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On a certain mixed boundary value problem for iterated Helmholtz equation

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ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE X X (1978)

Je r z y G0L4B (Krakôw)

On a certain m ixed boundary value problem for iterated Helmholtz equation

1. In the present paper we shall give the solution of the equation

(H) (A —C2)2%{xx, x z, x8) = 0

in the domain E f — {(a?,, x z, x8): x x > 0, x z > 0, x3 > 0} satisfying the boundary data

(a) = /i(« 2 j ж3),

(1) (Ъ) = / 2 (*^i ? #3) »

(c) [few(Z)+Diei^(X) ] | a.3=0 = /3(а?1,ж2), (a) ^ ( Х ) Ц„ 0 = 9i(v2 , Xz),

(2) (b) A D Xiu { X ) \ a^ 0 = д г { х х, х 8),

(c) \ h A u ( X ) - \ - ADX3u ( X ) ] la;3=o ==

where b is a negative constant, X — (xx, x z, x8), g{ (i = 1 , 2 , 3 ) being given functions.

In the sequel we shall cal this problem the (R-N-M) problem. To the construction of the solution of the (R-Ж-М) problem we shall use the convenient Green function.

2. Let X x — (xx, x z, x3) denote an arbitrary point belonging to E+, Y = (yx, y z, y3) an arbitrary point belonging to E z. Let X z = ( —x ly

^a)) X8 = ( x Xi x z, Я/3), X4 = (xXj x ZJ a?3), X s = (a?x, x zi a?3), X8 = ( x Xj x Zj x8), X-j — ( x X) x Zj x8) and X8 = {xx^ x Zt я^з)»

Let

(3) \ X j Y \ = r iy for y{ = 0 (j = 1, = 1 ,2 ,3 ) and

(4)

f) = (yx- x 1J)2 + (yz- x 2j)2 + (x8 + y8-{-v)2t

R% = ri for yi = 0 (j = 5, ..., 8, i = 1, 2, 3), Q) = (У 1 -% )2 + (У2- % ) 2 + <2 (j = 6 , . . . , 8 ) ,

(2)

324 J. Gol^b

where x rj, x2j denote the first and second coordinate of the point X i , v ^ 0, xz + Vz ^ < < °o. Й is easy to verify th at

B u = В 21 > - ^ 3 1 = P 4 1 ) B 51 —^ 6 i > P 71 ~ -®81 »

( 5 ) B l 2 = P }2 > - B22 - B3 2 ? B52 B S 2 f ^ 6 2 = P72 7

-®13 = В&31 P3 O)CO

= -®73> P 33 - ® 6 3 ) & 4* W II - ^ 5 3 and

(6) “®5 1 ^ 6 1 1 i?7l = - ^ 8 1 » P 52 - ^ 8 2 > - ^ 6 2 — P 72 ?

Qj = ^ 3 for t = II for £ — a?3+ 2/3 (i == 5 ,.

Let

(7) 7 (гу) = e~Cri.

Let

(8) G(X, Y) =.G 1(X > Y )-G ,(X , Y )-G 3(X , T)+G<(X, Y), where

Y) = i [Y(r*+1) + r ( r ,_ t )] + 2 / «toF ( W * > (* = 0, 1 ,2 ,3 ) .

n 0

Let W = {(X, Y): 0 < a < x{ < &, a < < 6, Ï = 1, 2,3}, where a, b are arbitrary positive numbers.

Let

oo oo

P ( X , Y) = J ehvV(f3.)dv = f ^ ^ з - г /з >V(ef)dt,

0 x3+vs

oo

Z ^ ( X , Y) = / (7 (F ,))* ,

0 1 2 3

g, r being non-negative integers for which 0 < p + q -f r < 4, j — 5, 8.

We shall prove the following

Le m m a 1. The integrals P { X , Y) Ppqr{X , Y) are uniformly convergent in every set W.

P ro o f. The majorant of the integrals P { X , Y) and I 3pqr(X , Y) is OO

the integral K 0 f ehvdv, where K 0 is a positive constant, d

From Lemma 1 follows

Le m m a 2. There exist the integrals F ( X , Y), I 3pqr(X , Y) and Ppqr{X , Y)

= D p q rP { X , Y) and the functions P { X , Y) are of the class Ü4 for X1 X2X3

(X , Y) e T7.

Le m m a 3. The functions G{ {i = 1, 2, 3, 4) satisfy equation (H) with respect to the point Y.

