ANNALES SOCIETATIS MATHEMATICAE POLONAE Seiies I: COMMENTATIONES MATHEMATICAE X X (1977) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO
Séria I: PRACE MATEMATYCZNE X X (1977)
K. K a p t u r k i e w i c z (Krakôw)
On the mixed problem for the Poisson equation in the quart-plane
1. In the present paper we shall solve the boundary problem for the equation
in the quart-plane E t = {X : x > 0, y > 0}, where h denotes the negative constant and {i — 0 ,1 ), F are the given functions.
We shall call problems (1), (20), (3); (1), (2J, (3); (la), (20), (3) and (la), (2X), (3) conventionally М г (i = 1 , 2 , 3, 4) problem.
2. For the solution of the problems we shall use the convenient Green-functions.
Let Y = ( s , t ) denote an arbitrary point belonging to E2.
Let
r\ = ( oc-s)z( y - t ) 2, r22 = ( x - s y + (y + ty, r\ = {x + s)2 + { y - t y , rl = (x-\-sy + (y + ty , r\ = (x — s)2-f (y + t + vY, r\ — (x + s)2-f
(1) (la)
Au( X) = 0 , X — (x, y), А и (X) = F { X )
with boundary conditions (2i)
(3)
Щ и {0, у) = My ) (i = 0 ,1), D yu(x, 0) + hu(x, 0) = / 2(^)
+ (y + 2 + v)2,
oo
0
K 7( x, t ) = x ( x2 + { y - t y ) ~ 1, X Q{K,t) = x( x2 + (y + ty) S
OO
K9( X, t ) = J ehvx (x 2 + (y + t + v)*) xdv, о
K10{ X, s ) = y ( ( x - s Y + y 2) S
7 — Roczniki PTM — Prace Matematyczne X X
98 K. K a p tu rk ie w ic z
К г1{Х,8) = 2/((Я + $)а + у2) S K 12{X,S) = In ((07-s)2 + У2),
K 1B{X,s) = ln((a? + s)2 + y2),
oo
K U( X, $ ) = J ehvl n ( ( x - s)2 + (y-hv)2)dvt
0 oo
K15( X , s ) = f ehvln[(x + 8)2 + (y + v)2}dvf 0
K i6{ X, t ) = ln (x2 + { y - t ) 2), K „ { K = In [ос2 -f- {y + ^)2) i
OO
R:ia( X , t ) = f ehvln(x2 + (y + v + t)2)dvr
0
K19( X , t ) = { y - t ) ( { y - t )2 + x 2)-\
K ,0( X, t ) *= (y + t ) ( { y - t )2 + æ2)~\
K21{ X , s ) = ( x - s ) [ { x - s )2 + y 2) - 1,
K22{ X, s ) = (a? + s)((a? + s) 2 + y2)~S
OO
K22{ X , s ) = J ehv( x - s ) ( ( x — s)2 + (y + v)2) - 4 v t
0 00
K2i { X t s) = J ehv{x + s)[(x — s)2 + (y + v)2) - ldv,
0
K \(X , Y) = l n f* (i = 5 , 6 ) and
OO
Hi{X) = 2GXf f2(s)Ki( X f s)ds- (i = 10, 15).
0
Let
6M X , Y) = G1{ K1( X 9 Y) + K 2( X f Yfl + O M X , Y), G2( X, Y) = Gx[K3( X, Y) + K t { X, Y)) + G2I 2{ X, Y),
where I t{ X f Y) = K i+4{ X, Y) (i = 1,2) and Gx = (27c)_1, C2 = 2hGx.
Let
W = {(X t Y): 0 < A < x < B , 0 < A < y < B, 0 < < < J?, 0 < s < B } t where A , B are positive constants.
Let
OO
l f „ mn(X , Y) = 0 , / \ D ^ ]n ^ i+i\dv (i = 1 ,2 ),
B
m , n , p , q being non-negative integers for which 0 < p - f g - f - m + w < 4 .
Poisson equation in the qvart-plane 99
Let e be an arbitrary positive number. Now we shall prove some lemmas which we shall use in the sequel.
Lemma 1. There exists the number B0(e) such that for every В > B0(e) and every point (X , Y ) e W we have Ifmmn < e (j = 1, 2).
