Application of Fisher-Riesz-Kupradze method to solving the second Fourier problem

ANNALES SOCIETAT1S MATHEMATICAE P O I O N A E Series I COM MENT ATIO N ES M A T H E M ' T I C A E XXVII (1988) ROCZNIKI POLSKIEGO TOWARZ YSTW A MA TEM AT YC ZN EG O

Séria I PRACE MATEM 'T Y C Z N E XXVII (1988)

Z. Do m a n s k i and Z. Ro j e k (Warsaw)

Application of Fisher-Riesz-Kupradze method to solving the second Fourier problem

Abstract. In this work we construct the solution o f the second Fourier problem using Fisher-Riesz-Kupradze method.

1. Introduction. Let Q be a bounded domain in Euclidean space Е ъ the boundary o f which (*) is a closed surface S of class C 2,A. W e shall denote the points of Q as x = (x ,, x 2, x 3), у = (у,, y 2, y3), . the point o f the surface S as £ = (<^, £2, £3), d = (Vu Ч2, Чз)> ••• The symbol |x-y| denotes the Eu­

clidean distance between points x and y: QT = Q x(0, 7); ST = S x(0, T);

= t^i (//), ot2(r}), x3(»/)] is the inner unit normal to surface S at //.

Let у be a function defined in the domain Q and let this function have partial derivatives dx/dxj, j = 1, 2, 3, uniformly continuous in Q. (x) will be the function defined on the surface S by

( X ) ( r j ) = lim x(x).

xeQ

De f i n i t i o n 1. Let a function /< defined on surface S be given and let (x) = ц. We define the derivatives Dxn , j = 1, 2, 3, of the function ц (see [3],

p. 15-20) by the formula

The function /r for which Dx p , j = 1, 2, 3, exist will be said to be differentiable in Hugoniot-Hadamard sense, simply in H -H sense.

О For the definition o f the surface o f class C 2,A see [3 ], p. 96-98, where L 2(B, X) = C 2 X.

2 — Prace M atem atyczne 27.2

216 Z. D o m a n s k i and Z. R o j e k

W e know that (see [3 ], p. 19)

(2 ) I o C j D ^ f i = 0 .

j = i

Instead o f using Dx. we shall use Dj. If the function g defined on ST has in H -H sense derivatives of order l ^ 0 with respect to space variables satisfying Holder condition with exponent h with respect to these variables, 0 ^ h < 1, and if the function g satisfies the Holder condition with exponent x, 0 ^ x ^ 1, with respect to time variable, we shall say that g e C l,h,x(ST).

II. Properties of the Poisson-Weierstrass integral and heat potentials. We know (see [5], p. 529) that for the equation

(3) A u - ~ = g

we have the fundamental formula

t

(4) ju(y, 0) v ( x - y , t)dy— J$g(y, x ) v { x - y , t - x ) d y d x -

(2 0 Q

1 r du

— f — v(x — rj,t — x) dS„dx + b'sdn„

f dv

+ f f “ (»7, t) — (x - rj, t - x) dSndx

os d n ti

Г0 for хфй, t > 0,

■=<U u (£, 0 for (^, î)eSt , lu(x, t) for (x, t )e Q T, where

(5) ',<X’ ,) = ( 4 J p eXP( “ ^ ) f O r , > 0 '

The integrals in (4) play the main role when initial value problems are concerned with (3). Some properties o f these integrals will be given.

De f i n i t i o n 2. The Poisson-Weierstrass integral with density q> is (see [11], p. 21-23; [9], p. 282-283) the integral o f the form

(6) I ( x , t) = $ (p (y )v {x -y , t)dy.

Q

Th e o r e m 1. I f the function (p belongs to Cq(Q), then the Poissonr-Weier- strass integral /(£, t) for set ST belongs to the class C 2,0;1(ST).

W e can prove this theorem using the classical method.

Application o f Fisher-Riesz-Kupradze method 217

De f i n i t i o n 3. The heat volumen potential with density g is (see [11], p.

23-34) the integral of the form

t

(7) J{x, t) = t) v { x - y , t-x )d y d x .

о h

Th e o r e m 2. I f g(-, t) eCq(Q) for every t e(0, T) and g(x, )e C ((0 , T)) for every xeQ, then J( -, •) e C 2,0:1 (5r ).

