γ ( log( d. (2

17 Boundary value problems in complex analysis II 233 The proof is as the last one but [3], Theorem 2 is involved rather than [3], Theorem. Theorem The Neumann problem for the inhomogeneous Bitsade equation in the unit disc w = f in D, ν w = γ 0, ν w = γ on D, w(0 = c 0, w (0 = c is uniquely solvable for f C α (D; C, 0 < α <, γ 0, γ C( D; C, c 0, c C if only if for D c + γ 0 ( d + f( (γ ( f( d The solution then is given as w( = c 0 + c γ 0 ( log( d + (γ ( f(( log( d ( ( ( 2 = 0 (22 (γ ( f(( log( d + f( ( 2 = 0. (23 Proof From applying [3], Theorem w( = c 0 (γ 0 ( ω( log( d ω( = c if only if (γ ( f( log( d d (γ 0 ( ω( ( + f(. (24 ω( (, f( (, ω( ( 2 = 0,

18 234 H. Begehr d (γ ( f( ( + f( ( 2 = 0. Inserting ω into the first condition leads to (22 on the basis of = c < d ω( 2 ( f( = c + = (γ ( f( d ( ( d ξd η (γ ( f( d + log( d d 2 f( ( ω( ( 2 = = ( ω( ( 2 ( 2 ω(d ( 2 f(, ( 2 f( where for = = = < ( 2 ω(d = = ω( d (γ ( f( f( ( 2 ω(d log( ( d ( ( d ξd η d d

19 Boundary value problems in complex analysis II 235 From = + k= k k = = log( d = (γ ( f( log( d (γ ( f( log( d. log( log( d = log( d 2 = + log( d = 0, k=2 k log( log( d k! k log( =0 + = k=2 k k (k k = log( (log( + = log(, log( d = = log( d log d = 0 the relation = ω( log( d ( (γ ( f( d log( + (25

20 236 H. Begehr follows. Similarly from ( = log( = ( log( = ( log( + =, d ( log( ( log( d ( = ( log( + log( 2 = log(, log( = log( = it follows d log( 2 =, ( ( = ( ω( ( = c + f( ( 2 (γ ( f( From (25 (26 the representation (24 follows. ( = d log( +. (26

21 Boundary value problems in complex analysis II 237 Theorem 2 The boundary value problem for the inhomogeneous Bitsade equation in the unit disc w = f in D, w = γ 0, w = γ on D, w(0 = c 0, w (0 = c, for f L (D; C, γ 0, γ C( D; C, c 0, c C is uniquely solvable if only if for = d γ 0 ( ( + γ ( log( d = The solution then is Proof The system f(, (27 ( 2 d γ ( ( + w( = c 0 + c + + ( 2 γ ( f(. f( ( 2 = 0. (28 γ 0 ( log( d d log( + w = ω in D, w = γ 0 on D, w(0 = c 0, ω = f in D, ω = γ on D, ω(0 = c, is uniquely solvable if only if d γ 0 ( ( + ω( ( 2 = 0 (29 d γ ( ( + ω( ( 2 = 0.

22 238 H. Begehr The solution then is w( = c 0 ω( = c γ 0 ( log( d γ ( log( d ω( (, f( (. Inserting the expression for ω into the first condition gives (27 because as in the preceding proof on = = ω( ( 2 γ ( log( d ( 2 f(. Also from ω( ( = c + ( 2 γ ( f( d log( + formula (29 follows. Boundary value problems as in Theorem 2 can also be solved for the Poisson equation. One case is considered in the next theorem. Theorem 3 The boundary value problem for the Poisson equation in the unit disc w = f in D, w = γ 0, w = γ on D, w(0 = c 0, w (0 = c are uniquely solvable for f L (D; C, γ 0, γ C( D; C, c 0, c C if only if d γ 0 ( ( = ( f( + (30

23 Boundary value problems in complex analysis II 239 d γ ( ( + f( ( 2 = 0. (3 The solution then is w( = c 0 + c + + ( γ ( γ 0 ( log( d d log( + f(( + log 2 log( log 2. Proof The problem is equivalent to the system It is solvable if only if w = ω in D, w = γ 0 on D, w(0 = c 0, ω = f in D, ω = γ on D, ω(0 = c. d γ 0 ( ( + ω( ( = 0 2 (32 d γ ( ( + ddη f( ( 2 = 0 the solutions are according to [3], Theorem 2 w( = c 0 ω( = c γ 0 ( log( d γ ( log( d ω( (, f( (. For the first problem [3], Theorem 2 is applied to w the formulas then complex conjugated. For (30

