On a certain limit problem for the rectangular parallelepiped1. In the present paper we shall solve the limit problem for the equation (1) P2u(X,t) =f(X,t),whereP - D2x+D2 + D2-Dt, P2

ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE XXII (1981) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXII (1981)

F. Baranski and J. Musiaeek (Krakôw)

On a certain limit problem for the rectangular parallelepiped

1. In the present paper we shall solve the limit problem for the equation

(1) P 2u ( X , t ) = f ( X , t ) ,

where

P - D2x + D 2 + D2- D t, P 2 = P (P ), X = (x ,y ,z ), in the time spatial rectangular parallelepiped

В = B1 x(0, oo), B1 = (X : |x| < a, \y\ < b, \z\ < c], a, b, c being positive numbers, with the limit data

(2) u {x,y,z, 0) = f 1(x,y,z),

(3) Dtu ( x , y , z , 0) = f2 ( x , у, z),

(4) Dx u( — a, y, z, t) = f 3(y, z, t),

(5) DxPu( — a, y, z, t) = f 4(y, z, t),

(6) Dx и (a, y, z, t) = f 5(y, z, t),

(7) Dx Pu (a, y, z, f) = f 6(y, z, t),

(8) Dyu(x, —b, z, t) = f 7(x, z,t),

(9) DyPu(x, - b , z, t) — fs{x, z, t),

(10) Dyu(x, b, z, t) = f 9(x, z, t),

(11) DyPu(x,b, z, t) = / ю (х , z,t),

(12) Dzu(x, y, - c , t) = /ц (x, y, 0,

(13) DzPu(x,y, - c , t ) = /12 (x, y, t),

(14) Dzu(x, y, c, t) = / i 3(x, y, t),

(15) DzPu(x, y, c , t ) = f l4(x, y, t) .

f (i = 1,.. 14) and / being given functions.

148 F. B a r a n s k i and J. M u s i al e k

We shall call problem (1)—(15) for the domain B, the (B - N ) problem.

In paper [1] the similar problem for the equation Pu = f is solved.

2. To the solution of the (B-N) problem we shall use the convenient aid function G.

Let us consider the sequences:

(16)

*o = xo — x, x\„ = 4an + x, Xjn + i ~ — 4an — x — 2a, x2„ = —4an + x,

*2n + i = 4an + 2a — x,

Уо = Уо = У, У2m = 4bm + y, y\m+l = - 4 b m - y - 2 b , y\m = -4bm + >’, ylm + i = 4am + 2 b - y ,

Zq = Zq = z, z\k = 4ck + z, z\k+1 = —4ck — z — 2c, z\k — — 4ck + z, z2k+i — 4ck + 2c — z,

(n, m,lc = 0, 1, 2,...).

Let Y — ( p , q , r ) denote the points for which p e [ — a, à], q e [ — b ,b ], r e [ — c, c], and let

00

Gi (x, t, p, s) = U0(x, t, p, s)+ U\ (x, t, p, s)+ £ P l ( x , t , p , s ) ,

» n = 1

where

Pl = U l +1 + U U 2+ U l + U ^ ,

<?r

00

G i(x, t, p, s) = U 0 + U l + X n= 1

where

P 2 = Ul + U l v . + U l ^

+ u

G2 Cv, t, q, s) Vo (>>, t, q, s ) + V l1 (y, t, q, s) + 1

n= 1

where

Qn = K.+1 + Jn + 2 + К? + Ki + 1

G2( y, t , q , s ) = V0+ V ? + X <2^, n = l or

Limit problem f o r the rectangular parallelepiped

where

<& = V ' + Vn\ v+ V n\ l + V„\1,

00

G3{z, t, r, s) — W0( z , t , r , s ) + W l1 { z , t , r , s ) + £ R l { z , t , r , s ) , n=l

where

R ,1 = ИА, + mV 2 + w n2 + « А ,,

G3( z , t , r , s ) = W0+ W ? + •£ RÎ,

n = 1

where

R2 = и; 1 + M'A ! + м;2+ , + и А 2, and

(х, t, р, s) = ( t - s ) “ 1/2 exp [( —4(r —s))-1 (p —x‘ )2], (I 7) t, q,s) = ( t - s ) ~ 1/2 exp [ { - 4 ( t - s ) ) ~ l (q-yi,)2] , W^{z, t, r,s) = (t s)~1/2 exp [( — 4 (t — s)) ~1 (r — zj,)2]

for i = 1,2, n = 0, 1 , and l/j, = t/0, K<j = V0, = Ж0 for i = 1, Let for t > s ^ 0,

G(X, t, Y, s) = Gi (.V, t, p, s) G2 (y, t, q, s)G3(z, t, r, s), Q ( X , t , Y,s) = (t — s ) G ( X , t , Yt s).

Now we shall prove some lemmas.