(3)

P ro o f. We shall give the proof for Gx. The proof for Gt (i = 2 , 3, 4) is similar. Applying Lemma 2 and the conditions ( d F — C'2)2F(ri ) = 0 (j = 1, 8), ( d F —О2)2 V (rj) = 0 (j = 5 , . . . , 8) we get

( A r - C ^ G ^ X , Y) = j M r - C 2)![F(r1)+F(ra)] +

+2 J eftv(dF-(72)2F(f8)<fo = 0 . 0

Ltcmma 4. The function G defined by formula (8) satisfies the homo­

geneous boundary conditions

0 ( Х ,Г ) 1^0 = 0, A ¥ G ( X , Y ) \ yi=0 = 0,

2 > *G (X ,Y )|*_0 = 0 ,

P T )iI/2=0 = 0,

[Ш (Х , Y )+ D y3<?(X, Г )]|У5. 0 = 0, Y [hAYG( X, Y) + AyDV3G(X, Г )]|Уз_0 = 0 .

P ro o f. Conditions (a) follows from the equality Gx{X, Y) = GZ( X, Y), G,(X, Y) = Gt(X, Y) for y x = 0 and A Y V{r}) = A r V(rj+1) (j = 1, 3, 5, 7), AyVirj) — A r V{rj+i) (j = 5, 7) for y x = 0. Using for y z = 0 the formulas

O y fik+l{X , Y) = в* < 7 яЛ | [Г (В „ +1,2)( i t№ !r ' + ^ (-B 8-*,sH-B8-;(.2) - 1] +

oo

+ 2 / e*'F(Æ,_M )(Æ,_fc2) - J*> (fc = 0 , 1 , 2 , 3) where = 1 for Jc = 0, 2 and ел = —1 for & = 1 ,3 and

^ r Y ( r , ) + J F F (r,+1) ] = 0 (j = 2 ,6 ),

^ 2[ ^ ^ ) + ^ г ^ +3)3 = 0 (j = 1 , 6 ) ,

^ y,[2lr F(r5)+ J F F(r8)] =0,

we get DV2G{X, Г )|У2«0 = 0 and A TDy G{X, Y) |У2=0 = 0.

We shall verify conditions (y) for the function Gx. Por the function G{ (i = 2 , 3 , 4 ) the proof is similar.

Using for 2/3 = 0 the formulas

DV3l V( rx) + V(r8)] = 0 , I>V3l A r V(rx) + A 7 V(rü)] = 0 2>WsI*(X, Y) = -Л 1 8(Х , Y ) - F ( r e),

D„3d F I 8(X , Y) = -7 ^ fI 8(X , Y) — d F F (r8) and

(4)

326 J. Golî*b

we get

[hGx( X , Y ) + B yf i A X , Ш ъ - о = 0 and

[ h A Y G x( X , Y ) + â r D vf i x{ X , Г )]|Уз=0 = 0.

From Lemmas 3, 4 follows

Th e o r e m 1. The fu n c tio n G d efin ed by fo r m u la (8) is the Green fu n c tio n fo r the p ro b lem (K-N-M).

3. In the sequel we shall prove th a t the function

3

(9) u ( X ) = 2 } [ u k(X) + vk(X)],

where

«.(X ) = ^ / / Л ( Г , ) [ 2 С 27)1<1в ( Х , Y ) - D , l A T e ( X , Г)]|„1. 0а Г 1,

s x

(9a)

Щ( Х ) = A J j M T {) { A T e ( X 9 Y ) — 2 C2G ( X , (i = 2 ,3 ) and

Si

d, ( X ) = - A j $ g x( 7 j D t l G ( X , Г ) | ^ . » о & N1

(9b) «1

v t ( X ) = —A J f g i i Y J G i X , Y ) \ ^ d Y , (i = 2 , 3 ) ,

8,

where

г

S x = { ^ i = (у2,Уз): У а > 0, y3> 0}, d Y x = dy2dy3, S 2 = {F2 = (Ух, Уз): Ух > 9 , у3 > 0 } , d 7 2 = d y xd y z ,

Я3 = {Fs = (ух, у 2): У х > 0 , у 2 > 0 } , d T z = d y xd y 2

and A == Л(87сО)-1 is the solution for the problem (R-N-M).