Proof. If p + q + m + n > 0, then the majorant of the integrals Ifpgmn are the finite sum of the integrals of the form
oo
J = G J diV\x + { — l ) <e|a|y + i + «|e(ri)-aedi> (г = 5, 6),
R
where 2c > a + b and G is a positive constants. Since (x + ( — 1)*‘в)2(^)~2
< 1 (i = 5, 6), (y + t + v)2(ri)~2 ^ 1 (i = 5, 6), thus the integral
OO t
J1(E) = C1A a f ehvdv,
R
Gx being the convenient constant and a = a-\-b — 2c, is the majorant of the integral J.
The integral J l {B) is arbitrary small for sufficiently great B.
00
К p-\- q + m + n = 0, then the integral G f ehvlnvdv is the majorant
oo R
of the integrals J ehvlnr2dv (i = 6 , 6 ) for В sufficiently great.
R
Corollary 1. The integrals
00
I lpmn(X , У) = / e»”BSSrin»?+1<it> (» = 1 ,2)
0
are uniformly convergent in every s%et W, there exist the derivatives
- D g y i ^ X , У) and У) = I ipqmn(X , Y) for j = 1 , 2 . Theorem 1. The functions
G*(X, Y) = Gx{ X, Y) + ( - l ) iG2( X, Y) (i = 1,2)
are the Green function for the problem (Jff) {i = 1, 2), in the domain E f with the pole X.
Proof. The functions G{ (i = 1 , 2 ) are harmonic functions of the point Y.
Indeed, the functions lnr2 (г = 1 , 2 ,3,4) are harmonic and by Corollary 1 we obtain
00
АтЩХ, У) = Oj/Ij - ï . ! ! , Y) + 01d TE 2( X, Y ) + 0 , f г * Ar K \ (X , Y) d v = 0
0
and similary AYG2{ X , Y) = 0 . Hence the functions G% (i = 1 , 2 ) are harmonic as the sum of two harmonic functions.
100 K. K a p tu rk ie w ic z
Moreover, we shall prove that the functions Gl (г = 1 ,2) satisfy the homogeneous boundary conditions
(4) (Dt + h)Gi( X , s , 0 ) = 0 ( * = 1 , 2 ) ,
(5) G1{ X , 0 , t ) = 0 ,
(6) DSG2( X, 0, t) = 0 .
Condition (5) is obvious. We shall prove condition (4) for the function Gx and for i = 1. The proof of condition (4) for the function G2 and for i = 2 is the same. By Corollary 1 and Theorem 1 for t = 0 we get
.D(G,(X,»,t)|i-o = С . Щ Е ^ Х , Т) + К г(Х, Щ + С . Щ ^ Х , У))!,„0
оо оо
= Сг / ^ Щ К И Х , Г))|(_ 0Ле = 0 2/ < ^ ЩК\ ( Х , Т))|,.0«г»
о о
= { - С2К 2( Х, Y) — G2hK5( X , Г))|<=0
= - С 2К 12( Х, s)-JiC2K u ( X , в);
because for t — 0 we have Dt { Kx( X, Y) + K2(X, Y)) = 0 and K 2(X, Y)
= K12( X, s), K 5( X, Y) = K U{ X , s). Moreover, for < = 0 we have hGx( X, 5,0) = 2hC1K 12( X , s) + JiC2K u ( X , s).
Hence
JiG^X, 5, 0) + DtG1{ X, s, 0) = 0.
We omit the simple proof of condition (6). ^
3. Under certain assumptions concerning the functions (i = 0 , 1 , 2 ) we shall prove that the functions
оо CO
(I) % ( X ) = / f %(s ) G ' ( X , s , 0 ) d s - j f n(t)D,eHX,0,t)dl
0 0
and
oo oo
(II) u2(X) = f f 2(s)G2(X,s, 0)ds + j ш а \ х , 0,t)dt
0 ; 0
are the solutions of the problems ( M{) (i = 1 ,2).
We shall prove now that the function % given by formula (I) is the solution of the problem ( Mx).