By formulas (16), (17) we get two following lemmas:

Lemma 1. Let X e B l3 t > s ^ 0; then

Г Dp{V\n+ U \ n+x) = Dp( U 22n + l + U 22n + 2) = Ofor p = - a , 2° flpW i„ + i + C/I„ + 2) = Dp(U 2, + l / i + l) = 0 for p = a, 3° JMKA + KjA . ) = D,(M22„+1 + K22„ + 2) = Ofor q = -i>, 4° D'{V2U i + K2A 2) = 0 ,(4 A + M22 +1) = 0 for q = b, 5° Dr(M A + W A + 1) = A .(M £ + 1 + W£ + j) = 0/or r = - c , 6° ВДМ'гА. + М'гАг) = Dr(M£ + M£ +,) = 0 /or r = c, (n = 0, 1,2,...).

Lemma 2. L « X e B , , YeBt , t > s ÿ 0; l/ien

7° lim ^ ж(С7|и+ С ^ и + 1) = lim Dx( U 2Xn + 1 + U 12n + 2) = 0 as x -» - a , 8° lim £>x(G 1„ + 1 + G ^ + 2) = lim Dx(G i„ + l/l„ + 1) = 0 as x - + a, 9° lim Dy( V 2n + Vjn +1) = lim Dy( V 2n+l + V2ln + 2) = 0 as у -> - b ,

150 F. B a r a n s k i and J. Mu s i al e k

1<T lim Dy( V2l„ +1 + + 2) = I'm ü, 11'2‘„ + 1'^_, ) = 0 as y - * b, 1Г lim + = lim D,(WÎ, + l + Wÿ„ + 2) = 0 as z ^ - c , 12° lim Dz(W2\+1 + Wi„ + 2) = lim DAW2{ + W 2{ + ,) = 0 as z ~ c , for n = 0, 1, 2,...

Let

S r = {k> ГУ- kl ^ b> kl ^ c},

S2 = {(P , r ): \p\ < a, И < cb s 3 = {(P,€): Ipl < a» kl < b}.

3. Using some lemmas dealing with convergence of certain integrals, we shall prove that the function и defined by the formula

15

(18) u ( X , t ) = X

i = i where

u A X , t ) = A H l f x{ y ) G ( X t t, Y,0)dX,

*i

u2{ X , t ) = l4 J J J h ( Y ) G ( X , t, Y,0)dY, h(Y) = f2( Y ) - A f A Y ) ,

t

u3(X, t) = A J J J/3k , r, s )G(X, t, - a, q, r, s)dqdrds,

o s x t

u4(X, t) = A $ $ $f4(q, r, s )Q(X, t, —a, q, r, s)dqdrds, 0 Si

t

us( X , t ) = A J J J f5(q, r, s )G( X, t , л, <?, r, s)dqdrds, 0 sx

м6(Х , 0 = A J

J J

t

щ (X , t) = A J J j /7 (p, r, s) G ( X , t, p, —b, r, s)dpdrds,

0 s 2

t

u8(X, t) = A J J J f8{p, r, s)Q{X, t, p, - b , r, s)dpdrds, 0 s2

t

Ug{X,t) = A j J J fgip, r, s)G (X , t, p, b, r, s)dpdrds, 0 s2

t

uio ( X , t) = A J J J/10 (p, r, s )6 (Z , t, p, b, r , s)dpdrds, 0 s2

t

wn ( X , t ) = A l l S f1 1(p,q, s )G( X, t, p, q, - c , s)dpdqds, 0 s3

Lim it problem fo r the rectangular parallelepiped 151

u12(X, t) = A j J J f l2(p, q, t, p, q, - c, s)dpdqds, о s3

u1 3(X, t) = A\ j j/ 13(p, q, s ) G ( X , t, p, q, c, s)dpdqds, о s3

ul4(X, t) = A J j J/14(p, q, s)Q{X, t, p, q, c, s)dpdqds,

0 S 3

= A j j j j f ( Y , s ) Q ( X , t , Y,s)dYds, 0 Ht

A = (4r.)_3/2,

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

4. Now we shall introduce some notations. Let

G = g,g2g3 = U0V0W0+ X l„

i = 1

where

h = U ⁰ F ⁰ (HV+ X K„‘),

^h = Uo»i(»,i1+ I el),

00 00 00

/3 = к ⁰ и' ⁰ (с/;+ X я.1),

h = Uo(v}+ Z el,)W‘ + Z

^Rl)<

00 00

h = v„(i/}+ X PJ)(WÏ+ Z ^w*),

n — X k = 1

00 00

= ^w 0 (u{+ x p D(^‘+ Z ei).

n = X m ~ 1

00 00 00

17 = (c | + x _{m = 1 n = l}

Z el)W + Z

5. We shall give some lemmas which we shall use in the sequel.

Lemma 3. I f the function J ( X , t ) satisfies the equation

(la) P u ( X , t ) = 0,

then the function t J ( X , t ) satisfies the equation

(lb) P2u ( X , t ) = 0.

We omit the simple proof.