Let

and

1 for j = 1, 4,

« 3 = — « 2 =

—1 for j = 2, 3 1

1 for j = 3 , ^ 1 for j = 5, 8 ,

—1 for j = 4, —1 for j = 6, 7.

(5)

Using the Green function and formulas (9) we get

4

ut (X) = 2ACx1j j f l ( 7 i) { ( h ) - ' 2 v ( R 2i_l,l)[Cf(Rll_l,lr I +

Si J=1

+ 2 0 (E 2i_1;1)- 2 + 2(JB2i_1,i)-3] +

oo 4

+ 2 / e’" 2 V ( B 1,_ l,l)l<?(S2t_IJ - l + 2C (R2i_hlr 1 +

о / - 3

+ 2 < 5 y -Mr 3]dt>}<Ü\,

4

(9a) «.(X ) = 2 X / / /.( F .j j f A ) - 1 ^ £'F(K i+!,2)[C! +

«2

oo 4

+ 2 0 (Rj+2,ï)~l ] + 2 / +

o i= 3

+2C(Æy_1,!) '1]dt))dFï ,

4

«.(X ) = 2 + / / / , < У,) {(Л)-1 £ 4 Vdi y,) [0s + 2С(Д„)-‘] +

«s /-1

oo В

+2 / в*4'-**' ^7 ^ [ С 2 +- 2<7(ei) - ‘] <ï«} d F,

*3 J-s

and

«MX) = - 2 X 0 ® ,// »,( 7 .) [(А)-1 £ F(Bii_1,1)(fiJJ_1, , r 1 +

i - i

oo 4

+ 2 / e‘ ' ^ F ( J J M_ i.1)(Æ y _ 1, , ) - , *>]<»F„

0 j - 3 4

(9b) «MX) = - 2 X / / 02(F 2) [(A)-1 ^ F (i W +

«2 /-1

oo 4

+ 2 / ek”2 ’»)(F (« I/ - 1.î) H d ? ’

0 2=-3

4

«.(X) = —

2

X J j g A Ÿ , ) [ { h ) - 1 2 4 V ( B » ) + J-l

oo 4

+ 2 / ^ 7 , { V( e i ) d t ] d T, . 7 — 3

(6)

328 J. G o ^ b

Let

1 "(Х ) = J J f i ( r l)x,{Kit)- nr ( B jl)d Y t , Si

I ^ r ( X ) = / / M Ÿ t) D \xi Y ( R ii)(Bii ) - nV Y i (j = 1, S),

S . 1 2 3

oo

</*(Х) = / [ / eh’ (Bji) - 'V ( Ë ]()d v\d Ÿ (,

S{ 0

OO

J & m = e*’ (ÆJ4)-*F(Æ j()tft>]dri (j = 5 , . . . , 8)

for i = w = 1, 2, 3 and let

= { ( « i ? г » 3 ) : 0 < « j < ж * < a 2 , < x m < 6 2 }

(m, i _ = 1 , 2 , 3 ; m y= i ), where %, л2, &i> b2 are positive constans.

Lem m a 5. I f the functions fi (i = 1 ,2 ,3 ) are bounded and measurable convenably in the sets S{, then the integrals I jt(X), I 3plqr(X), J jl(X) and J }pqr(X) are uniformly convergent convenably in the sets Z {.

P ro o f. We shall give the proof for the integrals I n {X) and J 13(X).

The proof in otliers cases is similar.

For sufficiently great number i?0 > 0

IIOTjI2 < (-Bxl)2 < 4 |0 7 a|2 for every X e Z x and jOT^I > jR0.

Let

K R = { Y x = (y2, yf): |0 P aK B,*,y%> 0}, M x = snp |/ х( FJI»

«i ;■ ...

H " ( X , Pj) = X i V i ^ U R n ) - » , H " r( X, f p = D H U{X, Fj)

z r 2 3

and let ’

^ - f f 1ЛСГг)|Я “ ( ^ , КВо.

/ / i/« ( ÿ i) if f “ ( x ,

Sinèe the function H l l ( X , Y x) is the analytic function of the point X for Y x e K Rq , thus the integral I \ l {X) is the analytic function of the point X e Z x.

Lq the sequel we shall use the inequality

(10) e Crj <-■(Сгх) a for a e ( 0 , e ) .

Applying the polar coordinates in the majorant of the integral I lf { X ) we

(7)

get inequality

oo

П1( Х ) ^ 2 п+2п М 1(С)-2а2 j Q-n~l dQ<e

R 0

for every X e Z 1 if R% > 2n+2n Ml a.1(e)~l {C)~2.