By (I) and G1 we have
7
U{x) = 2 j t (X),
2 = 1
(la)
Toisson equation in the quart-plane 101
where
J, (X) = ( - 1 ) <+I2G1 f M » ) K , ( X , s)ds (i = 1, 2; j = i + 1 1),
0 oo
J, (X) = ( - l ) i+102 j Ш К , ( Х , s)ds (i = 3, 4; j = i + 1 1),
0 oo
(г = 5 , 6 , 7 ) ,
for i — 5, 6, for i = 7.
Let
= { ( X ): |a?| < .4.; 0 < Л < у < £ } ; ТГ2 = {(X ) : 0 < A < æ < Б ; |y| < J.}
and let
J ,(Z ) = C4J fo(t)Ki+I( X, t) dt where
Ct = 4cx
4C2
Jipq{X) — f fo{t)DxyKj t)dtj
0
OO ^
J<m(X) = f f,(i)D§SZj(X,»)ds;
0
j f m {X ) = j |/0(*)1>г«^(х,()|й*
Й
for i = 6 , 6 , 7 ; j = « + 2 ,
00
= / l/»(*)-»№ (-r>»)l*
R
for i = 1, 2, 3, 4 ; j = i + 1 1 , P , g being non-negative integers and 0 < 3> + g < 2.
Le m m a 2 . I f 1 ° function f0 is continuous and absolutely integrable for t > 0, 2° the integral
f |/2(e)lne|de
a
is convergent for any a > 0, 3° the function f2 is continuous for s > 0, then the integrals Jipq(X) {i = 1 , . .. , 4) are uniformly convergent in the set W x and the integrals JipQ(X) (i = 5 , 6 , 7 ) in the set W 2.
Pr oof . The common majorant of the integrals Jfpa{X) (i = 5 , 6 , 7 ) is the integral
C j I/o(i)I'M
R
102 K. Kapturkiewicfc
and for the integrals Jfpa(X) {i = 1 , . . . , 4) the integral
oo
G f \f2(s)]ns\ds,
R
where О is a positive constant.
Corollary 2. I f the assumptions of Lemma 2 are satisfied, then there exist the derivatives
D^jjJiiX) for i — 1 , 2 , 3, 4 and X e W lf and
D ^ J ^ X ) for i — 5 , 6 , 7 and X e W 2, and
CO
DZ’ J AX) = J / ,( » ) Л « Я , ( Х , s)ds (i = 1 , . . . , 4 ; j = l l + i),
0
oo
D % J {(X) = f т О ^ К , ( Х , t)dt {i = 5,6,7; J = i + 2).
0
From Corollary 2 and Lemma 2 follows
Lemma 3. I f the functions f { (i = 0 , 2 ) satisfy the assumptions of Lemma 2, then the function ux is the harmonic function in the quart-plane E f .
P r o o f . Since G1 is the harmonic function with respect to the point X we get
OO 0 0
Лux(X) = (-X) s ) fo{t)@l (-^-i Ojt)dt^
о 0
oo oo
= f f2(s)Ax G1( X, s, 0) d s - J f^{t)Ax Gl{ X, 0 , t)dt = 0 .
о 0
4. In order to prove that the function ux satisfies the boundary con
ditions (2г) (г = 0 , 1 ) and (3) shall give some lemmas.
Lemma 4. I f the function f2 satisfies assumptions 2°, 3° of Lemma 2 and y0 > 0, then
4
g = lim JT Ji(X) = 0 as X -> (0+, y0).
t=i
P r o o f . By Lemma 2 and continuity of the integrals J{ { X ) {i = 1 , . . . , 4), we obtain
4
g = ^ « M o , Уо) = о .
г= 1
Poisson equation in the quart-plane 103
Let Wj = supl/J (i = 0 , 1 , 2 ) and let
f0{t) for i > 0, fo(t) =
0 for t < 0.
Lemma 5. I f the function f Q is continuous, bounded and absolutely integrable, for t > 0, y0 > 0, then
(a) lim J5(X) = / о Ы as X ^ ( Q + , y 0), (b) lim Jt(X ) = 0 (i = 6 , 7) as X - * ( 0 +, y 0).
P ro o f . Ad (a). By Krzyzanski’s book (x) we get
oo
J5(Z ) = j f0(t)K7{X,t)dt-+f0{y0) as I - > ( 0 +,2/0).