Lemma 4. Let a ^ 0 and let b be the positive number; then ab exp ( — a2) ^ {\bf12.

(19)

152 F. B a r a n s k i and J. Mu s i a l e k

We omit the simple proof.

Lemma 5. Let n > 2 and let bx be a positive number, b2 a non-negative integer. Further let t > s ^ 0, p e [ - a , f l ] , and let

T = ( t - s y ^ i p - x l f 2 e x p [ ( - 4 ( t - s ) ) ~ l i p - x i ) 2] (i = 1,2);

then

\T\ ^ C { t - s f +b*l2{ n - \ y A- 2b', C being a convenient positive constant.

P ro of. We have

T = (t — s)2+ib2(p — xll) ~2bl~4(p — х\,р ( t - s ) ' ibx

x exp (( —4(f — s))-1 (p — xlt)2)(p — xl,)2bl + 4'(t — s) ~ b l ~ 2 (i = 1,2).

By inequalities (19) and by ([1], vol. I. p. 452), we obtain the thesis of Lemma 5.

Lemma 6. Let n = 1, t > s ^ 0 and let bx be a positive number, b2

a non-negative integer and let p e [ — a, a]; then

m « V a p < t - s ) { p - x ‘. r u i- 2 (i = 1,2).

The proof of Lemma 6 is similar to the proof of Lemma 5.

Let

S = ( t - s y h ^ i p - x i P i q - y l P i r - z l P exp [ ( - 4 ( t - s ) ) _1 x

x( P - 4 ) 2] exp [ ( - 4 ( t - 5 ) ) -1 (^ -y i,)2] exp [ ( - 4 ( t - 5 ) ) -1 (r-z*1)2]

(i,j, l = 1,2) , b i being a positive number, b(- (i = 2, 3, 4) non-negative integers.

By Lemmas 5, 6 we obtain

Lemma 7. Let О 1 or ш > 1 or к ^ 1, t > s ^ 0, Y e B l5 X e B 1;

then (S’! ^ C (t-s )| p -x j,r °i or |5| < C (r-s )| q r-> irI)i or |S| < C (i- s )| r - 4 r ° i (i , j , l = 1,2) D, being a positive number.

I f n ^ 2 and m ^ 2 and к ^ 2, then

|S| « C(r —s)((n —l)(m —1)(A;—l))-b

where b3 is a positive number and b3 ^ 2, and C is a convenient positive constant.

6. Now we shall examine the integral m1. Let

H (X , t, Y, s) = (( —4(t —s))_1 (lj>-x) 2 + ( q - y) 2 + ( r - z f ) ) , K ( X , t , Y,s) = (! —s)-3/2 exp ( H ( X , t , y.s)),

and let

ul — Wl,l + ul,2>

Lim it problem fo r the rectangular parallelepiped 153

where

and

«i,i " / ( J J I / t I W . i . Y,0)dY, 7

Ml ,2 = X ^Ml,2 J= 1

"i.2 j = Г,0)^У (/'= 1... 7).

*1

Let

UP1*2.P*P* = A S S $ M Y ) ( D pxl ; % P3’p* K ( X , t , Y,0))dY,

в i

where

О < P1+ P 2+ P 3 + P4 ^ 4, p4 = 0 ,1 ,2 , р; = 0 ,1 ,2 , 3,4 (i = 1,2,3).

Now we shall prove

Lemma 8. I f the function f x is continuous in the set Blt then the integrals мРьР2>Рз>Р4 are 1осаЦу Uniformiy convergent in the set В and

P2u1A ( X , t ) = 0 f o r ( X , t ) e B .

P roof. The functions Dpx)yP2’t P3’P4 K ( X , t, Y, 0) are finite sums of the functions

Abl’b2’b3’b* = bl { p - x ) b2{ q - y ) b2( r - z f4 K ( X , t , Y, 0),

where C1 is a positive constant, b1 a positive number and b£ (i = 2,3,4) are non-negative integers.

Similarly as in Lemmas 5, 6, 7 we get the absolute majorant of the functions Ab l ’b2’b3’b4 of the form

Fj = N j t - B\

Nj and Bj being a positive numbers. Consequently the functions F} are the absolute majorants of the integrals u^ p2>P3f4 ancj those integrals are locally uniformly convergent in the set B. Finally we have

P2uul = А Щ Л ( Г ) Р2Х'ПК ( Х ^ , ¥ , 0 ) d Y = 0

B\

because P2x<t)K { X , t, Y, 0) = 0 for ( X , t) belonging to the set B.

Now we shall prove

Lemma 9. I f the function /j is continuous in the set Bt , then the functions ui,2j 0 = 1,...,7) satisfy equation (lb) in the set B.

154 F. B a r a n s k i and J. Mu s i al e k

P ro of. We shall give the proof only for the function и 1>2Д because for the remaining integrals u12)j 0 = 2,..., 7) the proof is similar.