The function II]}qr{X, Yf) is the finite sum of the functions 4 { ЯпГ Чу* - ^ ) Чу, - ъ) \

where p, ô2, <53 are non-negative integers, y e N and y > n + <5a + <53. Simi­

larly as for the integral I й {X) we get the majorant for the integrals Ipgr{X), J n (X), J LPlr(X) of the form

00 —n—a

2л К М Х f K + es) 2 e d g , 0

^ { С ) ~ аМ х j ehv[§ {б' + Ъ\) 2 Qd6]dv,

о 0

where В — B(p) > 0 and К being a positive constant.

From Lemma 6 follows

Th e o r e m 2. I f the functions f {, gt (i = 1 , 2 , 3 ) are bounded and measurable convenably in the sets Sit then there exist the derivatives of the integrals Щ{Х), v^X.) and

Dxp y 3 ^ ( X ) = A pxa,1 Х2Х3Y )

i f î ( X y F)] \yi~odYi,y Г

'r P ~ Q ~ r 11 * * * i

X 1 X2 X3 JJ Sx X i £Cn *ïo 1 2 3 - *

(i = 2 , 3 ) , 1 2 ‘"З

Dx W Vi(X) = - A f j gi( Y t)D w O( X, Г ) (* = 1 , 2 )

1 2 3

Moreover, the functions щ , vi are of class O4 convenably in the sets Z i and the function defined by formula (9) is of class C4, in the set Z = Z 1r\Z2n Z 3.

4. Xow we shall give the synthesis of the problem (B-U-M). By Theorem 1 and Theorem 2 we get

(11) ( A - C 2)2u( X) = 0.

_

^ow we shall prove th a t the function и defined by formula (9) satisfies the boimdary conditions (1).

(8)

330 J. G oiçb

Let

F t ( X ) = f J x i V { B li) (Mlf)~3d Y { (* = 1 ,2 ,3 ) .

■®2

Le m m a 6. F {(X)-*2iz as Х-+Щ , where X° is an arbitrary point of the set S { .

P ro o f. We shall give the proof only for i 1; for i = 2, 3 the proof is the same. By (3) and (7) we get

.F,(X> = J”J ®i [®î + (y2 — ®s)2 + ( y3 ®s)S]_SBexP { —C* [®i +

Applying in the integral F x{X) the change of variables (12a) y z — odz — Xj^QCOBPf y8 — #з = sOxQSiiKp

(0 < g < 00 > 0^ 99^ 2тс) ,

(12b) l = f l 2 (1 < v < 00),

OO

we get F 1( X ) = 2 n f e~CXlV(v ) ~2d v - + 2 n as a?1->0+. Let 1

Ii(Y<)

f i i Y i ) for r f e 8 if

0 for Y { e E 2\ 8 { ,

Ut (Bu ) = 7 ( ^ ) ( Д И) - 3,

X((X) = A l j j M r i )i,i U , ( B u ) d Y i (i = 1 , 2 , 3 ) ,

■®2 where Aj = (27t)-1.

Lemma 7. I f the fu n c tio n s f i (i = 1 , 2 , 3 ) are bounded a n d m easu rable eon venably i n the sets 8 { a n d co n tin u o u s a t the p o in ts X®, then K i ( X ) - > f i { X <l) as

P ro o f. We shall give the proof only for i —1; for i = 2, 3 the proof is the same. Let

d i X i T , ) = £ ( ? , ) - / , ( * ! ) and

Hence

L ( X ) T ^ U ^ d Y , .

•S?2

K ^ X ) = A 1/ 1(X?)X

By Lemma 6 we get A1/ 1(X®)X1(X)->/1(XS) as

(9)

Let B x, B 2 denote the circles with the center X x, X x and radii <5,

\ <5. By continuity of the function f x a t the point X° we have Л V f . e B ^ m X l , F x)| < | e .

e>0 <5>0

Let

£,(X ) = A l f f d ( X : , T l)xl U3(Bll) d Y l and Di

B2{X) = A x f f d ( X x, Y l)x1U3(Bxx) d Y 1.

