In this paper we solve the bicaloric equation(1) P2u(X) = -F(X),where X = (x,y,t), P2 = P(P), P = D2x + D2v-Dt = (Ax>v-Dt), in the qnarter-plane W — {(x, y f t): x > 0, y > 0, t > 0}, satisfying some of the limit conditions :(

Series I: COMMENTATIONES MATHEMATICAE X X (1977) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE X X (1977)

J. Musialek (Krakôw)

The bicaloric problem with mixed conditions in the quarter-plane

In this paper we solve the bicaloric equation

(1) P 2u(X) = - F ( X ) ,

where X = (x , y , t ), P 2 = P(P), P = D 2x + D 2v- D t = (Ax>v- D t), in the qnarter-plane W — {(x, y f t): x > 0, y > 0, t > 0}, satisfying some of the limit conditions :

(²) и (x, y, ⁰) = / х(я?, у),

(3) У ч 0) = / a ( » j У ) ч

(4) (Dx + h)u{0 , y , t ) = f 3{ y ,t ) ,

'(5) P ( D x + h ) u { 0 , y , t ) = {Dx + h ) P u { 0 , y , t ) = M y , t ) , (6) u(x, 0, t) = f 5{oo, t),

(7) Pu(x, ⁰ , t) = f 6(x, t), (⁸) Dvu(x, 0,f) = / 7(®, t), (9) JDyP u ( x , 0 , t ) = f 8{ x, t), h being a negative constant.

Let

(la) P 2u{ X) = 0.

We shall solve the limit problem (1)—(7) which will be called shortly the problem (B- D-M) and also problem (l)-(5 ), (8), (9), called the {B-D) problem.

Let

A t = - ( 4 ( t - » ) ) - 1( ( æ - f ) * + ( y - 4)*), A , = - ( 4 ( < - « ) ) - 1( ( æ + f ) 4 - ( ÿ - » l ) ,)>

A , = - ( 4 ( < —s))~‘ ( ( æ - f ) 2 + (ÿ + vY), At = — (4 (« -« ))-1((® + £ )г+ ( у + ч)2)>

A 5 = - ( 4 ( < - 5 ) ) -¹((ж+ | + ^{^ ) 2} + (²/ - ^ ) 2), A 6 = - ( i i t - s ^ d x - h i + vy + iy + y)2),

A 7 = - ( 4 t r ' d x - i y + iy-r})*),

A 8 = ~ ( 4 ^ ( ( Æ + É + ^ + ^ - r ç ) 2),

A 9 = - ( 4 ^ ) -¹((æ^{+ | ) 2} + (²/ - t ? ) 2), A 10 = - ( 4 ( « - s ) ) -¹(a7² + (²/-> ? )2), A n = ~ ( 4 ( ^ - « ) ) " 1 ((^ + '^у)2 + ( 2 / - ^ ) 2),

A i 2 = - ( 4 ( г - « ) ) “ 1 (ж2 + (^ + ??)2),

Ai3 = - ( 4 ( < - « ) ) - 1((ж + ^{^ ) 2} + (²/ + ??)2), A i4 = - * ) - ( 4 (t)-l ((x- ! ; ) * + y*), and

A i3 = - ( 4 ( « -s))~1( (^ + I + v) 2 + ²/2).

We shall prove in the sequel that the following functions are the Green functions for the corresponding problems

(^{1 0}) G<(X, Y) = Gx( X, Г) + ( - 1 № ( 1 , Y) (i = 1 , 2 ) , where

<?i(X, Y) = G1(exp(J.1) + exp(J.2)) + I x( X , Y);

G²(X , Y) = 0, (exp (X 3) + exp (X4)) + I 2(X, Y), and

00 ^ oo

I x( X f Y) = C2

J

J

о ⁰

Сг = ( ^ Г \ C2 — Ъ(2п)~1, Y = ( i, г], s).

Let

TTj = { { X, Y ): 0 < а г < х < a 2) 0 < а3 < у < a4-,

О < t x < t < t 2; 0 < s < t ; 0 < i < оо, О < г} < оо}, W 2 = {(X ): 0 < а5 < х < а6; 0 < а7 < у < а8; 0 < tY < i < Z2}>

and

TY³ = {(X ): |ж| < а9; |у| < а10}, where а{ (i = ¹ , . . . , ^{1 0}) and ix, ^£2 are positive constants.

We shall need some lemmas.

Lemma 1. The integrals 7*(Х, Y) (г = 1,2) are uniformly con­

vergent in W x.

148 J. Musialek

P r o o f. This follows from the inequalities exp(A5) < ¹, exp(A6) < ¹„ Let

OO

h v..k6(X , Y ) = <?2 J ehvDx%k4hs\ h s4 x

0

x exp( - (4 (t - s))-1 (x -f £ + v f (y ± y fjd v t

6

where Tc{ = 0 , 1 , . . . , 4 (i = 1 , . . . , 6) and 0 < ^ < 8.

г= 1

Le m m a 2 . The integrals I k k are uniformly convergent in W x.

P r o o f. The integrals I kV"k are composed of a finite number of summands of the form

J = Cx j ehv{t — s) J’(æ+ Ç + v)k( y±r j) le x p( Ai)dv

о

(i = 5 , 6 ) for j = ï - f h + p , p > 0, for which

Oû

J = Gx

J

0

{i = 5, 6, q = p + Z/ 2 + &/2).

We can give the function J in the form

OO

J = C j ehv( x + ^ + v ) 29( 8 ( t - s ) ) ~ 9e x p ( - { 8 ( t - s ) ) - 1(x- h | + v)2) x

о

x(4(tf — s ) ) ~ ik( x + l + ^)*exp( — (⁸(f — e))-¹(a?+ £ + v)2) x

X (4 (t - s ) ) ~ il( y± ry)?ex p ( - (4 (t - s ) ) ~ 1( y ± y f ] ( x + | + v)2qd v.

Applying three times under the sign of the integral the inequality (11) ^{f ( A )} = ^{А е ~ л} < ^{a ae ~ a} for ^A > 0,

where a is a positive constant, we obtain eventually OO

\J\ < ССгС2Сг j e^ia. + v r ^ d v^{o o ,}

о

where C = 2z+3a+i3ft, Сг = gae~®, Ca = { Щ * ке~*к, Cz = { \ l ) hle~*\ which implies our assertion.

F om Lemma ² follows

Lemma 3. F o r each (X , Y)eint Wj Йеге derivatives

I = 2>/1/2^з^4/5/б7ЛА, Г ) (г = ¹ ,²)

150 J. Musialek

and

00

I = C* f ^ 9Р х% кФ ^ г ,кФ exp[А{) dv (i = ⁶,⁶),

0

i.e., the differentiation may be performed under the sign of the integral.