Let us consider the integrals Pl>P2>P3>P4

“1,2,1 A \ \ \ M Y ) D Pl>P2>P3>P4_x,y,z,t f d Y ,

в 1

where ⁰ ^ P 1 + P 2 + P3 + P4 ^ 4; Pa = ^{0 ,} 1, 2; pt = ^{0 ,} 1, 2, 3,4 (i = 1, 2, 3).

By Lemmas 5, 6 the common absolute majorant of the integrals uu2P12’P3’P4 is the integral

(20) £ * М Ш ( С 2 + С3 I n~2)dY,

п= 1

b, C2, C3 are positive constants and M = sup l/J.

Consequently we obtain the condition

P 2u1>21 = A S f f f1(Y)PfX't)I1( X , t t Y, 0)dY — 0

в i

because P2Xtt)I t (X, t, Y, 0) = 0 for (X, t) belonging to the set B.

Now we shall give

Lemma 10. Let the function h be continuous in the set B1; then the function u2 satisfies equation (lb) in the set B.

The proof of Lemma 10 is similar to the proof of Lemma 9. Let H0(x, t, a, s) = ( — 4(r — s))_1 (a + x) 2 and let H l (x, t, a, s) = (t — s) 112 exp H 0. ч

We shall prove

Lemma 11. I f the function /4 is continuous and bounded for ( q , r ) e S 1, s ^ 0, then the function n4 satisfies equation (lb) in the set B.

P roof. Let

U4. — W4> 1 + U4 2 j where

u4>1 = A J j j /4(q, r, s )Hx (x, t, a, s) G2 G3 dqdrds,

0 S j t

U42 = A

J J J

0 s x

Let us consider the integrals

«4дР2’РЗ’Р4 = A J j \ f4(q, r, s ) D ^ 2’P3’P4(G2 G3 (x, t, a, s))dqdrds,

0 s l

Pl-P2>P3’P4 t

“4, 2

A J Я Л (9. G s)D2Z2,r ,P4{(G1- H l)G2 G3)dqdrds,

0 s1

where 0 ^ P1+ P 2 + P3 + P4 ^ 4, p4 = 0, 1, 2, p, = 0 ,1 ,2 , 3,4 (i = 1,2, 3),

Lim it problem f o r the rectangular parallelepiped 155

and

« 4,1,1 = A J }/ 4(q, r, s )Hi(x, t, a, s) G 2 G3 dqdr,

s i

« 4,2,1 = A } J/4(g, r, s ) ( t - s ) ~ ll2(G1- H 1) G 2 G3dqdr.

s i

Let M = sup|/4|. For the integral и4ЛЛ by Lemmas 5, 6 we have the estimation

l«4,i,il ^ C4(t — s)1/2 -> 0 as s->t, C4 being a convenient positive constant, and consequently

lim u4дд = 0 as s -> t.

Similarly by Lemmas 5, 6 for the integral w4;2,i we obtain the absolute majorant of the form

00

(21) C(t —s)1/2(C! (a + x)~2+ C2 Ya (mnk)~ 2),

« n,m,k= 1

C, C l5 C2 being positive constants, and consequently lim w4;2>1 = 0 as For the integrals и4ЛР2’РЗ’р4 and up4)2 2’P3’PA by Lemmas 5, 6 we obtain the common absolute majorant of the form (21).

Consequently we get the condition

t

P 2U 4,1 = A j f j f 4(q, r, s) P(x,t) (H i (x, t, a, s)G2 G3)dqdrds = 0, 0 Sj

because

PxA H , G 2G3) = 0, and

t

P 2 «4,2 = A\ j j f 4(q,r,s)Pfx,t)((GlL- H 1))dqdrds = 0,

0 s x

because

= 0. Finally we get the condition

Pfx,t)u4- — 0 for (X, t) belonging to the set B.

Similarly as Lemma 11 we can prove

Lemma 12. I f the functions f (i = 3,5,6) are continuous and bounded in the set St x (0, oo), the functions f (i = 7,8,9,10) are continuous and bounded in the set S2 x(0, oo), the functions f (i = 11, 12, 13, 14) are continuous

156 F. B a r a n s k i and J. Mu s i a l e k

and bounded in the set S3x(0, oo), then the functions щ (i = 3 ,5 ,_, 14) satisfy equation (lb) in the set B.

7. Now we shall examine the integral u15. We shall prove that u15 satisfies equation (1) in the set B. In order to prove this assertion we shall prove some lemmas.

Let

K x( X , t , Y,s) = ( t - s ) ~ 112 exp H (X, t, Y, s),

t

“ 15,1 =

0 Bx

» I5,2,1 = A M M f ( Y , s ) I j ( X , t , Y , s ) d Y d s (j = 0 Bx

Let

M = sup (\Dpf\, \Dqf\, \Dif\) (i = 0, 1,2).

We observe that P (Xt)K x ( X , t, Y, s) = —K ( X , t , Y, s).