E2\Dl

For the integrals L x( X ), L 2{X) holds the estimations

\Lx(X)\ < A x\ e f f x x U3(Bxx) d T x < \ s for x x< àx{e),

e2

\L2(X)\ < М Л-iz)-1 f f x x U3(Bxx)dFx лв lI°l t X xl < i d .

e2\ d2

Applying transformation (12a) we get

00

|£ !( Х ) | <2Ж1 f e x p [-C *1(e* + l),/8] ( l + s V 5,V f > < i i ! S1

for |XJ, X x\ < \ d and x x < «^(e), where sx = d/2xx. Finally we get

\L(X)\ < e for |X J , X x\ < I <5 and x x< ôx(e).

Let

Mt (X) = / / f ({7t)x 1 Un{Ru )dTit N, (X) = / / UARuldT, ,

Si Si

where Un{RH) = V( R l{){Ru)-" for i = 1, 2, 3; n = 1, 2.

Lemma 8. I / Йе functions f {, gt (i = 1 , 2 , 3 ) are bounded and meas­

urable convenably in the sets 8 , then

M i ( X ) ^ 0 and Щ Х ) - > 0 as £ -+ X \.

P ro o f. We shall prove the assertion for M x(X). For remaing M {(X) the proof is similar. Applying transformations (12a), (12b) by (10) we get l Mx(X)\^.2TzMx{C)~a(xx)3~n~ad v < e for x x< dx(e) and 2 —n < a < 3 —n.

Let

B ki(X) = f f fi( Yi)Xi Un(Bki)dF{ {n = 1 , 2 , 3 )

and г

Qki(X) = / / g A T j x t U ^ d T , , Si

7 — Roczniki PTM Prace Mat. XX.2

(10)

332 J. Gol^b where

3, 4, 7, 8 for г = 1 2, 3, 5, 8 for i = 2 2, 4, 5, 7 for i = 3

Lemma 9. I f the functions f {, g{ (i = 1 , 2 , 3 ) satisfy the conditions of Lemma 8 and x\ > 0, then

P ki{X)->0 and Qki(X)-+ 0 as X->X".

P ro o f. We shall give the proof for the integral P31 (X). The proof in others cases is similar.

By (3) and (10) get

|P31(X)| < / * , H + (ÿa + *2)2+(ÿa-® a)2]<“" - “)S< m . Since (y2 + x z)2> U xl)2 + yl for ^2 > i®27 thus

00 oo

|P M( X ) K 3I1( C r a^ l j Ay, / [ i ( 4 V + yi + w*](- n-°v’-d w < e

0 - x 3

as x x < Ôx(e) and a > 2 — n, where w = y8 — x3.

Let

A„ { X) = / / М Т , ) я , и п(Вы)Л7< (n = 1 , 2 , 3 ) and St­

B M(X) = gi ( 7 t)xi V1(Bu ) d 7 t, Sf

where к — 5 ,6 for i = 1 ; fe = 6, 7 for г = 2 ; fc = 3, 6 for i — 3.

Lemma 10. I f the assumptions of Lemma 9 are satisfied, then A ki(X)^-0 and B ki(X)->0 as X->X? (г = 1 , 2 , 3 ) .

P ro o f. We shall give the proof for the integral A^ ( X) . The proof in others cases is similar.

By (3) and (10) we have OO

IAb(X)I < [(ÿ1 + æ1)2 + (ÿa+ ^ ) 2 + ^ ] <‘ “' ”,/2# i # a , 0

where M s = s u p |/3(F 3)|. Since ym + ccm > p m for xm > {pm = £ < + s3

-fym, m = 1 ,2) by th e transformation p x = pcosç?, ^>2 = g sin<p we get

OO

|Хзз(Х)| ^ ■|■7Г-M3(C,) ax8 J” q n a~*~ldQ—>0 as x8—^0"**,

«1

where a > 2 — n and qx = H (^i)2 + (^г)2]1/2*

(11)

Let

-4*(X) = * , / / / <( r i) P „ ( 5 (4)d F < (n — 1 ,2 ,3 ) ,

S {

and

ВН(Х) - Xi J f Уг(Тг) 1Тг(Ён )с1Т{, where j — 5, 6, 7, 8; i — 1, 2, 3.

Lemma 11. I f the assumptions of Lemma 9 are satisfied, then Aj{(X)-> 0 and Bji(X)-> 0 as X->X°{.

The proof of Lemma 11 is similar to the proof of Lemma 10.