— OO
Ad (b). For J6{X) we have
OO
|J6(X)i < тхСгх f ( i y0 + t)-2dt->0 as I - > ( 0 +, y0) о
and for J7(X ) we obtain the estimation
IJ7(X)\ < 2тха гх J J ehv(%y0 + t)~2dvdt->0 as X -> (0 +, y0).
Lemma 6. I f 1° the functions / t- (i = 0 , 2 ) are bounded and continuous fo rt > 0, s > 0, the function / 0 г'$ absolutely integrable for t ^ 0, 2° Йе integral
oo
J |/2(2!)ln«|^
a
convergent for any a > 0 and oc0 > 0, then
4
(c) (D„ + A ) ^ ' j j (X)->./! (*0) as X-*(æ0, 0+) i= 1
and
7
(d) ( Д „ + й ) ^ ^ ( Х ) ^ 0 as X - * ( * 0, 0 +).
i=5
P r o o f . By (la) and Corollary 2 we have
4
<Д„ + Й ) 2 У Д ) = 2 B 10( X ) - h 3 ls( X ) - W B u (X) + B 12(X) + h*Hu (X) +
» = 1
+ 2Я П (X ) - № 1 3(X ) - W B 1S (X) + hB13 (X) + h*Blt(X) _ = 2 (H10(X) -\-H11(X))-*f2(oc0) as X- >( x 0} 0+) .
0) M. K r z y z a n s k i , Partial differential equations of second order, vol. I, Warszawa 1, p . 2 6 3 .
104 K. K a p tu rk ie wicz
Indeed^), lim2H10(X) = f 2(x0) as X->(x0, 0+) and
oo
l - H i i W K ^ / 2/ ( ( ^o) 2 + s2) - 1 ^ - > 0 as 1 ^(ж0,0+);
о
A being a positive constant.
Ad (d). By continuity of the integrals (i = 6 , 6 , 7 ) we get
7 oo
(7>„ + й ) У У т = 8C,x f f a( t ) ( y - t ) { x \ + { y - t f ) - l +
i = 5 0
oo
+ (y + i){x2 + (y + tf)-'dt + i 0 1h f f 0(t)(Kt ( X , t ) ~ K 1(X,t)]dt
0
oo oo
->8Схх0 f f 0{t)(t(xl + t2) - l - t { x l + t2) - l)dt + ±Cxh J /о^^о^ + г2)-1-
0 0
— x0{xl-\-t2)~l)dt = 0» By Lemmas 2 , . . . , 6 follows
Th e o r e m 2. Let the functions f i ( i = 0 , 2) be bounded and continuous for t > 0, s > 0, x0 > 0, the function f0 be absolutely integrable for t > 0 and
the integral
OO
/ \f2(s)lns\ds
a ,
be convergent for any a > 0, then the function ux(X) defined by formula (I) or (la) is the solution of the problem ( Mx).
5. W e shall give now the solution of the problem ( M2). By formula (П) we get
OO
(Па) щ ( Х ) = f f2(s)(2G1K lt( X , s ) + 2G2K 13( X, s) + G, Ku ( X , s) +
0 oo
+ C2K l5( X , s))* + J Ш (2 0^ ( X , t) + 2CxK „ ( X , t) + 2СгК ы{X,t))dt.
0
We shall prove that the function u2 defined by formula (Ha) is the solution of the problem ( M2).
Le m m a 3a. I f 1° the functions / г- (i = 1 , 2 ) are continuous for s ^ 0,
0, 2° the integrals
OO
f ifii^lnzldz (i = 1 ,2),
Poisson equation in the quart-plane 105-
are convergent for any a > 0, then the function u2 given by formula (lia) is the harmonic function in E f.
The proof is similar to the proof of Lemma 3.
We shall prove now that the function u2 satisfies the boundary con
ditions (2г), (3)
P r o o f . By (lia) we have
7
Dxu2(X) = ?
i=1
where
oo
Qi(X) = 2 C ,.J f2{s)Ki+n( X, s) ds (i = 1 ,2 , 3 , 4 ) ,
0 oo
Q,(X) = 2Сг f Ш К {+2( Х, t)M (i = 5,"6, 7),
0
and
= Cx for i - 1 , 2 , Ci = C2 for i = 3 , 4 . By Krzyzanski’s book (*) we get
limQs( X ) = f1(y0) as X -> (0 +, y0).