Now we shall prove

Lemma 13. I f the functions Dlp f , Dlqf , Dlr f (i = 0, 1,2) are continuous and bounded for Y e B x, s ^ 0, then the function m15>1 satisfies equation (1) in the set B.

P roof. By the Weierstrass theorem ([1], vol. I, p. 442) we have limJJ J f ( Y , s ) K l ( X , t , Y,s)dY= 0 as s->r.

в ^{1 »}

Let us consider the integrals

“ 'Æ? = A$ J \ \ f { Y , s ) D ix K 1dYds = -41 и î f ( Y , S) { - l ) ‘ D‘pK i dYds,

О в1 ОBx

< s‘:° = A f f f f f i K s I D ' K . d Y d s = A f f f f / ( K s M - i y D ' K ^ Y d s ,

0 B1 0 Bx

= A j j j j f f ^ X D i K ^ Y d s = A S H ! f ( Y , s ) ( - ! ) ' D ‘ Kt drds

О Bx 0 Bx

for i = 0, 1, 2.

For the integral u ^ f we have the estimation l«?Ü°l < C9t " 2, C9 being a convenient positive constant.

Integrating twice by parts the integrals mVsI?, Ui’f i and (i = 1,2), we can verify that those integrals are locally uniformly convergent in the set В and by (21) we obtain

= - 4 j J l ( / ( y , s ) K ( X , I , Y,s)dYds.

0 Bx

Lim it problem for the rectangular parallelepiped 157

By ([1], vol. I, p. 442) we get the assertion of Lemma 13.

Lemma 14. I f the function f is continuous and bounded in the set B, then the functions uX5t2J (j = 1,...,7) satisfy équation (lb) in the set B.

P roof. We shall give the proof only for the integral u15i2,i, because for the remaining integrals m15>2j 0 = 2,..., 7) the proof is similar. By Lemmas 5, 6, 7 we have

lim J f J f ( Y , s ) I xdY = lim f J J f ( Y,s)PiXtt)I 1dY = 0 as s - * t ,

в i в 1

and consequently we obtain

\ m f ( Y , s ) P l t I 1dYds = 0 for ( X , t ) e B .

0 B,

8. Now we shall give some lemmas concerning the initial conditions.

Lemma 15. I f the function f x is continuous in the set B x and (X , t ) e B , X 0e B x, then

(a) lim им = / i ( X 0)

as o,0) (j = 1,.... 7).

(b) lim « 1,2,j = 0

P roof. Ad (a). Let f x = / x for Y e B x, f x = 0 for Y e E 2\Bx and

«1.1 = A \ \ \ f x{ Y ) K { X bt, Y,0)dY;

въ

by the Weierstrass theorem ([1], vol. I, p. 442) we obtain assertion (a).

Ad (b). Integral (20) is the absolute majorant of the integrals uXt2j (j = 1,..., 7) and consequently we obtain assertion (b).

Lemma 16. I f the functions Dlpf x, D‘qf x, Dlr f x (i — 0, 1,2) are continuous and bounded in the set Bx, and (X , t ) e B , X 0e B x, then

lim Dt ultX = Afx(X q) when (X, t) -*• (X 0, 0).

P ro of. Integrating twice by parts the integral Dt ux x and applying the Weierstrass theorem we get the assertion of Lemma 16. Let

m21 = A t ^ \ h { Y ) K ( X , f , Y,0)dY, в 1

u2,2 = A t i S S h ( Y ) ( £ U)dY,

В1 i= 1

“ 2,2j = At S $ $ h ( Y ) I j d Y (/ = 1 , 7 ) . В1

Now we shall prove the following

Lemma 17. I f the functions f x, Dlpf x, Dlqf x, D\fx (i = 1 , 2 ) are continuous and bounded in the set B x, and (X , t ) e B , X 0 = (x0, y0, z0) e B x, then

158 F. Barari ski and J. Mu s i al e k

13° lim u2,i = 0,

14° lim u2,2j = 0, j = 1, •••, 7, 15° lim Dt u2A = h ( X 0), 16° lim u22 = 0, when ( X , t ) ^ ( X o,0).

P roof. Ad 13°. By the Weierstrass theorem we have lim и2л = 0, h ( X 0) = 0 as (X, t) -> (AT0, 0). By estimation (20) we obtain assertions 14°

and 16°.

Ad 15°. From the Weierstrass theorem we have

Н т Д и 2д = 0 , lim Dt

J

J

Bi

By Lemmas 15, 16 and 17 we obtain the following

Lemma 18. I f the functions Dlpj\, Dlqj \, Dlr f x (i = 0, 1,2) are continuous and bounded in B 1 and (X , t ) e B , X 0e B 1, then

lim Dt (u1+ u2) — f 2 (A 0) as ( X , t ) - > ( X 0, 0).