The proof of this theorem is analogous to the proof o f Theorem 1.

De f i n i t i o n 4. The heat potential of the simple layer with density ф is (see [10], p. 81-112) the integral of the form

t

(8) U (x, t) = j j i /ф1,t)v (x — rj, t — x)dSndx.

o s

Th e o r e m 3. I f the function ф is defined on ST and if ф (•, t) is differentiable in H -H sense at any value x e(0, T) and if the derivatives D j ф , j = 1,2,3, are bounded and integrable on ST, then the derivatives D jU , j = 1 ,2 ,3 , exist and satisfy the Holder condition with respect to space variables with an exponent arbitrarily smaller than 1.

P r o o f. If x<£S, then

dU r dv(x — rj,t — x)

(x, t) = J jФ{г}, т )--------dSndx

Яу . ^ J Яу

V X j O S U X j

Using formula (1), we obtain ÔU{x, t) = -

os

OS ° 4 j

dv (9) 75—(*, 0 = - т)Я;1*ЙчЛт-ЯаД»/Ж»/, T) — dS^dx.

V X j O S O S a n ti

Taking into account the equality £ ak (rj) = 1 and that £ ak(r])Djak(rj) = 0, we have

k= 1

i

k= 1

( 10 )

0 И = £ a*

( af c

•

k= 1

3

Using the equality £ ak Dk ^ v ) = 0 (see (2)), we get k= 1

3

£ a7l ° k (<** Ф и) - ф и ° к a * ] = о.

k= 1

(И)

218 Z. D o m a r i s k i and Z. R o je k

ij/DjV = Djiij/i^ — vDiiJ/: therefore from (10) and (11) we get

t t 3

(12) — ( |if/Dj vdSn dx = f {(Dji/s — c/iji/j X Dkcck) vdSndx +

as a s к = l

t 3 3

+ J I l aJ X ° k ( a k Ф » ) - x a k d j (a * Ф М d s n d x •

bs fc= 1 k=1

From the Stokes theorem and from the fact that S is a closed surface we see that the last integral in (12) vanishes. Therefore we can write (9) in the form

dU ' 3 1 dv

(13) — (x, t) = U ( D j t - o L j t X DkoLk)vdS4d x - \\^(г})ф(г1, x) — dS4dx.

dX; os k= i os a n n

The first of these integrals is the heat potential o f the simple layer with

3

density Djij/ — cCjift X and the second o f these integrals is the heat

к — 1

potential o f the double layer with density а/ф. Using the known boundary property of the heat potential of the double layer (see [11], p. 13-21), we get

ÔU

Therefore

(14)

dU

d x j

lim ^— (x, t) = (^ — )(£, t) = f f ( D , > - a 7> X DkoLk)vdSvdx -

OS k = 1

- jS aljф Y-dSчdx + faj(Ç)ф(Çt t) = l s- W} + ф.

0 S dnn

D j U = l j - X j У « „ h - l W j - a j £ ^{Щ Wt).}

k = 1 k = 1

The heat potential of the simple layer f and the heat potential of the double layer Wj satisfy the Holder condition (see [11], p. 6-10, 18-21) with respect to the space variable with an arbitrary exponent smaller than 1 with the previously given assumptions about function ф, therefore (14) we see that Dj U satisfies the Holder condition with the same exponent.

Th e o r e m 4. I f the density ф is a bounded integrable function in ST and satisfies the Holder condition with respect to the time variable t with exponent x, 0 < x ^ 1, and the condition

lim ф(п, t) = ф(р, 0) = 0, t - o +

then the heat potential of the simple layer U(Ç, t) satisfies the Holder condition with respect to the variable t with exponent x.

The proof o f this theorem is easy.

Application of Fisher-Riesz-Kupradze method 219

III. Problem. Using Fisher-Riesz-Kupradze method, we shall solve the following Fourier problem:

We seek the function u(x, t) satisfying in QT the equation

(15) du

Ли— — = g

dt У

and the following conditions:

(16)

(17)

lim u{x, t) = tp(x), l-*0 '

lim — du(x, t) = ф(£, t).

x -*£, dïlç x e Q

We assume that S is of class С 2Д,

g{-, t)e C l{ Q ) for any f e ( 0, T), g{x, -)e C ((0 , T)) for any x gQ, (peCl(Q),

il/eCl ’0:X(ST) and 0) = 0.