By Lemmas 7-11 for i = 1 we get

Theorem 3. I f the assumptions of Lemma 5 are satisfied and the function f x, is continuous at the point , moreover, x\ > 0, then the function

a defined by formula (9) satisfies the boundary condition (la).

Tn the sequel we shall nse the formulas for x 2 — 0

D ^ h ( X ) = A J j M Y J D ^ V W D ^ G H X , Y ) - A r D ^ e ( X , Y ) ] \ , 1. , d T l ,

«1

D ^ X ) = - 2 A C j f M Ÿ 2 ) l D X2'3r < H X , Y ) - 2 C 1D lt2e ( X , Y ) ] \ yt_ ad ? t , S3

D ^ X ) = - A J j g t d j D ^ G i X , T) s,

D4 v3(X) = - 2 A C J f g , ( 7 , ) D H0 ( X , Y ) ] ^ , 0d Y 3

«3

and the formulas

4

D4 ih (X) = 2ACx2j j f 2( T2){(hr' 2 s i V ( R j + v )[C2(Rj+2,2)- 1 +

s 2 J=1

+ 2 0 ( ^ .+2j2)-2 + 2(.Ki+2f2) - 3] +

00 4

+ 2 / «‘, 2 'ч ^ ( « а - , , . ) [ в ’ (А )- 1, . Г 1 +

0 j = 3

+ 2C(R2j_1i2)~2 + 2(R2j_12)~3]dv}dT2t

4

D X2V2( X) — 2 A C x 2J J ffz( -^2) j^(^) 1^ e2 ^(-^4-2,2) (Щ+2,2) 1

s2 j=l

со 4

+ 2 J е*’ У ^ 7 (Д„_1>5)(Л!,_1,5Г1Ж>]<*Г,.

0 J = 3

By the last formulas, by Theorem 1 qnd by Lemmas 7-11 for i — 2 we get

(12)

334 J. Gol^b

Theorem 4. I f the assumptions of Lemma 5 are satisfied and the function f 2 is continuous at the point X 2, moreover, x\ > 0, then the function и defined by formula (9) satisfies the boundary condition (lb).

By Theorem 2 we can change the differentiation and integration in formulas (9a), (9b) and consequently we get for x3 — 0

В Хзщ( Х ) + к и ^ Х ) = A f f / l ( Y l)I)Vl{2G2[hG(X, Y ) + D xf i ( X , Y ) ] - - [ h A YG.(X, Y ) + D XiA r G(X, Y ) ] } ^ ^ , B X3u, (X) + hu2(X) = A f ( f 2( T 2){AY [DX3G(X, Y) + hG(X, Y ) ] -

S2

- 2 C2[Dxf i ( X , Y) + W ( X , Y ) ] } ^ . , , ^ , DX3vl (X) + ht,l (X) = - A f f g t i f j D ^ h G l X , Y ) + O xf i ( X , Г )]|У1.0<гГ,,

51

D X3M X ) + hv2( X) = —A J J g 3( F 3) [ h G ( X , Y ) + B xf i ( X , Л ] l„2,o<*72 52

and

4

-D ^ 3m + to3(X) = - 2 A C ^ ( h r ' f f gs ( Y 3) s l U ^ R ^ d F , ,

S 3‘ 1

4

В Хзи3(Х) + Ы 3(Х) = 2AC%(h)-1J f f 3( Y3) ^ s i [ C 2n i (Rn ) +

s 3 i=i

+ 2CU2(Bj3) + 2Us(Bj3)]dY3.

By the last formulas, by Theorem 1 and by Lemmas 7-11 for i = 3 we get

Th e o r e m 5. I f the assumptions of Lemma 5 are satisfied and the function /3 is continuous at the point Y®, moreover, x\ > 0, then the function и defined by formula (9) satisfies the boundary condition (le).

We shall prove th a t the function и satisfies the boundary conditions (2). From Theorem 2 and formulas (9a), (9b) follows th a t for x x = 0

Au{(X) = А С * / / / < ( Г {) в ( Х , Х)\Щ, 0Л¥< (i = 2, 3), Si

Л«1,.(Х )= - A J J gi(Ts) Ax G( X, Y) '( i = 2 , 3 ) ,

Si 4

AUl(X) = - 2 A C sx 1f j f 1( 7 t) [ ( h r ' ] ? U 1(R2!_1.1) +

Si 1

oo 4

+ 2 /

eh°]? UARu-iX^Ÿi,

0 / = 3

(13)

Av,(X) = - 2 A C x 1J j g 1( Y i) { [ ( h r ' J Z l C ’ U A B v - i . , ) -

j = 1

—2(7Z72(-R2j-i,i) 2 U3(R2j_iti)] +

со 4

0 J = 3

- З г М Д у - м ) ] * } й Г а.