By Lemmas 7, 9 we get

Lemma 19. I f the functions f (i = 3,4, 5, 6) are continuous and bounded for (q , r)eS l9 s ^ 0, the functions f (i = 7, 8, 9, 10) are continuous and bounded for ( p ,r )e S 2, s ^ 0, the functions f (i = 11, 12, 13, 14) are continuous and

bounded for (p,r)eS3, s ^ 0 and (X , t ) e B , XqgBj^, then

lim Dt щ = lim щ = 0 (i = 3, 4,..., 14) as ( X , t) -* ( X 0, 0).

Lemma 20. I f the function f is continuous and bounded for YeB, s ^ 0 and (X , t ) e B , X 0e B ly then

lim m15>1 = lim u152j = 0 (j = 1,...,7) as (X, t ) -> (X 0, 0).

P roof. The integral

t

C ioM

J

о

C10 being a convenient positive constant, is the absolute majorant of the integrals m15>1 and u15t2J (j = 1 ,2,...,7 ).

Consequently we obtain the thesis of Lemma 20.

Lemma 21. I f the function f is continuous and bounded for Ye B 1 , s ^ 0 and (X , t ) e B , X 0e B 1( t > s, then

lim j H f ( Y , s ) I j d Y = z l i m f j $ f (Y, s ) ( t - s ) KdY = 0 (/' = l , - , 7),

«1 Bi

as (X, s)-+ ( X 0, t0).

P roof. By the Weierstrass theorem we get Lemma 21.

Lim it problem fo r the rectangular parallelepiped 159

Lemma 22. I f the functions D p f t D^ ft Dlr f (i = 0,1,2 ) are continuous and bounded for Y eB x, s ^ 0 and (X , t ) e B , X 0e B x, then

17° lim f J f J f ( Y , s ) D t K 1dYds = lim f j J J/(7, s) Ay K x dYds = 0, as

0 e t 0 B x

( X , t ) - + ( X o,0),

18° lim J j' $ $ f( Y, s) Dt IjdYds = 0 (j = 1,2,..., 7), as (X, t) -> ( X 0, 0).

0 Bx

t

P roof. Integrating twice by parts the integral s )J y X 1 dYds, by

0

Lemma 6 we get assertion 17°. By the majorant (21) we obtain assertion 18° of Lemma 22.

By Lemmas 20, 21, 22 we obtain

Lemma 23. I f the functions Dlpf , D^ ft Dlr f ( i = 0 ,1,2 ) are continuous and bounded for Y e B x, s > 0 and ( X , t ) e B , X 0e B l5 then

lim n15 = lim Dt ul5 = 0 as (X, t) -* ( X 0, 0).

9. Now we shall verify the boundary conditions (4)-(15). At first we shall give some lemmas concerning the boundary property of the functions щ (i = 1,2,..., 15).

Lemma 24. Let the functions f u h be continuous in the set Bx and (X , t ) e B , Xo e » t > t0 > 0; then

lim Dx Щ = 0 0 = 1, 2) as (X ,r)-> ( ± a , y o.> zo, to), lim Dy щ = 0 O' = 1, 2) as ( A , t ) - (x0, ± b ? zo> to), lim Dz щ = 0 0 - 1,2) as (X, t)-> (x0, y0, ± c , t0).

Prc>of. We shall give the proof only for the assertion:

19° lim Dx ux = 0 as ( X , t ) (a> Уо» zo> A>)>

because the proof for the remaining conditions is similar.

Let us consider the integral

u\,0’° = A j f f f x( Y ) D x GdY= A f f f f 1( Y)(Dx Gl ) G 2G3dY

b x b x

By majorant (21) the integral u\,0’° is locally uniformly convergent at every point (a, y0, z0, t0). Hence from Lemma 2 we obtain 19p. This proves our lemma.

Let

мз’0,0 = A i $ S M q , r , s ) D ,

0 Si

G|p= -„dqdrds = u\$'° + u ff î 0.

160 F. B a r a n s k i and J. Mu s i al e k

where

will’0 =

j j J

j j J

+

O Si о

x(t — s)~512 exp [( — 4(r — s))-1 (x + a)2] exp [ — 4(r — s))-1 x x (q — y)2 + (r — z)2] dqdrds,

ui;0,° = A

J J J

0 st i= 1

Let

« 3:5:° = A S JJ f 3(q,r, s) Dx Ij dqdrds {j = 1 , 7 ) . 0 Si .

Lemma 25. Let (X , t ) e B ; then

t

I = A

j J J

-0 0 1 . 2

P roof. Applying the change of integral variables

(22) q — y — 2R(t — s)112 cos v, r — z = 2R(t — s)ll2sinv and x + a = 2(t — s)1/2w, we obtain the assertion of Lemma 25.

Let /3 = / 3 for ( q , r ) e S i , s ^ 0, f 3 = 0 for ( q , r ) e E 2\Si, s > 0, /3 = 0 for ( q , r ) e S i , s < 0. Therefore

t 00 00

* — A

J J j

— 00 — 00 — 00

Let M = sup |/3| and d (q ,r ,s ) = f 3(q, r, s ) - f 3(y0, z0, t0).