Lemma 4.

P x ^ ( X , Y) = - ( t - s r ' & i X , Y).

Pr oo f. It is enough to prove this for each summand of Gx{X, Y), which is of the form дг(Х, Y) = exp(^41). We shall show that

Р х дЧХ, Y) = - ( < - * Г У № Y).

Indeed,

4„ M X , r ) = № + - W ( X , Г)

= - ( < - » Г У ( - * , Г) + (4(* — s))-2((æ — ^)2 + CJET, Г) and

D,g'(X, r> = (4(« —«))-*((*—£)* + (» — Г) . Hence

(Лх , у - Щ д х№ , Y ) = - ( « - в ) -1 вхр (Л )-

Theorem 1. The functions Gl ( X, Y) (i = ¹ ,²) are the Green functions for the problem (B-JD-3I) and (Б-D) respectively.

P r o o f . We shall first prove that the functions G{ (X, Y) satisfy the homogeneous equations

Р 2г О*(Х, Y) = P 2x G{ ( X, Y) = ⁰ (г = 1 , 2 ) , where P Y = AS>n + Ds.

By Lemma 3 and (10) for i = j

P\ G' {X , Y) — P x G1{ X, Y) = P 2x GAX, Y) —P x G2(X, Y) = 0

since

(^{1 2}) P 2x G1( X , Y ) = C lP 2x exV( Al) + ClP 2x exV(A2) + C2f ehvP 2x exV(A,)dv = 0

0

and

-^ х е х р (-(4 (< -5 ))-1(ж + л)2 + (2/ + &)2) = 0 , a, b, being functions independent of x , у , t.

Similar identity may be obtained for G2( X, Y). We must still prove that the functions Gl{ X , Y) (i = 1 , 2 ) satisfy the boundary conditions;

(13) {Dt + h)0*(X, ⁰, y, s) = ⁰ (i = 1 , 2 ) ,

(14) P b h)Gi( X , 0, rj, s) = (Z)| + Ъ,)Рх &*(Х, 0, r], s) = 0 , (15) &l ( X , f^{, 0} , s) = 0 ,

(16) PyG1( X , f , 0 ,s) = P x & ( X , £ , 0 , e ) = 0 , (17) Д , 0 * ( Х , £ , О , * ) = 0 ,

(18) D nP YG2( X , f , 0, s) = JDnP x G2( X , f , 0, *) = 0.

First we shall prove (13). By Lemma ⁶

oo oo

(19) D * ( J eÜB(J.i)dt^{? ) f e 0} = J e ^ i ^ e x p ^ d ®

0 0

oo

= — exp (A {) — /& J еА®ехр(Аг.)Й⁰ (г = 1 1 , 1 3 ) о

and oo

(19a) Gk( X , 0, rj, s) = 02ехр.(Д-) + ^ 2 / eA” exp(J.y)d??

о

(Л = 1, i = 10, j = 11; j = 13, h = 2, i = 12).

By (19) and (19a)

D sGk( X, Y ) ^ 0 = — (2 (t —e)) L?exp(A,) + (2 ( / - s ) ) ^ e x p ^ ) +

oo OO

y (d s f ehvexp{Ai)àv)è=Q = ( - е х р ( ^ ) - Л

J

о о

(j = 1 1, S = 10, i = 5, к = 1; j = 13, s = 12, i = 6, h = 2),

whence •

(Dt + h)Gt ( X , 0 , f i , 8 ) ^{= 0} (f = ¹ ,²) which proves (13).

Next

Р г в*(Х, Г) = P x G‘ ( X , Y) = ( t - s y ^ Q ^ X , T) + ( - l ) ’ G²( X, Y))

which implies together with (13)

(г = 1 ,2),

(-Dj + A J P ^ X , X) = ( ( - s ) " 1 (X>{ + ft)(GMX, r i + t - i y e ^ Y , Y)),_„ = 0 (t = ¹,²), so we have proved (14).

152 J. Musialek

The proof of (15) and (17) is immediate, and this of (16) is similar to this of (13). To prove (18) we apply Lemma 4 and condition (17) which gives

D nP x G2{ X , 0, s) = (il - 8) - l DnG*{X, f , 0, 8)

= ( t - s ) - 1DvG2( X , Ç , 0 , 8 ) = 0 .

Now we shall prove that under some conditions imposed on the func­

tions (i = 1 , 6), the function

7

(20) u ( X ) = £ u t( X) ,

1 = 1

where

00 oo

« , ( * ) = j f Ш , г , ) Г 1в ' ( Х, S,v,0)d(dfi;

0 0 oo oo

Щ(Х) = / j W , r t)e1{ X , t , v ,0)d(dn,

о 0

and

k(£,v) — j (£, y, 0) = f 2(£,v) — du1(£,T},0),

t OO

Щ( Х) = - / / M v , s ) ( t - s ) - 1Gl ( X, 0, r}, s) dri ds>

о 0 t oo

= / / ЛОь Vi 8)dvjd8;

о 0 / oo

«s(-Ï) = f f f s ( ( , s ) D„ ( t - s ) - , G1(X, ^, 0, s)dSds,

0 0

£ oo

«.(-У) = / f f 6(S,s)V^e1( X, S, 0, s)di ds;

0 0

t oo oo

щ(Х) = f f f Х ( Т ) в 1(Х, Y)dY,

0 0 0

is a solution of the problem (B-D-M), and that the function

7

(20) V { X ) = £ v { ( X) ,

1 = 1

where

OO OO

u,(X) = f f ; , ( f , v)t~'02(X, f , n, 0)dtdv;

0 0 oo oo

U2( X) = f

J

о 0

*(f,4> = M £ , 4 ) i > . U A t , v , 0 ) = M £ , v ) - A V 1( t , v ,0),

t oo

v, (X) = - / / Мг ) , » ) { 1 - » г ' < ? ( х , о , ч , » ) ач а 1 ,

о 0 t oo

U,(X) = f f ft (V, s ) e 1{X,0,ri,s)dridS-,

0 0 t oo

u5w = f j m s, t ) ( t - s ) - ' c p ( x , e,o,s)dids-,

о 0 t oo

U„(X) = f f / , ( ( , . »)<P(X, S,0,s)dSds; U,(X) = « , ( Z ) ,

0 0

is a solution of the problem (B- B).

