ROCZNIK1 PO LS Kl EGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXVI (1986)
Eu g e n i u s z W a c h n i c k i
(Krakôw)
On the limit properties o f the solution o f the Dirichlet problem for the equation (A — c2) u( x, у) = 0 in the half-plane 1. In paper [6] a solution of the Dirichlet problem for the equation {A —c2) u( x , у) = 0, c > 0, in the half-plane Æ+ = {(x, y): |x| < oo, y > 0] has . been given. This solution is of the form
u(x, y) = /(/; x, y) = —^ = I f ( x - s ) H ( s , y)ds = f * H ( -, y)(x),
where
12 1
H (s, y) — - c y-
71 \/},2 + s
2 K 1 (C\/y2+S2)
and K v is the MacDonald function of index v ([7]).
The purpose of this paper is to present some limit property of the function / (/;x, y) as y 0+. Such a problem in the case of Laplace equation was investigated in papers [1 ]— [3], [5].
2. Let
соbe a function defined in
[0, oo).If (i) co(0) = 0,
(ii) со is a continuous and non-decreasing function in .[0,
oc),(iii) for every t ^ 0 and every positive integer n
со (nt) ^ n2co(r),
then we call со a function of the type of 2-nd modulus of smoothness.
It is easy to show that there exist positive constants C1?
C2 such thatl
(1) Cj
t2
^ co(0 ^ C2 j~y^d z,0 < t < j . t
Moreover, there exists a function со* satisfying conditions (i)— (iii) such that
5 Co(f) ^ co*(f) ^ co(f),
0 ^ t<
oo,and the function
co*(t)/t2is a non-increasing one in (0,
oo).180 E . W a c h n i c k i
Let Щ L„ denote the set of all functions f e L p(R) (1 ^ p < oo) for which
\\f(- + h ) - 2 f ( ' ) + f ( - - h ) \ \ p * : a > m,
where со is a function of the type of 2-nd modulus of smoothness satisfying the condition
m o
h ^
J *
о We prove
Th e o r e m 1.
I f f е Щ Ь р
(1 ^p
<oo), then there exist real numbers M > 0 and a ^ 1 such that
(2) II/(/; % y)—f (* )llp < M ^ ° ^ ~ d t + y J ^ - d t + y j.
о у
Proof. It is known ([6]) that I ( \ ; x, y) = exp( — cy), hence
(3) II/(/; -, y)-/(-)llp < »/(/; -, l; -, y)llp+
+ll/(i;% y)/(-)-/(-)llp
= ll/(/; -, y ) - H i;-, y)/(-)llp+ll/llp|exp(-cy)-i|
<11 /(/;-, y ) - / ( i; - , y)/(-)llp+c3y,
where C3 is a positive constant.
The kernel H(s, y) of the singular integral /(/; x, y) is a positive one and it is an even function with respect to s. Hence, by Holder-Minkowski inequality ([2]) we obtain
(4) l|/(/; -, •, y)f(')\\p l r
UJ (/ (x - s )- / (x ))# (s , y)ds
p \i/p dx
^/bz J ^/271 (/ (x - s) - 2/ (x) +/ (x + s)) H(s, y) ds
p \i/p
dx j
< U Iu , . | / (x -s )-2 / (x )+ / (x + s)|pdx y/p H{s,y)ds
J l K
J
\ y / 2nJ /
a)(s)H(s, y)ds.
We have the following ([6])
H( s, y) < ~y{ y2 + s2) 1, J > 0, |s| <
oo. V n
Since ([7])
К 1 (z) ~ /— exp( — z) asz->oo,
thus there exists a constant a ^ 1 such that
H(s, y) ^ M t ys~3/2exp( — cs) for s ^ a , where is a positive constant.
Hence
1 f , u f y^(s) 1 f yw(s)
J b . J л J s2+y2 n ] s 2 + y2
О О у
+ M 2y со (s) s 3/2 exp ( — cs) ds < — f ds+—
71 J S 71
О у
'co(s) ds +
+ M 2y co(s)s 3/2 exp ( — cs)ds,
where M 2 is a positive constant.
It is easy to show that 00
Jco(s)s~3/2exp( — cs)ds ^ C4, .*
a
where C4 is a positive constant,
Hence, by (3), (4) and (5) we obtain (2).
From Theorem 1, in the case where to(h) = Khx, 0 < a ^ 1, К is а positive constant, we have the following
Co r o l l a r y.