By continuity of the integrals Qi(X) (i = 1 , 2 , 3 , 4) we have
4 4
l i m y f t ( X ) = V « ( (0,y,) = 0 as X ^ ( 0 + , ÿ o).
г=1 г—X
Moreover,
oo
\Q6(X)\ < 2mxCxx j (t-\-ly0)~2dt-+0 as a?->0 + o
and
\Q1( X) \ ^ 2 mxC2co f f ehv{t + %y0)~2dvdt->0 as x-+ 0 + . Et
The proof of the condition
\im{Dy -\-h)u2{X) = / 2(#0) as X- > ( x 0} 0+) is similar to the proof of Lemma 6.
Finally we get
Le m m a 7. I f the functions fi (i = 1 , 2 ) are bounded for s ^ 0, t > О
and satisfy conditions 1°, 2° of Lemma 3a and x0 > 0, y0 > 0, then the function u2 satisfies the boundary conditions
(e) lim {Dv + h)u2(X) = f 2(x0) as I-> (æ0, 0 +), (f) limDxu2(X) = /i(a/o) as I - » ( 0 +, y 0).
106 K. K a p tu rk iew ic z
By Lemma За and 7 we obtain
Theorem 3. I f the functions f t (i = 1 , 2 ) satisfy the assumptions of Lemma 3a and 7, then the function u2 given by formula (lia) is the solution o f the problem { M2).
6. We shall give now the solution of the problems ( M{) (i = 3, 4).
Under certain assumptions concerning the function F we shall prove that the functions
v{ (X) = ui(X) + Ti (X) (г = 1,2),
where the functions щ are defined by formulae (I) and (II) respectively And
T{(X) = f f F ( Y ) G i( X1 Y)dsdt (i = 1 , 2)
Are the solutions of the problem ( М{) (i = 3, 4),
»
7. We shall prove some lemmas concerning the integrals T{ (i = 1 , 2).
Let
StiX) = 2Cx f f F(Y)lnrïdsdt (i = 1 , . . . , 4 ) ,
And
St(X) = 02f J J F ( Y ) e hvlnr2i dsdtdv (i = 5, 6).
Let the function / be bounded and continuous in E2 and f { Y ) = F ( Y ) for Y e E f and let
\f{æ+QCÔS(p, y + Q8m<p)\ < Fx{q)9
\f(X+QOOS<p, - y — V+QSin.(p)\^F2{Q,V),
îoiæ2 + y 2 > <52> 0 , v > 0 , 0, 2тс, 6 being a positive number.
Let
S?«(X) = f f F ( Y ) D ™ K t( X, Y) dY,
where p , g = 0 ,1 ,2 ; 0 < ^ + g < 2 ; г = 2 , . . . ,6.
Let K R denote the circle with the centre X and the radius R.
We shall prove now
Poisson equation in the quart-plane 107
Ti-rmma 8. I f the functions F { (i = 1 , 2 ) are continuous and bounded for q > 0, v > 0 and the integrals
00 OO 00
*i = / F i(Q)e№Q\dQ, s2 = J F 1(Q)dQ, s3 = J F 1(Q)Q~1dg,
a a a
OO 00 OO 00
s4 = f j F 2{Q,v)ehvQ\biQ\dQdv, s5 = J f ehvF 2(g, v)dgdv,
a a a a
oo oo
s6 = J J ehvF2{Q,v)Q~1dgdv
a a
are convergent for any a > 0, then the integrals Sf9 (i = 2 , . . . , 6) are uni
formly convergent for x > О, у > 0.
P r o o f . It is sufficient to verify the thesis for the integrals 8f 9(X) { < =2, . . . ,6).