By the last formulas, by Theorem 1 and by Lemmas 7-11 for i = 1 we get

Th e o k e m 6. I f the assumptions of Lemma 5 are satisfied and the function is continuous at the point X \ , moreover, > 0, then the function и satisfies the boundary condition (2a).

In the sequel we shall use the formulas for x 2 = 0 В ^ А щ ( Х ) = -~ACi f j f l ( Y 1)DHDX2e ( X , Y ) \ l,l, 0d T 1,

«1

1>,гАщ (Х) = ^ C4/ / /3(F ,)D l2e ( X , Т)|„з„0<гг3,

«3

= - a ^ g l ( 7 1) DVlD4 Ay G( X , Y)\n . ad Y 2,

^ ^ ( X ) = - A f f f f ^ r j D ^ A y e i X , T)|„3. 0dF3 S3

and the formulas

4

= -2 A C 5oo2 J J /2( F 2) [(^Г1^ , 417г( ^ +2>2) -

со 4

- 2 / e,"’2 ’^ P 12)-1,2) * ] d F ! ,

0 j = i x

4

^ Л 1)2(Х) = - 202/ / f f2( F2){(A)-1 2 '4 [C,2P l №.+2,2) -

^2 i = 1

2C U 2 ( R j + 2 , 2) — 2 ? 7з ( - Кл - 2 , 2 ) ] +

со 4

+ 2 ' / e4’ 2 1 •)![C2Еч{B2S- l,2) - 2 C U 2(E2j_,,2) -

0 j = 3

- 2P a(Æ2J_1>s)]<fo}dFs.

By the last formulas, by Theorem 1 and by Lemmas 7-11 for i = 2 we get

(14)

336 J. G o H b

Theorem 7. I f the assumptions of Lemma 5 are satisfied and the function g2 is continuous at the point X®, moreover, x\ > 0, then the function

и satisfies the boundary condition (2b).

By Theorem 2 we can change the differentiation and integration in formulas (9a), (9b) and consequently we get for xz = 0

h d u ^ + D ^ A u ^ X ) = A f f - A r )[hAy e ( X , Y) +

*i

+ПХзА г в ( Х , Y m i ^ o d F ,, bd u 3( X ) + D ^ A n 3(X) = A J J f s( T 2)(Ar — 2C2)[hAFG ( X f Y) +

s 2

X D XzA YG ( X , Y)] ly2 = 0^ ^2? h A v ^ + B ^ A v ^ X ) = —A J f g ^ T j D ^ H A r e i X , Y) +

Sl +DX3AYG(X, Y)]\Vim.0d Y lt h Av , ( X) +DXiAvz{X) = - / / g2( T z)[hAr G(X, T) +

s 2

+DX3A y G(X, Y)]ly2=0d F2 and

4

Ы щ ( Х ) + О н Лщ(Х) = - A C ^ ^ h r ' j j M f , ) ^ 4 U l (Bj, ) d r 3,

« 3 j = 1

4

b â v , ( X) +DX3Av3(X) = -2AC æ 3(h)~l j j g3( 7 z) ] ?

S 3 j = l

- 2 C U 2(Ej3) - 2 U 3(Bj3)]dY3.

By the last formulas, by Theorem 1 and by Lemmas 7-11 for i = 3 we get

Th e o r e m 8. I f the assumptions of Lemma 5 are satisfied and the function g3 is continuous at the point X3, moreover, ж® > 0, then the function и satisfies the boundary condition (2c).

By Theorem 3-8 and (11) follows

Th e o r e m 9. I f the functions f {, (i = 1 , 2 , 3 ) are bounded and measurable convenably in the sets S{ and continuous convenably at the point X®, moreover, ж® > 0, then the function и defined by formula (9) is the solution of the problem (B-JST-M) (x).

i1) F. B a r a n s k i, On certain boundary-vaine problem for the equation (Л — O2)2 u (x , y , z) = 0 , Boezuik Naukowo-Dydaktyczny, Zeszyt 51, p. 17-24.

Prace Mat. 7, Krakow 1974, Wydawnictwo Naukowe W.S.P.

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