ss SS On the Neumann problem in an ^-dimensional hall-space

ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I: PRACE MATEMATYCZNE X I (1967)

ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE X I (1967)

E.

(Kraków)

On the Neumann problem in an ^-dimensional hall-space

In the present paper we give a solution of the problem of Neumann in the ^-dimensional half-space. The assumptions concerning the density of f(s) are more general than in the paper [1] (see [1], p. 185-192).

Let

denote the distance of the points X = {xx, ..., xn) and

ń

— ((? ^, • • •, i) and let d$ — ds-^ ... ds^ ^.

Let f(8 ) be a continuous function defined in the whole (n—^ -di­

mensional Eucliden space E n_u T = (<x, ..., <n_1} 1), \T\n = p2 + . .. + + < »_i+ l)n/2 and dT = . .. dtn_x. We shall prove that, under some assumptions on this function, the solution of Neumann problem in half- space is given by the formula

(1) * ( * ) = -

SS

^{f{S )d S}

2_L/»2-|ł»-l Ln- 1

1

ss

(2— n ) a

f{S)dS П-2

_ i

where an = S S dT

E „ _ 1

\T\ That means that if xn > 0 then the function

u(X) satisfies the conditions 1° u(X) e<72, 2° Au(X) = 0, 3° if X ^ (x\, ...

...,ж°_!,<)) then we have limn* (-^) — /(ж?, ..., a?$Li)-

We begin with two definitions and two lemmas. We shall consider two classes of functions, К and L.

De f in it io n

1. К is the class of functions f ( X ) defined in E n_x and satisfying the following conditions: 1°/ ( X) is continuous in E n_1, 2° there exist a positive constant rx and a function co(r), where v = (s2- f ... +sL_1)1/2

>7*!, such that if r is outside C(Kr ) = {S: sj| + . ..

then

\f{S)\ < co(r), 3° co(r) is a nonnegative nondecreasing function such that oo

the integral f r ~ 2(o(r)dr is finite.

1 8 0 E. Ś l i w i ń s k i

Definition

2. L is the class of functions defined for S e E n_i and satisfying the following conditions: 1° f(8 ) is continuous in E n_x, 2° there exists a nonnegatiye nondecreasing, continuous function Q(r) defined for r ^ 0 such tllQjtl f

? 6V6ry р

)

Г of J)OHlfS (

_j)y (^/ij •••? Уть_l) ^ -^ть_

and r = \ X Y I the inequality

!/(®1} •••, ®n_x)—/(y i, •••, y»_i)l < Q(r)

00

is satisfied, 3° the integral f r~2 Q(r)dr is finite, 4° there exists a continu-

ous nonnegatiye function rp(oc) defined for

> 0 such that

(

) = 0 and Q{a-b)

o(a) Q(b) for a > 0, b > 0.

Lemma 1.

I f f ( 8 ) e K , then the integrals

A = < S S ^ T W d S ’ Jo = ss

Ет.— Л

(X i-S iffiS )

M + 2j

dS

converge uniformly in any cylinder 4-.

< В 2, у < xn < A, where rj, A and В are arbitrary positive numbers, and i = 1, .. . , n —1, j — 0,1.

Proof. Let r — («Х + . . . + sn-i)1/2- We choose two numbers r0 and В such that r

^ 2B and the inequality

(2)

f(JS) < eo(r) if r ^ r 0,

is satisfied. If r > r0, then we have

Q =

+

^ r —B ^ r - \ r = |r , and, in view of (

), we obtain

SS

^{f(S) г dS <} w SS

^{rn+2j dS, ™(r) ЛС?}

C being a positive number. Let us introduce spherical coordinates (T) 6q = rcos^x ... cos

?n_ 1sin

?„_1, .. . , sn_l = rsiiupx in the right-hand integral which is transformed into the following

SS C4E

)

oj(r) r w( r ) r n 2

= - p * * -

r

)(r)

*■ = °>J тг+и *>

R

00

Cx being a positive constant. We have assumed that J r 2oj(r)dr <

. в

Then for every e > 0 there exists

such that if R >

then

% h SS

0(^ri)

№ и 4-2 j dS < e.

Neumann problem in an n-dimensional half-space 181

In order to prove the uniform convergence of the integral J 2 we consider the inequality \Xi —s*| < r-\- (ж2 + -.. - f x \ _ i)1/2 < r-\-B < |3r. Hence if r > r0 we obtain

\J'2\ = SS

^dS «SS _C(KR) ^{~Zn+2T dS w(r)rJ}

and, by introducing the transformation (T) we obtain finally

\-K\ sj С^П-> f

R

co(r)

dr < e.

Lemma 2. I f

/($ ) is continuous in

and

then the function

(3) « д а = — o o — sr-

ап лп-1

Q dx

converges to f { x \ , a?$Li) when {xx, ... , xn) -> {x\ , ..., a£_i, 0).

Proof. Lemma 1 imphes the continuity of u(X ) in the half-space xn > 0. Moreover, we have

We transform the integral (3) and obtain substituting Sj — х г = ^ х п,

Sn—i X n _ j = tn_j X n,

?

* ( v = ~ - S S

E n.— 1

f (^1+

^\Xn , ..., xn_ x tn_ xхъ

- - jrj" dT.

Then

g(X) = v ( X ) - f ( x 0 1,...,x ° n^ )

[/ (*«1 ~b > • • • ) ^n-l + tn—l<^n) / (*«1 ) • • • > *«7l—l) ]

Ern—1

- S S

and

«~ss

dT

1 f* i2([(a?x—Жх+^1Жи)2 + >• • + {xn- i — +hi-i^n)2]^2)

Eyi,—i

|Tf dT

< 1 O O ^ ([([а?

— a??| + 1<х|Д?п)2 + --. + (|а?п-

—Д?п-х1 + l<n-i|a?n)2]1/2) Rr£

^ „ O O |T|W

E™

1 8 2 E . Ś l i w i ń s k i

Let e be an arbitrary positive number. If the coordinates of a point X satisfy the inequalities \хг — ж$| < e, \xn_i~Xn_i\ < s, 0 < xn < e, then we have

*£?([(\x1 —x\\ + |Ь1 Ж »)2~К • • 4” (\®n-i—®n~il 4~ Ihi-il^n)2]1^2)

< Q(ne\T\) < (p(ne) Q(\T\).

In view of the convergence of the integral

^ S S

i

Q(\T\)

I T\n dT we obtain the estimation

\g(X)\ <

cp(ne) an

^(\T\)

\T\n dT = Сг <p(ns)

<*n

where Сг is a positive constant. In fact, if we apply the transformation (T), then we have

J = C2f

0

о ( ( г Ч 1 ) ' у - 2

(r2 +1)"'2 C2 = const,

f Q((r3 + l ) ll2)rn- 2 ^ r a((r2+ l ) ,/2) ( r 4 l ) < n- 2>'2

J ( i ^ + i r d r < J Р + т р dr r O ^ + v n dr

J r2+ l

D 1

r Q(r)

^ < p ( 2 ) j - y - d r ( B > 1).

R

So we conclude that g{X) -> 0 if X -> (x\, 0).

Th e o r e m.

I f f ( X ) is continuous and belongs to К r\ L for S ^ E n_1 then the function u( X) defined by formula (1) is a solutio7i of the problem of Neumann in the half space xn > 0.

This follows from Lemma 1 (since the function Q2~n is harmonic).

R eferences

[1] О. Л. С о б о л е в , Уравнения математической физики, Москва 1954.