Let us first solve the problem (B-D -M ); let

6 6 oo 00

u, (X) = = L G‘ / J /■(*>n)s.\dsdr,,

г = 1 î = l 0 0

where

с < _ \ ° 1 for i = 1 , 2 , 3 , 4, 1

j

K\ = r ^ x p ^ ) ( i = 1 , 2 , 3 , 4 ) , and

OO

K\ =

J

A

let also

u

« U ( Z ) = c[

J

0 0

where p , q = 0 , 1 , . . . , 4, r = 0 , 1 , 2 ; 0 < p + g + r < 4 .

In the sequel M will denote a quantity greater then sup |_F|,sup \Afx\r SUP Ш (* = 1 , 6).

Lemma 5. Let the functions (i = 1, 6) be continuous and bound­

ed; then the integrals u\pqr {i = 1, . .. , 6) are uniformly convergent in W2.

P r o o f . We shall prove this in the case’ i = 1 ,3 (the proof in the=

remaining cases is similar).

There exists а й > 0 such that for I2 + rf ^ R2 we have 4 4_ 1 ( £ 2 + »?2) < ( £ - x ) 2 + ( r i - y ) 2 < 4( | 2 + г72)

154 J. Musiaiek

for each (x , y ) e W 3. Let OO00

J f = £ 1 / f \ f A t , v ) t - 4 * - ! ) 4 y - v r \ * M ) c i t d v , R R

where k , m , j are non-negative integers satisfying Zc + m < j . The inte­

grals u]pQr are composed of a finite number of summand of the form J i

= c, / /

J / <ri(^2 + »?2)i(fc+m )exp(-(16<2)-1(|2 + ^2))d|^.

R R

Introducing the polar coordinates

I = QCOStp, Г} = ^»sinç9, R Q < OO, 0 < Ç5 < тг/4 we get

op л/4

jf<iU0J J (^ГУ*+т+1^(-(1бУ~У)^<р < e

R

0

for R > R0(e) and every X e W 2 (M0 = 4 :0 ^ ).

This inequality implies the uniform convergence of the integrals u\pQr in W 2.

Let

0 0 0 0 CO

J f = 0 2 f f f l/i(£, y ) e hvt j ( x + £ + v ) k( y - r ) ) m \ e x p ( A s) d v d £d r } .

"

0

‘The integrals are composed of a finite number of summands of the form

00 00 00

J

/ / / /i(£>

+

0 0 0

Jc, m, j satisfying the same conditions as in J f . The integral J f satisfies

00 00 00

|Jf|< 402i f J

J J

0

Introducing again the polar coordinates, we obtain for R > R 0(s) and X € W 2

00 тг/4

l</?KC4 J / (f{r¹e<*+m+,,exp(-(16y-Y)de<üf>,

R

0

where C4= — 4C2Mh x. The integrals u\pqr are uniformly convergent in W 2.

This implies the assertion of our Lemma 5.

Lemma 6. Under the assumptions of Lemma 5 the derivatives Dxpyqtru\ (X) (i = ¹, . .., ⁶) exist and

6 oo oo

D xpvqtr U x { X ) = C i

J f

i=l 0 0 where

c = I 0i /<w » = 1 , 2 , 3 , г ~ | 0² for i = 5 , 6 , and 0 ^ p + q + r < 4 , p, q , r = 0 , 1 , é.

From Lemmas 4 and ⁶ and from (12) we can easily deduce the Lemma 7. Let the function f x be continuous and bounded] then the func­

tion ux satisfies the homogeneous equation (la).

P r o o f . By Lemma 5, (12) and Lemma 4

OO oo

ргщ(Х) _{= o J} f Аа,ч)Ргх (г

0‘(Х, s,v,o))didt, _{= o,}

0 0

since

P\Ql ( X , f , , , 0) = P x [Px Gl ( X , I , t,, 0» = P x ( r¹ e¹ ( X , f , ч , 0)) = 0 which implies that

Р ^ Г 'в Ч Х , f , , , ⁰)) = Р ^ Р ^ г ' е ц х , f , , , ⁰))) = ⁰. •

L e m m a 8 . Let the function f x be continuous and bounded and let x0 > 0 ,

Уо> 0 ; then the function ux satisfies the initial condition (2a) limux(X) = f x{x0, y0) as X - > { x 0, y0, 0+),

P r o o f . We shall check that lim ^ (X ) = Уо)

0

Let

for i = 1 ,

for i = ² , as X ^ { x 0, y0, 0+).

Ш ^ г , ) 1°

for ( Ç, f ) ) €W0, for (£, r))eE2\ W 0, where W0 = {{Ç, y): £ > 0 , rj > 0}. Thus

+ oo + o o

»!( X)=o

f f Mt:,n)K]didri

— OO —00

156 J. Musialek

and by the Weierstrass theorem (see Krzyzanski(1), p. 454) limtti(X) =Л( ж0»Уо) as Уо» 0+)- Next, we prove that lim ^ (X ) = ⁰ as Х->(ж0, yQ1 ⁰+).

Indeed,

00 00 Oû oo

К С Х Ж О Д / f = СгМ f f Г ¹е х р ( - ( 4 « ) -¹(ж+^)2) ^X

0 0 ' 0 0

x exp( — (±t)~l(y — rj)2)dtjdrj.

Introducing new variables in this integral

x + i = 2 t i z1, У — y = 2tlz2 we obtain

oo 00 _ 2 _ 2 °0 2

\ul(X)\ ^ — ±CXM J J t ^ e Zie Z2dz1dz2^. C5 J e Zldzx-> 0

0 Aj Aj

as X —>(я?0, Уо, 0+), where = œ(2ti)~1 and ^{0 5} =p — 4(71Ж (я)4.

Let us check also that limit*(X) = 0 when X -> (x0, y0, 0+). Indeed,

oo oo oo + 0 0

|«5(X)| < MC2 f f K\d^drj = G6 J j t- 1 exp (A9)

0 0 0 —oo

where (76 = — h~lC2M. Arguing similarly as for the integral u\ we deduce th atwü(X) - > 0 as X -> (#0, y 0, 0+).

The proof that u\(X)->0 (i = 3 , 4 , 5 ) is similar to the case i = 2 , 5 . This proves our Lemma ⁸.

Lemma 9. Let the function f x be continuous and bounded and let x0, yQ, t0 be positive numbers ; then the function ux satisfies the homogeneous boundary conditions

(4a) (5a) (6a) (7a)

\im(Dx -\-h)u1(X) = 0 lim(Dx + h)Pux(X) = 0

lirrn ^ X ) = 0 lim P % (X ) = 0

as X - » (0+, y0J t0),

as X —>(a?0, 0+, tf}.

Æ) = v P r o o f . By Lemma 6

OO 00

lim {Dx + h)ux{X) = f J Л(|, rj)lim (L>x + h)t~1G1{ X , £, vj, ⁰)dgdrj

p) M. K r z y z a n e k i , Partial differential equations of second order, vol. I, W ar­

szawa 1971.