24 240 H. Begehr ω( = ( 2 = [ [ ω( ω( d f( ] f( has to be evaluated. For = ω( d = = + = = < ω( d γ ( f( f( log( dd ξd η d d so that ω( ( = 2 ( f( +. For ( = < ω( = c γ ( f( log( d d ξd η

25 Boundary value problems in complex analysis II 24 needs some modification. From (log( = = + + = + log( (log( d (log( d ( log( log( from which also log( = 0 follows, = log 2 = log 2 + log( d = log 2 log( log 2 d d log( ( from what = = + (log 2 log( ( +

26 242 H. Begehr is seen, < ω( ( = c + = log 2 ( γ ( d log( + f(( + log 2 log( log 2 follows. Remark Instead of this constructive way the proof can be given by verification. From (32 w ( = d γ 0 ( ( f(. This obviously coincides with γ 0 on D if only if (3 is satisfied. Similarly from w ( = c γ ( log( d ( f(, w ( = d γ ( f( ( 2 it is seen that w coinsides with γ on D if only if (3 holds. The other two conditions are obviously satisfied. 2 The inhomogeneous polyanalytic equation As second order equations of special type were treated in the preceding section model equations of third, forth, fifth etc. order can be investigated. From the material presented it is clear how to proceed what kind of boundary conditions can be posed. However, there is a variety of boundary conditions possible. All kind of combinations of the three kinds, Schwar, Dirichlet, Neumann conditions can be posed. And there are even others e.g. boundary conditions of mixed type which are not investigated here. As simple examples the Schwar problem will be studied for the inhomogeneous

27 Boundary value problems in complex analysis II 243 polyanalytic equation. Another possibility is the Neumann problem for the inhomogeneous polyharmonic equation, see [4, 5], the Dirichlet problem, see [2]. Lemma 3 For <, < k N 0 2 ( k + ( + k+ = ( k+ ( + k+ k ( + k. (33 Proof The function w( = i( + k+ /(k + satisfies the Schwar condition w ( = i( + k in D, Re w( = 0 on D, Im w(0 = ( k+ k + (+k+, so that according to [3], (33 w( = i ( k+ k + (+ k+ i 2 ( + + ( + k. This is (33. Corollary For < k N 0 ( 2 ( + k = 0 (34 2 ( + + ( + k = ( k+ k + ( + k+. (35 Proof (34 (35 are particular cases of (33 for = 0 =, respectively. Theorem 4 The Schwar problem for the inhomogeneous polyanalytic equation in the unit disc n w = f in D, Re ν w = γ ν on D, Im ν w(0 = 0, 0 ν n, is uniquely solvable for f L (D; C, γ ν C( D; R, c ν R, 0 ν n.

28 244 H. Begehr The solution is n n c ν w( = i ν! ( + ( ν ν + ν! ν=0 ν=0 ( + ( n f( + 2(n! + f( + γ ν ( + ( + ν d ( + n. (36 Proof For n = formula (36 is just [3], (33. Assuming it holds for n rather than for n the Schwar problem is rewritten as the system n w = ω in D, Re ν w = γ ν on D, Im ν w(0 = c ν, 0 ν n 2, having the solution ω = f in D, Re ω = γ n on D, Im ω(0 = c n, n 2 n 2 c ν w( = i ν! ( + ( ν ν + ν! ν=0 ν=0 ( + ( n ω( + 2(n 2! + ω( ω( = ic n Using (35 ( n 2(n 2! 2 + γ n ( + d ( f( + + f( +. γ ν ( + ( + ν d ( + n 2, ( ω( + + ω( + ( + n 2 c n = i (n! ( + n + ( n (n 2! ( = γ n ( ( + n 2 d

29 Boundary value problems in complex analysis II ( + + ( n 2(n 2! < + f( ( + n 2 d ξd η + ( n 2(n 2! < f( 2 follows. Because ( + = ( + n 2 d ξd η ( 2 ( 2 = + ( 2 ( = 4 ( ( = 2 + ( ( ( = + = + ( 2 (

30 246 H. Begehr similarly ( 2 ( 2 = ( 2 ( 2 = 4 ( + ( ( = 2 + = + = + ( + + ( 2 2 ( + + applying (33, (34, (35 2 = + 2 ( ( = + [ n ( + n ( n n 2 = + 2 ( ( + n 2 + ( + n + ( n n = + n ( + n, ( ( + n 2 ( + n 2 ( + n ] ( + n 2

