The mixed boundary conditions oî Dirichlet-Neumann type for the Laplace equation in the positive quadrant of the plane

(1)

ROCZNIKI POLSKIEGO TOWARZTSTWA MATE MAT YCZNE G O Séria I: PRACE MATEM AT Y CZNE X V II (1974)

J an Mtjsiâlek (Krakôw)

The mixed boundary conditions oî Dirichlet-Neumann type for the Laplace equation in the positive quadrant of the plane

In this paper we solve the Laplace equation in the region D — {(xx ? > 0, x 2 > 0} with Dirichlet condition an one half-line ah the boundary of D and with the Neumann condition on the another one.

In the paper [2] we solved only the Neumann problem for this region.

The Green function by means of which we solve the problem, is constructed by aid of the symmetric images.

1. Let X = X xx(xx, x 2), Y = Y ( y x, y 2) denote two different points in D.

Let X _n = ( — x x, x 2), X x_x = (xx, — x 2) be symmetric images of X in the axis Oy2 and Oyx, respectively. The point X _ x_x = ( —xx, — x 2) is the symmetric image of X _ n in the axis Oyx or of X x_x in the axis Oy2.

Let

rxT =

Y X

n =

[ ( Уг - о е х)2 + (у2- х 2)2Т * , r z h = Y X _n = [{Ух + осх)2 + (у2- х 2)2Т * , ГГЛ = * £1 - 1 = [(У1-®1)* + (Уа + ®2)2Г*, r l L i = X X -i-i ^— [(2/1 + ^i)2 + (Уг + ^г)2] *

and let denote l x the half-line x 2 = 0, xx > 0 and l 2 the half-line xx — 0, æ2 > 0.

Theorem 1 . The Green function harmonic in D and satisfying the boundary conditions

(1) G(X, Y) II о II о for Y e l

(2) dG(X, Y)

= 0 for Y e l is of form

дух 2/1 = 0

(3) G(X, Y) = —logru —logr_ 11 + logr1_1-flogr._1-_1.

P ro o f. The function G(X, Y) of form logrlx-f .ff(X, Г),

(2)

where

H (X , Y) = - l o g r .n + lo g ^ .i + lo g r .j.!

is harmonie with respect to Y. The function G (X, Z) satisfies condition (1). For Y e l xwe get

iogru = lo g r^ j and logr_n = lo g r ^ ^ . For Y e l 2 we have by (3)

dG(X, Y)

\x\

f y i 2xx

У1 = 0

2а?! 2a?i 2a?i

+ (У2- ^ ) 2 ocl + (y2~ x 2)2 ^x\+ {y2 + ^x2f ^X?x+ {y2 + ^X2Ÿ = 0.

2. Using some lemmas dealing with convergence of certain integrals we shall prove that the function

(⁴)

OO OÛ

и (хг, х 2) = ~ - f /1(2/1) dyx- f f 2(y2)G(X, Y)

2tz J d y2 y2=0 2tcJ ^Vⁱ=0^2/27

where G(X, Y) is defined by (3) solves in D the Laplace equation and satisfies the mixed boundary conditions

(5) lim^(a?i, a?2) = ^{f x(x°x)} as (a?i, a?2) ->■ (a?J, o),®; > o , (6)

Indeed

lim ---- = /du 2(^2)

da?i 2 as (a?i, a?2) -► (0, ^x2). & V ©

= 2a?2{[(2/i-a?i)2 + a?5] x + [(2/i + ®i)2+ ^ ] x}.

a/2=°

Hence by (3) the function и {xx, a?2) defined by (4) takes the form 4

(4a) u (x x, x 2) = £ Ii(X ),

i = l

where

OO CC Г

I x(X) = -± fx (y x m y i-o o 1f + 4 T 1dy1,

7Z J 0 CC c00

I 2(X) = — / i W ^ i + ^ + ^ r 1# ! ,

7Г J 0

1 00

1 з(Х) = “

2

^ J /

2

Ы ^ [ ^ж 2! + (

2

/

2

- ^ж

2

)

2

]Й

2

/2,

0 OO

I

⁴

(X) = - ^ i J /2(2/2)l°g[a

^?2

+ (2

^/2

+ #

²

)

²

]d2/2.

№ dy*

(3)

Let a > 0 and let

Wx (a, A, B) = < —В , B} x <№, -4.), TF2(a, -4, B) = {a, A} x < —В, B>.