OO 00

= < - * / / f 1( l , v ) ^ ( I > x + h)(O1{ X , l , n, 0 ) ~ e 2( X , ( , v ,0))dldr1 0 0

oo oo 0 0

— t 1 f f f i d , r])\im(Dx +h) G2( X, Ç, ⁷⁷, 0)d£dr) = 0 0 0 æ_>0 +

since, as can be seen,

+ £, rj, 0) = 0 as X->-(0+, t/0, <0) for i = 1 ,2 which proves conditions (4a).

The proof of (5a) is similar. To prove (⁶a) observe that

lim fl^ X , I,»?, ⁰) = lim(Gx(X , |, ⁷⁷, ⁰) - G a( X, f , *⁷, 0)) ^{- 0} as y-> ⁰+ , whence by Lemma 5

00 00

limwx(X) = <7X

J J

о ⁰

The proof (7a) based upon Lemma 4 is similar:.

Now we shall prove some lemmas dealing with the function u%.

L^{e m m a} 10. Let 1° the functions f { (i = 1 , 2 ) and Afx be continuous and bounded, ²° / x(£, ⁰) = /¹(⁰,¹⁷) = ^ /¹(⁰ ,⁷⁷) = A , / X( £ ,⁰) = ⁰ , and 3° *(£, 0) = £ (0,77) = 2W 0,iy) = D ri1c(i, 0) = 0; t o

lim ^ (% + w2) = /2(®0, y 0) «s X -> (#0,7/0, 0+).

P r o o f . First we prove that

(o>x) limzl% = ⁴ /i(æ0,t/0) as Х -> (ж0, t/o, ⁰+).

By Lemma ⁶

Аиг = f j f iiëiV)* ^Z'yG'iX, Ç,rj} 0)d£dr}

0 0

/ ^{00 00}

= / / / i (£, £, V, 0)dÇdrj 0 0

00 00

= / / A ( f , 4 ) - » t G ‘ ( - ï . î , 4 , 0 ) d f ^ +

о о

oo oo

+ /

0 0

158 J. Musialek

Integrating by parts and applying 2° we obtain

oo oo

Aux =

j

о 0 where

00 OO 6 6 oo oo

*, = / / лш,п)^к\^йп = y r <; Tt = f f АШ ,п)К\ащп.

О о г=1 ï = i о о

%

То prove (coj) we shall verify that 4Л(Юо,Уо) for i = 1 lirn l7,- =

0 for г = 2, . .. , 6 as X ^ { x 0, y0, 0+).

Let

A f A i , v )

Af i t t i V) for (£, W0,

0 for {£,fj)€E2\ W 0.

By 1°, applying the Weierstrass theorem we obtain

limTj = lim f

j

— OO —00

Next

00 oo

\Тг1 < M

j J

oo oo

<

^J

1(я?+

0 —OO -- *

Introduce now variables y — y = 2 tiz1, æ + £ = 2ti z2, then

\T2\ < M

J J

J

кл —oo

as Х->(я?0, y0, 0+), The integral K 5 satisfies

00 OO " OO OO 00

|T5| < M

J J J

J

0 0 —oo 0 —oo

Arguing similarly as for T2 we obtain T5->0 as X - > (x Q, y0, 0+). In the case i = 3 , 4 , 6 the proof that T{->0 is similar to the previous cases»

By M

(*>*) W , v )

It is easily verified that Au\ — Dtu\ (i = 1, 6) which implies

(<*>3) Aux — Dtux.

By (i = 1 , 2 , 3 ) , 3°, 4°, Lemma 6 and by the Weierstrass theorem

OO 0 0 0

Щ щ + и2) = Dtul X D tu2 = Au1 + L>t \t

J J

0 0

oo oo

= [zl%+

J

£,

oo oo

+ tI}t(f J k(Ç, 7i)t~1G1(X, if Г), 0)d£dr)}\

о 0

- >A f i(x

0

0

= 4Л (a?o > Уо) ^{+ / 2} (®0 » Уо) - Afi ^{(® 0} ? Уо)

= /а(®о»Уо) as 1 ->(ж0,2/0,0+).

Similarly as Lemmas 7 and 9 were proved making use of (12) and Lemmas 4 and 6 we can prove

Le m m a 11. Let the function k(tj, rj) be continuous and bounded; then the function u2 satisfies the homogeneous boundary conditions (4)-(7) as well as equation (la).

The lemmas to follow concern the function u3.

Le m m a 12. Let e > 0 be arbitrary, then there exist positive constants R

and т5 > 0 such that

oo t m к

I —

J

R t - ô

for each X e W 3.

P r o o f. Introducing now variables rj = 2 (t — sŸz we obtain

oo t t oo

1 = 2 J f (t — s ) 2 e x p (—z2){t — sŸdzds = 2 f ( J e x p ( —z2)dzj(t — s) lds,

k0 t - ô t - ô k0

where l = \{m + l) — k > 0 and k0 = R(2(t — «)*)_1, whence

t 00 t

2

J|J

t—ô k r .%0 ^{t - i}

Multiplying and dividing the integrand by R21 and applying inequality (11) we get

t t

I ^ M 1 ^ {R 2{ t - s ) - l)l^ x p [ - R 2[ 4 ,{ t- s )) - l) d s ^ M 1G Jds = M ^ d K e ,

t - ô ■ t - ô

when R > R0{e) and for each X e W 3, with М г = 2R~21.

160

Lemma 13. Let the functions fj ( j = 3, . . . , 6) be continuous and bounded;

then the integrals

t oo

V, = / j f l (n , s ) D i { ( t - s ) - 1Gl ( X , 0 , v ,s))d^ds,

0 0 t oo

F s = J 7 f l (V, s ) D l ( ( t - s ) - se 1( X , 0 , v ,s))drids,

0 0 t oo

v 3 = f f Mv i s)B t( { t - s ) - 1G1(X , 0, r},s))drjds,

о 0

where i = 1 ,2 , j = 3, . .. , 6 are uniformly convergent in W 3.