31 Boundary value problems in complex analysis II 247 = + [ n ( + n ( n n ( + n + ( n n ( + n ] then ( n 2(n 2! c n = i (n! ( + n + ( n (n! + ( n 2(n! = + n ( + n, ( ω( + + ω( + ( + n 2 γ n ( + ( + n d ( f( + + f( + ( + n. This proves formula (36. Theorem 5 The Dirichlet problem for the inhomogeneous polyanalytic equation in the unit disc n w = f in D, ν w = γ ν on D, 0 ν n, is uniquely solvable for f L (D; C, γ ν C( D; C, 0 ν n, if only if for 0 ν n n λ=ν + ( n ν ( λ ν γ λ( ( λ ν (λ ν! f( ( n ν (n ν! d = 0. (37 The solution then is w( = n ν=0 ( ν + ( n γ ν ( ( ν ν! d f( ( n (n!. (38

32 248 H. Begehr Proof For n = condition (37 coinsides with [3], (34 (38 is [3], (35. Assuming Theorem 27 is proved for n rather than for n the problem is decomposed into the system n w = ω in D, ν w = γ ν on D, 0 ν n 2, ω = f in D, ω = γ n on D, with the solvability conditions (37 for 0 ν n 2 ω instead of f together with d γ n ( f( = 0 the solutions (38 for n instead of n ω instead of f where ω( = γ n ( d f(. Then for 0 ν n 2 ω( ( n 2 ν (n 2 ν! where = ψ ν (, = = = = γ n ( ψ ν (, d < ( n 2 ν (n 2 ν!( ( n ν (n ν!( ( n ν (n ν!(. The last equality holds because ( n ν d = ( ( = ( n 2 ν d = 0. f( ψ ν (, d ξd η, ( n ν d (n ν!( ( n ν d ( (

33 Boundary value problems in complex analysis II 249 Thus for 0 ν n 2 n 2 λ=ν + ( n ν = n λ=ν + ( n ν ( λ ν γ λ( ( λ ν (λ ν! This is (37. For showing (38 similarly ( n 2 ω( (n 2!( with = ψ n (, = = = ( n 2 (n 2!( (n!( + (n!( ω( ( n 2 ν (n 2 ν! d ( λ ν γ λ( ( λ ν (λ ν! f( ( n ν (n ν! γ n ( ψ n (, d ( n 2 (n 2!( Hence, w( is equal to n 2 ν=0 ( ν γ ν ( ( ν ν! < d = 0. f( ψ n (, d ξd η ( = ( n (n!( ( ( n ( ( n d = ( n (n!( ( n ( d = ( n (n!(. d + ( n ω( ( n 2 (n 2!

34 250 H. Begehr = n ν=0 ( ν i.e. (38 is valid. γ ν ( ( ν ν! d + ( n f( ( n (n!, Acknowledgement The author is very grateful for the hospitality of the Mathematical Department of the Simón Bolívar University. In particular his host, Prof. Dr. Carmen Judith Vanegas has made his visit very interesting enjoyable. References [] Begehr, H.: Complex analytic methods for partial differential equations. An introductory text. World Scientific, Singapore, 994. [2] Begehr, H.: Combined integral representations. Advances in Analysis. Proc. 4th Intern. ISAAC Congress, Toronto, 2003, eds. H. Begehr et al. World Sci., Singapore, 2005, [3] Begehr, H.: Boundary value problem in complex analysis, I. Bol. Asoc. Mat. Veneolana, 2 (2005, [4] Begehr, H., Vanegas, C. J.: Iterated Neumann problem for higher order Poisson equation. Preprint, FU Berlin, 2003, Math. Nachr., to appear. [5] Begehr, H., Vanegas, C. J.: Neumann problem in complex analysis. Proc.. Intern. Conf. Finite, Infinite Dimensional Complex Analysis, Appl., Chiang Mai, Tahil, 2003, eds. P. Niamsup, A. Kananthai, [6] Bitsade, A. V.: About the uniqueness of the Dirichlet problem for elliptic partial differential equations. Uspekhi Mat. Nauk 3 (948, 6 (28, 2-22 (Russian. [7] Haack, W., Wendl, W.: Lectures on partial Pfaffian differential equations. Pergamon Press, Oxford; Birkhäuserr, Basel, 969 (German. Heinrich Begehr I. Math. Inst., FU Berlin Arnimallee Berlin, Germany begehr@math.fu-berlin.de