Let t > t0 > 0 and let e denote an arbitrary positive number and b = Ь(е) a convenient positive number depending on e. Let (y — y0)2 +

+ (z —z0)2 < b2/4 and let

Z i = {(q,r,s): t0- b < s < t < t0 + b , ( q - y 0)2+ ( r - z 0)2 < b2} , Z2 = {(q,r,s): - 0 0 < s < t0- b , ( q - y 0)2+ ( r - z 0)2 < b2}, Z 3 = {(q, r,s): - с о < s < t, ( q ~ y 0)2 + ( r - z 0)2 > b2}.

Now we shall prove the following

Lemma 26. I f the function f\ is continuous and bounded for (q , r ) e S i , s ^ 0 and (X , t ) e B , t > t0 > 0, y0e ( - b , b ) , z0e{ — c,c) and (X , t ) e B , then

lim u\fi’° = /3(y0, z0, t0) as { X , t) -» ( - a , y0, z0, t0).

P roof. By Lemma 25 we have

t 00 00

Мзл’0-/ з(У о ,20Ло) = A

J

j

Lim it problem for the rectangular parallelepiped 161

By continuity of the function /3 at the point (y0, z0, t0) there exists a positive number b = b(s) such that \d(q,r,s)\ < e for (q, r, s ) e Z l . We have the inequalities

K ,’i ,0 -/з (Уо, z0, t0) I ^ A$SS\d(q,r,s)\Dx Kdqdrds +

+ 2A M

J

J

J J J

and

A J J J d(q, r, s)Dx Kdqdrds < e.

Applying the change of the integral variables (22) we obtain the inequality I J J J DxKdqdrds\ ^ C(t — t0+ b ) ~ 112 (x + a),

z 2

00

I J J J DxKdqdrds\ ^ C | exp ( — v2)dv

z3 _ь_

x + a

and consequently we get the assertion of Lemma 26.

Lemma 27. I f the function /3 is continuous and bounded for ( q , r ) s S l , s ^ 0 and (X, t)eB, t > t0 > 0, y0e ( - b , b ) , z0e ( - c , c), then

2OP lim = 0 and

21° lim DxPu3(X, t) = 0 as (X , t) - » ( - a , y0, z0, t0).

P ro of. Ad 20°. By Lemmas 5 and 7 we have K i ’,9K C l t (x + a) a = 1 ,2 ,4 ),

Cn being the positive constant, and by Lemmas 1, 5 and 7, and by uniform convergence of the integrals m3;2j 0 = 3, 5, 6, 7) at every point ( — a, y0, z0, t0) we get the conditions

lim ut,2’j = 0 (j = 3,5, 6,1) as (X, t) -+ ( - a , y0, z0, t0) and finally we obtain assertion 2CP.

Ad 21°. By Lemma 9 we have Pu3( X , t ) = 0 for (X , t ) e B . Hence we obtain condition 21°.

Now we shall prove the following

Lemma 28. Let the function J4 be continuous and bounded for (q , r ) e S l , s > 0, and (X , t ) e B , x0e( — b,b), z0e ( — c,c); then

22° lim Dx u4(X, t) — 0,

23° lim Dx Рщ ( X , t) = /4 (y0, z0, t0) , as ( X , t ) - ^ ( - a , y0, z0, t 0).

162 F. Barariski and J. Mu s i al e k

P roof. Ad 22°. By uniform convergence of the integral

г

A J J J /4(4* r, s)Dx Q ( X , t, —a, q, r, s)dqdrds

0 Si

at the point ( — a,y0, z 0, t 0) and by Lemma 2 we obtain assertion 22° 4 Ad 23°. By Lemma 11 we have

t

Р щ ( Х , t) = A J J J/4(g, r, s)G(X, t, —a, q., r, s)dqdrds

0 Sj

and similarly as in Lemma 26 we obtain assertion 23°.

Let

u uj = A J S J f i DxPx,t Ki,jdsdSj,

0 Sj

where for / = 1, i = 5, 6,

К 5Л = G ( X , t , + a , q , r , s ) , К 6Л = Q { X , t, a, q, r, s) for j = 2, i = 7, 8, 9, 10,

Kl t 2 = t; p, - b , r, s),

K8,2 = Q ( X , t ; p, - b , r , s ), K 9;2 = G ( X , t; p , b , r , s ), K io .2 = Pb b, r, s),

for 7 = 3, i = 11, 12, 13, 14,

K 1U3 = G (X , t; p, q, - c , s),

K i 2,3 = P, - c , s ), К 13>3 = G (X ,t; p , q , c , s ),

^ 14,3 = G P* 4, c, s).