P r o o f. We shall prove this for the integral Vlf the proof for the remaining ones runs similarly. Let ns observe that Vx is composed of a fi­

nite number of integrals of the form

t oo

V L ==Gi f J f j ( v , s ) ( t - s ) - k ( x ± r ] ) m e x p { A n) d r ] d s

(n = 1, ..., 4, j = 1, ..., 6) or

t oo oo

Hm = c2 f J f fj(y , s)ehv{t~ s )~ k(x + £-\-v)mex-p(An)dr}dsdv ( n = 5 , 6) ,

0 0 0

where m, Jc are positive integers and m < k . Transforming V\m into the form

t oo _ m m j.

v i m = C i f f f j ( y , s ) ( t - s ) 2 ( а ? ± £ Г в х р ( - ( 4 ( < - в ) ) _ 1 ( < в ± £ ) 2 ( < - в ) 2 ) x о 0

X exp( — (4 (t — s))-1 (y ± rjf) dr) ds and making use of inequality (11) we obtain

oo t m к

W i j < М С ХС J f ( t - s ) 2 e x p ( -( 4 (t — s ) ) - 1( y ± y ) 2j dr)ds о 0

and from Lemma 12 it follows that the integrals V\m are uniformly con­

vergent.

Analogously

j OO CO

\Hkm\ ^ C 2MC f j J ehv(t — s)~zexp( — (4(tf — s))~\(y±r))2]jdr)dsdv

0 0 0 t oo

= C7

J

о 0

where 0 7 = —h~1C2MC.

J. Musialek

By Lemma 12, H * is uniformly covergent, which concludes the proof of Lemma 13.

L^{e m m a} 14. Let f 3 be a continuous bounded function; then

t с о 6

lim

J j

0 0 i = l

OO

where = (/ —s)-1 exp(J.;) {i = 1 , . . . , 4), = f ehv(t — s)~l exp(A i)dv {i = 5, 6 ).’

P r o o f. It is enough to prove that

OO

g = lim j \fsirj, s)\Nxdr} = 0 as 0

OO

(for the integrals / |/3 (r), s )X i\dr]-^0 (i = 2 , . . . , 6 ) the proof is analo- o

gous). Let us observe that

OO OO

g < lim M j {X) dtj < lim M J (t — s)~1exp( — (é(t — s))~1(y — y)2jdy

0 0 1

as s —> t , for exp( —(4(£ —s))-1 (x — |)2) < 1. Upon the change of variables

(21) y - r j = 2 ( t - s ÿ z ,

we obtain

g < lim ( —2ill) J (t — s) i e z2dz < lim ( — 2Ж)

J

= lim 2Ж (t — s)~i e~k2 <C lim2My~2 (t — s)i ( t —s)~ly2exp[ — (4:(t — s))~1y2'}

< 2<7JIlimy_2(tf — sÿ = 0 as s->t, yo > 0, where &2 = y(2(t — s)*)-1, since by (11) we get

/ ( J - s r ' e x p f - ^ - s ) ) - 1^2) < C.

Lemma 15. Let the function / 3 satisfy the assumptions of Lemma 13.

Then the function иг satisfies the homogeneous equation (la).

Proof. By Lemmas 5 and 13

t OO

Лиг = - J J f 3(i}, s)(t — s)~l Ax G1(X , 0, rj, s)drjds1

о 0

П — Roczniki PTM — Prace Matematyczne X X

162 J. Musiaïek

and

OO

Dtu3 = lim — f f z{rj,s){t — s ) - 1G1{ X , 0, y, s)dr} —

0

t oo

- / / 0, ??,s))<b?<fe.

0 0

We have

OO

lim J f s{ r ) , 0 ) ( t - s ) ~ 1G1(X , 0, y, s)drj — 0 as

0

Applying Lemma 14 we obtain t OO

Pw3 = Л«в — = —f f f 3(y, s)Px ((t — s)~1G1(X , 0, fjt s))drjds = 0,

о 0

since by (12) and Lemma 4

P A. ( ( ( - « ) - < ? ( * , 0, v , *)) = - Р * х в ' ( Х , Y) = 0.

Thus Pu3 = 0 implies P 2ua(X) = 0.

L^{e m m a} 16. Letf Й е function / 3 continuous and bounded and let xQ> 0, y0 > 0 ; Йе?г the function u3 satisfies the homogeneous initial con­

ditions

(2b) lim'Wa(A) = 0

(3b) lim Dtu3(X) = 0

P r o o f. We first prove (2b). Let

t oo

as X ^ { x 0, y0, 0+).

я* = Cl f f fa(Vt ( » = 1 , 5 ) ;

then

< oo

IW1! < (7xJf J J (£ —$) 1exp(A10)d^ds

о 0

t oo

= GXM j j {t — s)-1 exp( — (4(t — s))-1a?2)exp( — (4(t — s))—1 (3/ — rj)2)dr]ds.

0 —oo

Upon the substitution (21), and since exp(— (4(t — s))-1 a?2) < 1, we get

t oo

IA1! < 2CXM J J (t — s)~i e x p ( —z2)dzds = — 2(75(£)*->0 as 0+ . Similarly

t oo

|W5| <

*exp( — (4(<

s)) ^ jx

0 —oo

x exp( — (4(< — s))-1 (y — rj)2}dyds

whence

|N5| < 08 (<)*-> 0 as <->0+ ,

where Cs = —2С1М{тс)к~1. The integrals N* {i — 2, 6) admit similar estimations. This proves (2b).

Now we shall prove condition (3b).

In this case it is enough to estimate the integrals L i -= DtN* for i = 1 , 5 . Then

3

L, = D,N4,„, =

г= 1

where

OO

Щ ^ 0 = Oilim

J

< oo

X 2le-o = - C , f f f 3 ( f j , s ) ( t - s ) ~ 2e x p ( A 10) d y d s , о 0

t oo

X3lf=0 = (I)-1^! J J /8(iy,

* о ⁰

Now

oo

l i 1! < C XM {æ)~2lim J (< —$)- 1ж2ехр( —(4(2 —s))-1 #2) x

— 00

x e x p ( — (4(2 — 8))~1{у — г))*^г) as s-+t.

Introducing in this integral the substitution (21) we get by (11) IX1! < C9cc~2\im(t — sÿ = 0 as s->2 {x0> 0), where 09 = СгС М (rz)K Thus

limX1|{e0 = 0 as s-+t (x0> 0).

Similarly,

t oo

|X2| < (^.Ща?)"4 J J (t — s)_2a?4exp( — (4(2 — s))-1a?2) x

0 —oo

x exp( — (4(2 — s))_1(y — rj)^dr]ds.

Applying again the substitution (21) we obtain by (11)

|.L2| < C10( a r 4(<)3/2-*0 when 2-^0+ (a?0 > 0), where C10 = СгСМ, whence lim X2|^==0 = 0 as X -> {x0, y 0, 0+).

164 J. Musialelc

For L 3 we obtain

t oo

|Z3|< Oi-Mé"1 J J ( t - s ) ~ 3 ( x2 - \ r ( y - r i ) * ) ( x * + ( y — ^)2)~2exp(|J.10) x

- ' 0 —oo

x exp(|A10)dî?ds.