Applying to the integrals 8f 9(X ) (i = 2 , 3 , 4 ) the transformation s = ±æ+eeosç>, t = ± t / + psin<p
and to the integrals (i = 5 , 6 ) the transformation 8 = gCOS<p, t = ± y — V + Q8Ïn.(p,
where p>£>0> 0, 2rc, we can easy verify that the common majo
rant of the integrals 8P9(X) {p + q = 0; i = 2 , 3 , 4 ) is the integral sx and for the integrals 8P9(X) (p + q = 0 ; i = 5 , 6 ) is the integral s4, for the integrals 8f 9(X) (i = 2 ,3,4 ; p + q = 1 ) is the integral s2, for the integrals 8$9{X) (i = 5 , 6 ; p + g = l) is the integral s5, for the integrals S%9(X) (i = 2, 3 , 4; p- j - q = 2) is the integral s3, for the integrals 8f 9(X) {i — 5 , 6 ; P + q = 2) is the integral s6.
By Lemma 8 follows
Co r o l l a r y 3 . There exist the derivatives Df^S^X) and
» % ( f f F { Y ) K t( X, Y ) d x ) = / / F { Y ) D g K , { X , Y ) d Y
< Et
(i = 2 , . .. , 6 ; 0 < 2> + g < 2).
Let
SAX) = 8\(X) + 8l(X), where
8ЦХ) = 0, / / F ( Y ) K A X , ¥ ) d Y , Kr
8\ ( X ) = G 1 f f F { Y ) K1( X , Y ) d Y .
e?\k r
108 K. K a p tu rk ie w ic z
We shall prove
Le m m a 9 . I f the assumptions of Lemma 8 are satisfied and F e C x{ Ef ), then the functions Tt (i = 1 , 2 ) satisfy equation (la) for X e E f .
P r oo f . For T x by Corollary 3 we have
A8((X) = 2 0 ; j j ] ? ( Y ) A x K (( X , Y ) d y = 0 (i = 2 , . . . ,6), К
because K t{ X, T) (i = 2 , . . . , 6) are the harmonic functions and \XY\
> ô > 0, <5 being a positive number.
From Rrzyzanski’s book(1), p. 328, we have
A8x{X) = AS\{X) + AS\{X) = F ( X ) . Hence the function Tx satisfies equation (la) in the set Ef .
Similary we can prove that the function T2 satisfies equation (la) in the set E f .
8. Let Z = E f kjZx\jZ2, where Z x == { X: x = 0, у > 0}, Z2 = { X: x > 0, у = 0}.
We shall prove
Le m m a 10. I f the functions F i (i = 1 , 2 ) are continuous and bounded for q > 0, v ^ 0, the integrals s{ (i = 1 , 2 , 4, 5) are conver gent for any a > 0, then the integrals S f X ) (i = H , . . . ,6) and its first derivatives are localy uniformly convergent for every X e Z.
P ro o f . By Krzyzanski(x), p. 326, and from the assumptions concern
ing the functions F { (i = 1,2) we get the thesis of Lemma 1 0.
Le m m a 11. Let the assumptions of Lemma 10 be satisfied; then the functions Tj (j = 1,2) satisfy the homogeneous boundary conditions
(a) limВ {хТ{+1(Х) = 0 as X -> (0 +, yQ), yQ> 0 (i = 0, 1) and
(b) Н т(Н у + Ь)Т{ (Х) = 0 as X-+(æ09 0+), x0 > 0.
P r oo f . W e shall give the proof for the function Tx; for the function T2 the proof is similar. By uniform convergence of the integrals (i = 1, . .. , 6) and its first derivatives in the set Z follows the continuity of these functions in the set Zxu Z2 and consequently we get
6
UmTx(X) = £ 8t(0 , y0) = Jf F ( Y ) G X(0, y0, s, t)dY = 0
*= 1
as X -> (0+, у0), Уо> 0 ,
Poisson equation in the quart-plane 109
and
+ = j j F { Y ) ( D v + h)G1(xQ, 0 , s , t ) d Y = 0 as X -+{x0, 0+), x0 > 0 because Ox(0, yQ, s, t) = {Dy + h)Gx(xQ, 0 , s, t) = 0.
By Theorems 2, 3 and Lemmas 8-11 we obtain
Th e o r e m 4. I f the assumptions of Theorems 2, 3 and Lemma 8 are satisfied, then the functions v{ (i — 1 ,2) are solutions of the problems (i = 3 , 4 ) respectively.