Lemma 1. Let the function f x{yx) be Lebesgue integrable and let for every rectangle Wx(a, A, B) exist a constant N (W X) > 0 such that the integral

/ \fi{yi)Vy\dyx

N

converges ; then the integrals

r dk

Ц (Х ) = (т г)-ч J

о *

and

!(X) = (7i)~1x 2 J /1(y1) ^ r [ ( y i + ®i)2 + ^ r 1^ 1»

w/iere & = О, 1, 2 ; i = 1 , 2 ; s = 0 ,1 , .. ., 7 exist and are almost uniformly convergent in the half-plane — oo < x x < + oo, x 2> 0.

P ro o f. The proof will be carried only for 1ЦХ), h = 0, i = 1, s = 0; the proof in remaing case is similar because those integrals admit an analogous majorant.

Let <TX = (yx — x x)2-\-ot%. By the inequality of the triangle: \ yx < dx

< 2y x. For y x > N (W X) and for each (xx, x 2)e Wx(a , A, B). Hence / l/i(2/i)I {.{Vi + ^ ] l dyx

N

OO

< j \МУ1)\Х21(У1-Х1)2 + 4ТЧ(У1-Хх)2 + Х1]-Ыух

N oo

f lfi(yi)lyr*d yi-

N

Lemma 2. Let f z{yz) be Lebesgue integrable and let for each rectangle

^2(a, A, B) exist a constant N (W 2) such that the integral

N

/

converges. Then the integrals

\ M y ^ o g y 2dy2

I f dk

Ц (Х ) f2(y2)-^gr{log[x2x + (y2-hx2f]} d y 2,

0 i

OO £

If(Z ) = - i - J f 2(y2) - ^ { l o g [ x î + (y2- x 2)2]}dy2,

n *

(4)

where к = 0 , 1 , 2 ; г = 1, 2 ; р = 0 ,1 , ..., 7 exist and are almost uniformly convergent in the half-plane x x > 0, — oo < x 2 < + oo.

P ro o f. We will prove thesis for the integrals I\{X) and 1\{X) for i — 1 ; к = 0 ; p = 1.

Let d2 — x\-\-(y2 — x 2)2. By the triangle inequality \ y 2< d < % y 2 for y 2 > N (W 2) and for each (xx, x2)e W 2{a, A, B).

Hence

OO

l f ( X ) = f lf 2(y 2)llo g[x21 + (y2- x 2)2]d y 2

N

OO

< 2

j

\f2(y2)\logy2dy2 for N (W 2).

N

For the integral I\(X) we obtain ^

OO

I f( X ) = 2

j

\f2{y2)\oc1 [x\ + (y2 + x 2)2Y 1dy2

Noo oo

< 2

J |/

²

Ы1«1Й + 2/2Г

¹

%2<

²

J

^|/2

(2/2)1 l°g2/2^2

for N (W 2) sufficiently large and for each (a?x, x 2)e W2{a, A , B). This majorant applies also to the remaining integrals.

Theoeem 2. TJnder the assumptions of Lemmas 1 and 2 the function u (x1, x 2) defined by (4) or (4a) is harmonic in D.

P ro o f. By Lemmas 1 and 2 the integrals If (X), Ц (Х ), I f (X), I f (X) converge almost uniformly, whence we can interchange the integration with the differentiation.

Therefore

*» OO

СО Г

A^u{^{x u x}2) = — fi{ y i)f i(y i)A [(y 1 ± x 2)* + 4 ]~ 1dy1±

7C J

0 00

00 c

± — /2(2/2)^ {l0g[^ + (^2 ± ^ 2)2] } ^ 2 = <>•

7C J 0

The Laplacjan A under the sign of the integral vanishes, since the involved

function is harmonic. 7

We shall prove now that the boundary conditions (5) and (6) are satisfied.

Lem m a 3. Let the function f 1{y1) satisfy the assumptions of Lemma 1 and let it be continuous at x% moreover, for each Ô > 0 let

f ^2I/ⁱ(2/ⁱ) - /ⁱK)I[(2/ⁱ —^i)2 + ^ r ^ 2 / i- ^ 0

(5)

as xx x\, x2 -> 0 ;

i i W Л Ю to » ^2) -> К j o).

P roo f. Let

0 for y x < 0, Л Ы =

/1(2/1) f o r y ^ O . Hence

+ 00

/ £ /* ^—

i 1{ X ) = — f A y J K v i - ^ + x Y l t y x.

7t J It is known ([1], p. 270) that

x« +00

(8⁾

+00 +00

— 71 J

f

^{[(2 /}ⁱ - % ) 2 + ^ ] _ 1 # i = — fЯ J ds

1

+

^6*2 ⁼⁼¹ ^.

Whence +00

T /*

(9) — I fi{x\)i{yi~ X i? + ®lYl dyl = Л К ) . 7C J

— 00

Let ns write I x{X) as

+ OO

СО г

Ii(X)=J(X) + -± fMmyi-Xif+rtT'ày,

TZ J where

+ 0 0

со г

J ( X ) = - 1 [ Л ы - Л Ю З И л - ^ + ^ Г 1# !-

7C J By (8) and (9)

i . m - A K ) = j ( X ) .