By (11) and since (^{x 2 À { y}—t?)2)2:^ # 4

t 00

|£3| < 4_1O10 J J a?_4e x p ( - ( 8 ( « - s ) ) _1( y - r j ) 2)^ d s .

0 —oo v

Introducing now variables

(22) v j - y = ( S ( t - s ) Y z

we get

t o o

|L3[ < 4-1 O10æ- 4 J (8(tf —s))*ds j exp( — z 2) d z = ; G 11x ~ i (t)3l2- + 0

0 —oo

as t—>0+}

where C tl = 3(2)~3/2(79 and x 0 > 0. Thus Ы т . Щ ^ ^ 0 — 0 as X - + ( x 0 , y 0 , ) Let us consider now the integral

L, = D,N*\e_0 = £ q( , i= 1 where

Qi — lim(72 J J fa(r), s)ehv(t — s) 1e x p (A 11)dvdr) as s~>t,

о 0 t oo oo

Q2 = G2 j J J Mrj,8)ePv(t — 8)-2exp{A 11)dvdr)d8, and

t oo oo

Q3 = 4 / J VbOl 8)(Pv(t — 8) 3((x + v)2 + ( y - r i ) 2)ex p (A 11)dvdr;ds.

0 0 0*

Then

(X ) OO

IQil < (7?Jflim

J

■ S_>< 0 -00

X exp( — (4ь(t — s))-1(²/ — rj)2}drj.

Arguing as above we obtain

OO

IQil < 0 7a?~2lim J (t — s)i exp( — z2) d z = 2 G 7x~2lim(t — s)i = 0 as s-rt

thus lim^i = 0 when s-+t {xQ > 0). Next

t OO -

IQal < 2C1x~4 j (t — sÿds j exp ( —z2)dz = C12x~4(t)3l2-+0

0 — oo

£IS t— Xq^> Oj

where 0 12 = 3(тс)*С7. Thus П тф 2 = 0 as X-»(a?0J y0, 0+) (a?0> 0 ) . Finally,

t 00 00

1 « . К 0 И| f J ^ ( t - s)->((x + vŸ + ( y —nf Y [( x + vY + ( y ~ n Y Y ^

0 — oo 0

x exp( A 1X) dv dr} ds, where C13 = é~'C2M. Since [(x + v)2 + {y — »?)2)~2 < a?-4, applying (11) we get

t 03 oo

1ФзК 0 0 l3J f f ehvx~4e x p ( - ( 8 ( t — s j j -'x 2) x

0 — OO 0 •

x exp( — (8 (t — s))~l {y — r])^dvdr}ds.

Making use of the substitution (22) and since exp( —(9(t —s))-1 #2) < 1

t oo

|Q8| < 0 14a r4 J f ( t - s ) i exj)( — z2) d z d s = C u (n)ix~4(t)3/2->-0 as i-> 0+ ,

0 — oo

where Cu = — (2)~i CC2Mh~1. Thus lim$a = 0 when X -+ (x0, y0, 0+) (#0 > 0). This proves that Dtu3-> 0 as Х->(ж0, y0, 0+) (ж0 > 0).

L e m m a 1 7 . L e i i > 0, x > 0; i/h m

t oo

^ (X ) =

J

— 0 0 —00

P r o o f. Since

OO oo

Q(X) = 4

J(-a7(a?2 +

=

J

+

- O O -OO

Applying the substitution т/ — у = tx, we get

OO

G (X) = 4 J (1 + i2)-1 di = 4tt.

— 00

L e m m a 1 8 . Let the function / 3 be continuous and bounded and let Уо > 0, t0 > 0 ; then the function % satisfies the mixed boundary condition

(4b) Iim(2>e + fc)w8(X)* = / 8(y0,<o) as X -> {0 +, y 01t0).

166 J. M usia lek

P r o o f. By Lemma 5

t 00

Dxu2 = - J f s)(t s) lDxG1( X , O t Г], s)dr}ds.

Now

DxG1 {X, 0, rj, s) —F xGx(X , 0, rj, s) I)xG2(X , 0, rj, s), where

DxGx{X, 0, rj, s)

oo

= — Cxx(t — s)_1exp(JL10) — O2exp(J.10) — C2h j ehvexj)(Axx)dvf 0

Dx0 2{ X , Q , rj, s)

OO

= —Cxæ(t — s)~1ex.Tp(A12) — C2ex.Tp(A12) — C'lh j ehvexp(Alx)dv, 0

moreover,

Ш Ц Х, 0, ff, s) = hGx(X , О, Г), s) - hG2( X , О, ц, s)

00

= 2Cxhex-p(AX0) + iiC2 J ehvexp(Axl)dv —

Hence

Thus

where

— 20x h exp {A X2) — W 2 J ehv exp ( A X2) dv.

0

(Dx + h)0‘ (X , 0, ij, s) = DxQ '(X , 0, n , S) + hG'(X, 0, v , s)

= — ^ ( « - « Г Ц е х р ^ м ) —exp(J.12)).

(Dx + h)u9(X) = F x(X) + F 2(X),

i oo

F x(X) = 0 XJ J f 3(4 , s )o c (t-s )-2exip(Aw)drjds, 0 0

t oo

F 2{X) = - C x f f /3(97, s ) æ ( t - s ) - 2ex p (A 12)d4ds.

We shall prove that

(*) limJ71(X ) = M y 0, t Q),

(P) lim-P2(X) = 0 as X~>(0+, y0, t0).

Let us prove first (a). Consider four regions:

Zx = (0 b « ): Vo~à < П < 2 /o + <5, t0- ô < s < t < t0+ 0, t > t 0},

Z % = {(*h 8) : y 0+0 < rj < 00, \y — y2\ < é ~1ô ,t0- ô < s < t < t Q+ d , t > t 0}, Z3 = {{y,s): - o o < r ) < y 0- ô , \y - y 01 < ^{4 - 1} Ô Л - ô < s < t < <0+ à , t > t0} y Z4 = {rj, s): — 00< 7] < + 00, 0 < s < <0 — <5, t > t0} 1

where Ô is positiv number. Then

Z = {(rj,8): — 00 < ^ < + 0 0, — 0 0 < 8 < t } = Z1yjZ2\jZiuZA.

Now

t 00

F 1( X ) = C 1

J J

— OO — OO

£ 00

+ ^1/»(УоЛ) / / Æ (« -s )-2exp(.410)<fyds

— СО — со

whence by Lemma 17

$ = F 1(X) /з(Уо? ^0)

< oo 4

= Cl f f (f3(y, s) /3(2/0> * o )W < -s r 2exp(J.10)<fyds = £ s *i

— 00—00 <■=!

where

/ / (Л(»?, » ) - / 3(2/o, <e))®(*- « Г 2exp(A 10)dr)ds (i = 1, 4).