S o lu t io n o f th e p ro b le m . Using the basic formulas (4) we can write the solution (if it exists) in the form

t

(18) u(x, t) = j’ (p (y )v (x - y , t)dy— Ц д {у , z ) v ( x - y , t - z ) d y d z -

b on

1 1 dv

- J $Ф(г}, z ) v ( x - r j , t - T ) d S 4dT+ JJ/to, t — z)dSndz,

0s 0 s

where the known function / (g, z) is the solution of the functional equation

(19) i i f ( g , z ) ^ - ( x - r i , t - z ) d S t,dz = F ( x , t ) ( x f Ü , t e(0, T)).

os <4

In (19) we denoted

t

(20) F (x , t) = - (< p (y )v (x - y , t)d y + f f g (y , z ) v ( x - y , t-z )d y d z +

b о n

t

+ I \ф(т], z )v (x —rj, t — z)dSndz.

o s

Th e o r e m 5. The functional equation (19) has exactly one solution be­

longing to class C l,h:/(ST).

P r o o f. Let us assume that (19) has got a solution. Using the properties of the*heat potential o f the double layer (see [11], p. 15), we get the integral

220 Z. D o m a n s k i and Z. R o j e k

equation

(2 1 ) 1 dv

- V ( L t ) + JJ/fo, t t - x ) d S ndx = F d , t).

Л Я Я

OS u n r\

The integral equation (21) is a Volterra equation with a continuous right side; it has exactly one continuous solution (see [12], p. 129-135). Taking into account the assumptions, we infer on the basis of Theorems 1-4 that the function F(Ç, t ) e C 1,h’x(ST) and F ( Ç , 0) = 0, W e shall show that the solution of the integral equation (19) is o f class C 1,h;X(ST).

W e shall use the following lemmas.

Le m m a 1. I f a function F is continuous and bounded on ST, it satisfies the Holder condition with exponent A with respect to variable t and F(Ç, 0) = 0 for ÇeS, then the solution of the integral equation (21) satisfies the Holder condition with the same exponent A with respect to variable t.

We get the proof o f Lemma 1 by showing that any o f the terms o f the Neumann series (which is the solution o f (21)) satisfies the Holder condition with respect to variable t with exponent A, and that the sum o f the Neumann series satisfies the Holder condition with respect to variable t with expo­

nent A.

Lemma 2. I f the function F ( - , ' t ) e C 1,h{S) for any te(0, T), then the solution f { - , t) of the integral equation (21) belongs to class C l,h(S) for any t e {0, T).

Lemma 2 is proved by application o f a localization principle based on the appropriate partition of unity (see [1], p. 63-88).

The function F(£, t) satisfies assumptions o f Lemmas 1 and 2, therefore on the basis o f these lemmas we conclude that the solution o f the integral equation (21) is o f class C 1,h:X(ST). W e shall show that the solution / o f the integral equation (21) is a solution o f the functional equation (19).

Let us suppose that the function / is not a solution o f the functional equation (19). Let us denote

where the function / (rj, x) is a solution o f the integral equation (21). W e have Ф(х, t) f it ) for хфй. It is easy to check that

0 S u n f)

(22⁾

(23) И т Ф ( х , t) = 0 ,

Application of Fisher-Riesz-Kupradze method 221

(24) НшФ(х, 0 = 0,

x$Q

(25) lim Ф(х, t) = 0.

Ijcf -►+ GO

Looking for the solution o f problems (22), (23), (24), (25) in the form o f the heat potential of the double layer with continuous and bounded density p

* dv

(26) Ф{х, t) = j jp(rj, t) — ( x - r j , t~z)dSqdz

o s d n n

and using condition (24) and the property of the heat potential of the double layer, we get an integral equation o f the form

1 dv

(27) ~ ? p ( £ , f) + J j> (? 7 , t) — - t - z ) d S vdz ⁼ 0.

0 S a n n

The equation (27) has exactly one solution p(rj, т) = 0. Therefore, Ф{х, t) = 0. The contradiction we arrived at ends the proof of Theorem 5.