Let e > 0 ; continuity of f x at xx implies existence of <5 > 0 such that

\yx-x\\ < 6implies

Now where

I/i(2/i) - /iK)I < £/2 - J ( X ) = J x(X) + J 2(X ),

J x{X) = — _7Г _J

Г

^[/ⁱ⁽²^/ⁱ^{) - /}ⁱ^{K )][(}²^/ⁱ^{- ^} ⁱ⁾²

+ ^]

I3/X—

CO c

J 2(X) = —, [ Л Ы - Л Ю Ш ^ - ^ + ^ Г 1^ ! .

7C , J

(6)

Thus applying (9) we obtain

^ ^«/

|î/x - a : J | < ( 5

£ 00o Г+ oo „ e

< 7 7 — [(У 1-Я 1) +®?] = — .

2 7Г J 2

—00

It follows that J 1(X )-+ 0 as xx -> a?2 -> 0+.

By our assumption aq being sufficiently close to x\ and x2 being sufficiently close to zero, we have

|J2(X )| < £/2, whence

| J|< e.

Lem m a 4 . Let the functions /1(2/1) and /2(2/2) satisfy the assumptions of Lemmas 1, 2, 3 ; then

(a) I 2(X )-> 0 as ⁰+ , ^x2> ⁰,

(b) \1 §{Х} -\-1 ^Х)~\ — 0 us a?2 —> 0+, a?2 —> 0. P ro o f, ad (a) The integral J2(X) satisfies the inequality

00

I2( X ) < — f |/i(2/1)l(2/î + ^ + ^ r 1# i

7C J 0 /£» 00

TC J

о

The last integral is independent of x2.

ad (b) By Lemma 2 the integrals I 3(X) and I4(X) converge uniformly for х г > a > 0, —

00

< a?2 < + 00, a being an arbitrary positive number, whence those integrals are continuous functions and for хг = 0 we have

i , ( X ) + i t ( X) = ~ f

/ .M P o gtâ + sD -io gtâ+ ÿî)] =

⁰

.

0

Lemmas 3 and 4 imply that the boundary condition is satisfied. Let us prove now the boundary condition (6). By Lemmas 1 and 2

du дхг

4

У ^ Х ) ,

1

(7)

where

SAX) = f /^Уг){Уг-^хШУ

¹

- ^

^{1 )2}

+ (^Т

²

Луи

1 ' 0

OO

®z(X) = ~ f ^

=~ J

М У 2 ) [ ^ 1 + ( У 2 - 0 0 2)2Т 1^У2,

1 0

OO

= - ^ —“ = — — f Л (2/i)(У1 + æi) {.(Ух + x i f + ^]~uOC-i CO *) 2dyx,

0

d l (X) x r°

^4W = —^ ---- = - ^ J - J / а Ы Й + ^ а + ^а)8] - 1^ -

1 о

Lemma 5. Let the function /1(2/1) satisfy the assumption of Lemma 1, let /2(2/2) be continuous at x 0 and satisfy the assumptions of Lemma 2, and let for every ô > 0

f 001 ^1/2

^(2/2)

^{- / 2}

^(a>

²

) I O ? + (2/2 - ^ a

)2]-1

# 2 - >

0

! Î/2 —•*'21 > ^

as x2 -» a?2, #i -> 0, a?i ■> 0 ; йетг

(c) S 3(X )-+f 2(x°2) as (я?и a?a) (0, a$);

(d) $4(.X) -> 0 as (a?!, a?2) -> (0, a?2);

(e) [^ i(X )+ ^2(X) ] - > 0 as (®j, a?2) -> (0, x2).

P ro o f of (c) is similar to this of Lemma 3, and this of (d) and (e) to this of Lemma 4.

Lemma 5 implies the boundary condition (6). Lemmas 1-5 and Theorem 2 imply .

Theorem 3. The assumptions of Lemmas 1-4 being supposed, the function defined by (4) or (4a) forms the solution of the Laplace equation

in the quadrant x x > 0, x 2 > 0 with boundary conditions (5) and (6).

References

[1] M. K rz y z an ski, B o w n a n ia rozniezkowe czqstkowe rzçdu drugiego. Czçsc I, W ar

szawa 1957.

[2] J. M usialek, F u n k e ja Greena oraz rozw iqzanie z ag a d n ien ia N eu m an n a i D iricM eta, Zeszyty Naukowe AGrH, Krakow 1970.

The mixed boundary conditions oî Dirichlet-Neumann type for the Laplace equation in the positive quadrant of the plane