W e must show that

limfl = lim (-Fj(X )-/з (2 /0, <0)) = 0 as X -> (0 +, y0, <0)- Let e > 0 be such that

1/зОЬ«)~/з(2/о,*о)1 < e for (rj, 8)*ZX\

then by Lemma 17

t 00

\8г\ < Ог j j x{t — s) 2exp(A10)drjds = e.

—00 —00

Next

00 t 00

|$2| < 2CXM f

J

V + ô t 0 - ô y+i<5

168 J. Musia lek

- Transforming the variables г) — у -j- tx we obtain

00

I I < —8CXM f - ( t z-}-l)~1dt->0 as x->0+ ,

*3

where k9 — d(4x)~\ Thus the integral S2 is arbitrary small. Analogously we obtain lim $3 = 0 as ж-^-0+ .

Now

OO <Q — d

|$4| < 2

— oo 0

Introducing the now variables (21) and by exp( — (4(i — s))_1a?2) < 1 we get

oo t q — (3 t q ~~ Ô

[Si \ ^ 4 M C 1x J J (t — s)~3/2exj)( — z2)dzds = —C5x J (t — s)~3/2ds

— oo О 0

< - С 5х ( Г * - 0 ~ * ) ^ 0 as x-+0+ .

Therefore $4->0 as x-> 0 +. From the above estimations it follows that lim F ^ X ) = / 3(y0, f0) ш X - + { 0 +, y 0, t 0),

i.e., (a) holds.

To prove (b), observe that

t 00

\FZ\ ^.CxMx J J (t — s) 2 exp (A12)drjds 1

о 0 t oo

= CxMx j j { t - s ) ~ 2(i1A y ) i e x ^ { - ( 8 ( t - s ) ) - 1(yAr))2) x

0 0 t

x exp( — (8(t — s))-1(^ + ²/)2)(t/ + ^)~4exp( — x 2(4(t — s^^drjds.

Since •

( < - s ) - 2(2/ + 7?)4e x p ( - ( 8 ( t - s ) ) - 1(2/ + ?y)2) < G and since exp( — (4(« — s))-1#2) < 1 and < y, we have

t 00

\F2\ < G 15 j j ex-p(-(8{t-~s))~1(y + r))2)df]dsy

о 0

where C1S = C1MC(%y0)~4'. Introducing a new variable by (22) г

t

|F2| < C158*(7c)*o? j (t — sÿds = x{t)3l2-> 0 as x-+0+ , t-^t0> 0,

0

where (716 = — 3~1(2)2l3GxCM(y0)~i (v:)i . Therefore (^) is satisfied.

Lemma 19. Let the function / 3 be continuous and bounded and let xQ > 0, y0 > 0, t0> O', then the function uz satisfies equation (la) and the

homogeneous boundary conditions :

(5b) \imP(Dx -\-h)uz{X) = lim (D x + h) Pu Z(X) = 0 as X -> (0 + , y0, t0),

(6b) limw3(X ) = 0

(Tb) limPw3(X ) = 0

P r o o f. Similarly as in the proof of Lemma 6 we can show that as X->(œ0, 0+, t0).

t 00

Р*Щ(Х) = - f j M n , s ) P m t - s r ' e ^ X , 0 , v ,s))drjds = 0

thus u3 satisfies equation (la). By Lemmas 5 and 14 Puz{X) = 0 whence (1)х + Ь)Риь(Х) = 0 which implies the boundary condition (5b).

By uniform convergence the integral uz is continuous at у — Or whence

t oo

lim itj(I) = — j j M V , s)(t — s ) - 1G1(xQ, 0, «о, 0, rj, s)dr)ds — 0

о 0

since G1(x0, 0, t0, 0, rj, s) = 0. Thus we have proved condition (6b).

Finally, (7b) follows from the identity

t OO

Pu3(X) = - / / Ш , s)Px ( ( t - s ) - lG '(X , 0, v , s))dnds = 0 0 0

which, in turn, is a consequence of (6b).

Lemma 20. Let the function / 4 be continuous and bounded and let x0 > 0; then the function u^ satisfies equation (la) and the limit conditions : (2c)

(3c) (4c) (6c) (6c) (7c)

lim^4(X ) = 0 lim Dtu4(X) — 0 lim (DxJrh)ui (X) = 0 lim (Dx + h)Pu4 (X) = ft(y0, t 0)

limM4(J ) = 0 lim Pu4(X) = 0

as X ^ ( x 0, y0, 0+), y0> 0,

as X -^ (0 +, y0, t0), y0 > 0,

as X -+ (x0, 0+, t0), t0 > 0.

P r o o f. Making use of Lemmas 5 and 12 we can cheek the initial conditions (2c), (3e) similarly as (2b) and (3b) of Lemma 15. The bound­

ary conditions (4c), (6c) and (7c), result from Lemma 8, analogously as (4a), (6a) and (7a). The proof of (5c) is similar to this of (4b) in Lemma 17.

By Theorem 1 and Lemma 17

t oo

' Р*щ (Х) = / / f i {n , s ) P 2x e 1( X , 0 , n,t)d nds = 0 0 0

since, as follows from Theorem 1, the function G1(X , 0, rj, t) satisfies (la).

170 J. Musialek

L^{e m m a} 2 1 . Let the function / 5 he continuous and bounded and let x 0 > 0, y0 > 0, t0 > 0; then the function u5 satisfies equation (la) and the homogeneous limit conditions :

(2d) limw5(X) — 0

(3d) lim Dtus{X) = 0

(4d) lim (Dx + h ) u 5{X) = 0 (5d) lim (Dx-\-h)Pus(X) = 0

(6d) lim P u s(X) = 0

and the boundary condition

as X-*(a?0, y0, 0+),

as X -+ (0 f, y0, t0), as X -* (# 0, 0+, t0)

(7d) 1гаш5(Х) = f 5{xo, as X -+ {x0, 0+, Q .

P r o o f. Making use of (12), (19) and Lemma 6 it is easy to prove (as in the case of the functions % ,% ,w 3) that

t CO

P *«sm = / j M h ^ 0.

о 0

The proof of conditions (2d) and (3d) is similar to this of (2b), (3b) of Lemma 15. The proof (5d) and (7d) is similar to this of (5b) and (7b) in Lemma 18, and the proof of (4d) is similar to this of (4a) in Lemma 8. Taking into account the function

t CO

us(X) = 2 C X f f f 5( £ , s ) ( t - s ) ~ 2y e x p (A 14)d£ds +

0 Ô t o о oo

+ / J /s ( f , s ) { t - s ) ~ 2ehvy e x p (A 15) d ^ d s . 0 0 0

The proof of (6d) is analogous to this of Lemma 17.