Because the solution o f (19) belongs to class C 1,h,x(ST) the Lapunov- Taubers theorem is valid for the function и given by formula (18) (see [2], corollary, p. 142). Taking the above into account, it is easy to check that the function и is a solution o f problems (15), (16), (17).

C o n s t r u c t io n o f th e s o lu tio n . Let S1 be an arbitrary closed Lapunov’s surface which is the boundary of a domain Q1 which includes Q in its interior.

Let us write

Q\ = Q 1 x (0, T ); SlT = S1 x (0, T).

Let us take a countable dense set o f points {(£*, rf) } , k , i = 1, 2, ..., on the surface Sj and let us consider the set of the functions

ГКО rtpk , 4 J for T < t h

(28) r { C - r j , f -т) = < dn„

[ 0 for f ^ ^t< T, k, i = 1, 2, ...

Ordering the set (28) in a certain way, we write (29) Г (i ki - 4, t, . - t) = r t((/, ^t), 7 = 1 ,2 ,...,

and consider the space L 2(ST) with the norm

||ф|| T)\2dS„dx)1/2.

os

Le m m a 3. The set of functions {T j( -, •)}, j = 1, 2, ..., is linearly indepen­

dent.

222 Z. D o m a r ï s k i and Z. R o je k

Lemma 4. The set of functions {T j { -, •){, j = 1 , 2 , . . . , is complete in the domain L 2{ST).

The proofs o f Lemmas 3 and 4 are similar to those in [7], [8].

Subjecting the set {//•(*, •)] to the process of orthonormalization, we get the set \cûj(-, •)] o f orthonormal functions. The elements of the set [to/-, •){, 7 = 1 ,2 ,. . ., are linear combinations o f the elements of the set {ГД-, •)) and

vice versa (see [4], p. 72-73), i.e.,

CO 0 = £ A k j T k ( t i , t); Гj(rj, r ) = £ Bkjcok(rj, i ) .

k = 1 k = 1

We denote Fourier coefficients of the function / ( •, •) with respect to the set

\a)j(\ •);, j = 1, 2, ..., by

T

Ф] = ï I f(r i, ?)Mj(th ^dS^dx, j = 1 ,2 ,...

0s

Putting x = c J, Л ; t = t(j into equation (19), multiplying this equation by Ar and taking the sum from r equal one to r equal j, we get

T

Jо s

JJ/to, ^t) £ Arjr r(tt, r)dS„dx = £ Arj F(£kr, g r = 1

and therefore

Ф,

= I

r = 1

Because F (x , r) is a given function the coefficients Arj are the normalization constants, therefore all Fourier coefficients Ф, are determined. Since f e L 2{ST), we have

(30)

Let us write

(31)

lim Ц/- X Фко)к|| = 0.

fc= l

/ * ( > 7 , 0 = X Ф к ю к (>1, т), l

(32) и * / * , /) = f ^t) —dv(.y- ; ; , t — x)dSndx — F( x, t) .

о s dn„

Th e o r e m 6. For any x e Q and for any t from the interval (0 , T) and for any £ > 0 there exists a N 0 such that for N > N 0 we have

(33) \u(x, t) — uN(x, t)\ < 8.

P r o o f. W e have

dv (34) \u(x, t ) - u N(x, 01 ^ JJ|f i n , t) - f N(r],t)|

os dn.^{■ (x} tj, t -t) dSqdx.

Application of Fisher-Riesz-Kupradze method 223

Using the estimate dv

dn,( x - r j , t -t) ^ M [ ( t - z ) vab- 2vT \

where 0 < v < 1, a = inf|x — rj\ and M is a positive constant, we get

f/ e S

(35)

И

r6 — 2v

(1 — v) = M\.

On the basis of (30) we can choose an N such that

(36) IJ If ( 4 , T ) - / w(4, t)|2dSndx W f - r w 1 S)

os M i

Using the Schwarz inequality in (34) and taking into account (35) and (36), we get (33).

On the basis of Theorem 6 we get the solution of problems (15), (16), (17)

1 N dv

(37) u{x, t ) = lim J j ( X 4>kcok(ri,T)) — ( x - r i , t - T ) d S 4d T - F ( x , t ) . N ~*OD Cl S k= 1 Xln„

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