Now we shall prove the following

Lemma 29. I f the functions f (i = 5,6) are continuous and bounded in the set St x(0, 00^{) ,} the functions f (j = 7, 8, 9, 10) are continuous and bounded in the set S2 x ( 0, 00⁾ and the functions f (i = 11,..., 14) are continuous and bounded in the set S3x ( 0, 00^{) ,} and (X , t ) e B , y0e( — b,b), z0e ( — c,c), t > t0

> 0, then lim Dx ut = lim DxPut = 0 (i = 5,..., 14) as (X, t) tends to the point ( - a , y0, Zq, t0).

P ro of. From Lemma 2 and by uniform convergence of the integrals Uij (i = 5,..., 14; j = 1,2,3) at every point ( — a,y0, z 0, t 0) we obtain the thesis of Lemma 29.

Similarly as Lemma 29 we can prove the following

Lemma 30. I f the functions f (i = 3,4, 5,6) are continuous and bounded in the set Sl x(0,oo), the functions f (i = 7,8,9,10) are continuous and

Lim it problem fo r the rectangular parallelepiped 163

bounded in the set S2x (0,cc), the functions f (i = 11,12,13,14) are continuous and bounded in the set S3x (0, oo), and (X , t ) e B , X 0e B x, t > t0 > 0, then lim Dxu5 — /5 („Vo> zo> fo)> lim DxPu5 = 0, lim Dxu6 = 0, lim Dx Pu6 = /6 (y0, z0, t0) as ( X , t ) -> (a, y0, z0, t0), lim Dxu{ = lim PDxut

= 0 (i = 3, 4, 7, 8, ..., 14), if ( X , t ) tends to the point (a, y0, z0, t0), lim Dy u7

= f 7(x0, z0, t0), lim DyPu7 = 0, limDyWg = 0, limZ^Pitg = f 8 (x0, z0, t0) as ( X , t) -* (x0, —b, z0, t0), lim DyUi = lim й уРщ = 0 (i = 3,..., 14, i Ф 7, 8) as the point ( X , t) tends to the point (x0, —b , z 0, t 0), lim Dyug — f g(x0, z0, t0), UmDyPug = 0, UmDyuxo = 0, lim DyPuX0 = /10(x0, z0, t0) as (X, t) -> (x0, b, z0, t0), lim Dy ut = lim Dy Рщ = 0 ( i ; = 3,4,..., 14; i ф 9, 10) as (X, t) tends to the point (x0, b, z0, t0), lim Dzuxx = f lx (x0, tQ)> DzPuxl

= 0, lim Dzuxl = 0, lim DzPu12 = /12(x0, yo,t0) as { X , t) ^>(x0, y0, - c , t 0), lim DzUi *= lim DzPut = 0 (i = 3,4,..., 14; i Ф 11,12) as (X, ?)->(x0, Уо, — c, t0), lim DzuX3 = /13(x0, y0, t 0), lim Dz Pul3 = 0, lim D2w14 = 0, lim DzPul4.

= f i4.(x0,yo,tb) as ( X , t ) - + { x 0, y o , c , t 0), lim Dz щ = lim Dz Рщ = 0 (i

= 3,4, ..., 14; i Ф 13, 14) as (X , t)^>{x0, y0, c, t0).

By Lemmas 2 and 13 we obtain the following

L^emma 31. I f the functions Dlp /, Dlq f, Dlr f (i — 0, 1,2). are continuous and bounded for Y e B x, s $> 0, and (X , t ) e B , X 0e B x, t > t0 > 0, t/ien lim Dxm15 = lim DxPuxs = 0 as(X, f) ( - a , y0, z 0, t0) or (X,t)^>(a,y0,z0yt0), lim Dyu15 = lim DyPul5 = 0 a s { X , t ) - * ( x o, - b , z 0, t 0) o r ( X , t ) - + ( x 0,b,z0,t0), lim Dz u15 = lim DzPuX5 = 0as'(X , t ) - > ( x 0, y0, - c , t 0) o r ( X , t ) - > ( x 0,y0,c,t0).

10. By Lemmas 1-31 we get the fundamental

Theorem. I f the functions f \,Âf x, f 2 are continuous and bounded for Ye B x, the functions f (i = 3,4, 5, 6) are continuous and bounded in the set Sx x(0, 00^{) ,} the functions f (i = 7, 8, 9, 10) are continuous and bounded in the set S2 x(0, 00^{) ,} the functions f (i = 11,12, 13, 14) are continuous and bounded in the set S3 x (0, 00^{) ,} the functions Dlp /, Dlq /, D\ f (i = 0,1,2) are continuous and bounded for Ye B x, s ^ 0, then the function u ( X , t ) defined by formula (18) is the solution of the ( F- N ) problem.

References

[1 ] J. H a c h a j, On the Fourier Problems fo r the rectangular parallelepiped, Technological University O f Cracow, Politechnika Krakowska im. Tadeusza Kosciuszki, Zeszyt Naukowy No. 6, Krakow 1978, p. 59-82.

[2 ] M. K r z y z a n s k i, Partial differential equations o f second order, Vol. I, Warszawa 1971.

2 — Roczniki PTM — Prace Matematyczne XXII