Le m m a 2 2 . Let the function f 6 be continuous and bounded and let x 0, y0, t0 be positive numbers ; then the function u6 satisfies equation (la) and the homogeneous limit conditions :

(2e) lim%(JT) = 0

as X->(a?0, yQ, 0+),

(3e) limD<% (Z ) = 0

(4e) ]im{Dx + h)u6(X) = 0

as X -> (0 +, y0, t0), (Be) lim (Dx + h)Pu6(X) = 0

(6e) Ииш6(Х) = 0 as X-+(Xo, 0+, t0),

and, additionally, the boundary condition

(7e) \im.Pub{X) = f b{Xo,tQ) as X-*(a?0, 0+, tQ).

P r o o f. Similarly as for functions щ (i = 3 , 4 , 5 ) with, the aid of Lemmas 6 and 13 we can prove, that the function u 6satisfies equation (la).

The proof, that the homogeneous initial (2e), (3e) are fulfilled is similar as that for conditions (2b), (3b) in Lemma 15. Conditions (4e), (5e), (6e) may be proved similarly as (4a), (5a), (6a) in Lemma 8. To prove condition (7a), we have from Lemma 6

t oo

J 4 ( X ) = / / m , s)Px { D ^ ( X , f , 0, s))d(ds

0 0 t oo

= / / m , > ) D n[Px (CP(X, f ,0

о 0 t oo

= / / / « ( f , s ) D „ ( ( t - s ) - le 4 X , f , 0, »))d{<fe.

0 0

Then

t oo

Р щ (Х ) = 2CX f f f 6( i , s ) { t - s ) ~ 2y e x p {A u )d£ds-\-2Gz x

о 0

t oo oo

x f f f f 6( £ , s) ( t — s ) ~ 2ehve x p ( A n ) d v d £ d s 0 0 0

and analogically as in Lemma 18 we finish the proof of Lemma 22. Now consider the integral u 7. We shall prove

Lemma 23. I f the fu n c tio n F is c o n tin u o u s and bounded, then the integrals

t oo oo

< r(X ) = f f f F ( Y ) B xVirQ \ X , Y)d£drjds

0 0 0

fo r p , q = 0 , 1 , 2 , 3 , 4 , r = 0 , 1 , 2 , 0 < p + # + r < 4 , are u n ifo r m ly convergent in every set W z .

P r o o f. Let us consider now the integrals

t oo oo

ÜB'S = f f f D ^ O ' i X , Y )d Y

t - ô R R

for some positive В and <5. The integrals UR6 are composed of a finite number of summands of the form

< 00 00

Ur? =

J

t - 0 |||>Л |»?|>Д

t со со со

U km lv = f f f f {t — s ) ~ k( æ + £ + v) m ( y ± r j ) lehve x p ( A i) d £dr ] ds d v t - ô ^о | £ | > Д \n\>R

{i = 5, 6),

172 J. M u sialek

where Jc,l, m are non-negative integers and besides m + Similarly as in the proof of Lemma 4 we obtain the thesis of Lemma 23.

Le m m a 24. I f F is a function continuous and bounded and x0, y0, t0 are positive numbers, then the function u7 satisfies equation (1).

P r o o f. By Lemma 22 we have

t oo oo

= - f f f F ( Y ) A XfVG '{X, Y)dY

0 0 0

and

( oo oo

DtGl {X , Y) = lim J f F (Y )G 1( X , Y)dÇdv + J j J F { Y ) D tGl { X , Y ) d Y .

e_y< 0 0 0 0 0

Let us consider now the integrals

OO oo

B* = Gx J f exp (A{)d£dij (i = 1 , 2 , 3 ,4 ),

— OO — 00

oo oo 00

B j =

<72

=5,6).

0 — oo —oo

Using in the integrals B l the change of variables

(23)' g ± x = 2(t — s)i z1, r )± y = 2(J —s)*²2 we obtain

OO oo

Вг = Gx j J 4(< — s)exp ( — z\)exp( — z\)dz1 dz2->-0 as s->t.

— OO — OO

Bj (j = 5 , 6 ) satisfy the inequality OO OO 00

B j

< 02

J

=5,6).

0 — oo — oo

Using in the above integrals the change of variables (23) we have

00 oo

Bj < 0 2( — h~l) i{ t — $)

J J

— OO —00 oo oo

Because lim f f F ( X )G 1( X , Y)d£dr) = 0 as s->t, we have

о ⁰

t oo oo

Ptt7(X) = Jw7( X ) - D ^ 7(X ) = f f f F ( Y ) P x ( X , Y)dY un

t oo oo

= f f f F ( Y ) ( t — s)~1G1( X , Y)dY.

0 0 0

where

Furthermore we shall prove that

Р 2щ {Х ) = Рщ = - F ( X )

and this will be the end of the proof of Lemma 23. Indeed, by Lemmas 6 and 13

4m,« . = - / / F ( Y ) A xJ ( t - s r le ' ( X , Y))dY and

< OO OO

D , « „ = l i m - f J F ( Y ) Nididt]--lJ f J F(Y)£>, ((t- s)- lGl(X, Y))dY

0 0 0 .0

(i = 1, 6).

s-*<

By virtue of the Weierstrass theorem lim J f F ( Y )N 1d^dr] = F (X )

oo oo 0 0

as s->t and the other integrals / J F ( Y ) N id£drj^-0 when s->t for i — 2, 0 0

. . . , 6 , because of the following estimations:

oo oo oo oo

iLTi = ! J ^ F { Y ) N id^dr]\ ^ M f f ( t - s ^ e x p i A J d g d r j

0 0 — oo — oo

(t = 2 , 3 , 4 ) , '

OO OO 00 oo oo

M

2 = j J J F( Y).yfd£«fy| < Ж J J J

0 0 0 0 0

{i = 5 , 6 ) .

Using in the integrals the change of variables (23) we obtain successively M !

<

J J

0

fc4 - o o

0 0 OO

M 2 < —h lM

J J exp(

as

k ± — o o

where fc4 = ж (2(2 — 5)*) \ Hence lim — f f F ( Y ) N idÇdr) = F (X )

8—*t 0 0

(i = 1, ..., 6), From these considerations we obtain the equality:

t OO 0 0

Р 2щ ( Х ) = Р щ = Ax>yu0~ D tu0 = - f J f F ( Y ) A x>y( ( t - s ) ~ l x 0 0 0

< oo oo

Y ) ) d Y - f J J F( Y) Dl( { t - s ) - ,G4 X , Y ) ) d Y - F ( X ) ,