I f f e L p(R)
( 1^ p <
00)and
then
11/(4- h ) - 2 f ( • ) + / ( - h)\\p = 0(\h\% 0 < a ^ 1,
II/(/;*, y)-/(-)llp =
0 (y l 0(y\\ny\)
for 0 < a < 1,
for
ol= 1
as у — ■» 0+.
182 E . W a c h n i c k i
3. Now we consider (in a way) the inverse problem to the problem from Theorem 1. First, we prove an inequality of the Bernstein type for the integral /(/; x, y).
Le m m a
1. I f f e L p(R), 1 ^ p < oo, then ( 6 )
dx2 I if', x, y) < MAf\\Py~\
where M 3 is a positive number.
Proof. It follows from the results of [ 6 ] that
dx2 /(/; x, y) = f (s ) к % R
d_2_
dx2 H( x — s, y)ds.
Applying the formula ([7])
y -z v K v(z) z~v К
v + 1(z) we get
(7) ^ / ( / ; x, y) = — \ f { x - s ) s 2{cQ) 5K 3{cg)ds c6y
сАУ f { x - s ) { c e ) K 2(cQ)ds,
where
q2= s 2 + y 2.
In view of properties of the convolution ([2]) and (7) we have d2
dx 2/( / ;
у)
с у
^ Wf Wp^- ^i C
|s 20 3K 3{cg)ds + Ip 2
K 2{cq)ds cz у
2^ 2||/||_ —I c s2
q3 K 3(cg)ds +
q2K 2(cg)ds By the formula ([7])
00
J t2m+1(t2 + z2) ~n/2K n{ a J t 2+ z 2)dt = K „ - m-i(az)
о
for m > — 1, a > 0, z > 0 and n = 1, 2, we obtain i l
ôx2 i (/; y) ^ C 5\\f\\py - l>2K 3l2(cy), where C5 is a positive constant.
Since ([7])
K 3/2(z) = J ^ { l + l / z ) e x p { - z ) , z > 0,
then (6) is true.
We shall prove
T
heorem2. Let f e L p(R), 1 ^ p < oo. I f
(8) \\I ( f- , y) - f ( - ) \\P ^co(y),
where co{y) (y > 0) is a function of the type of 2-nd modulus of smoothness in L p metric, then 2-nd modulus of smoothness co2(f, t) of the function f in L p metric satisfies the inequality
l
(9)
co2 ( f ,r)=SC6r2 0 < f < 1/2,
t * where C6 is a positive constant.
Proof. We apply the Bernstein method. We have m- 1
f ( x ) = /(/; x, i ) + X [/(/; x, 1/2j +l ) - I ( f ; x, l/ 2 ')]+ / (x )-/ (/ ; x, 1/2")
j= ifor every integer m ^ 2.
Hence
f (x + h) — 2f (x) + / (x — h) = (x, h) + A 2(x, h) + A3(x, h), where
A i (x, h) = /(/; x + h , % ) - 2 I ( f ; x, i) + /(/; x - h , {),
m— 1
A2( x , h ) = z {[/ (/ ; * + &, 1/2J + l) — I ( f ' , x + h, 1 / 2 J)] —
— 2 [/ (/ ; x, i/2j +i ) — I ( f ; x, 1/2-0] + + [/(/; * - A , 1/2J + x) —/(/; x - A , 1/20]}, Л3(х, A) = [/ (x + A) — / (/; x +A, 1/2")]- 2 [/ (x )- / (/ ; x, 1/2-)] +
+ [/ (x — A) — / (/; x — A, 1/2")].
184 E . W a c h n i c k i
We notice that
/(/; x + h, i ) - 2 /(/; x, !) + /(/; x-/i, i) ft
/2ft
/ 2^ / ( / ; x + f i + t j , $)dtl dt2.
— ft/2 —h/2
Hence, by Holder-Minkowski inequality and by Lemma 1 we get
ft/ 2 h/2(10) \\AA',h)\\P -
«/ v
■h/2 -h/2
d2
dx 2I { f ‘> x + tx + t2, -j)dt! dt2
< h 2 x, i) ^ C 7 h2,
where C7 is a positive constant.
From the assumption and Hôlder-Minkowski inequality if follows that
(И ) 1Из(‘> h)\\p ^ 4co(l/2m).
In order to estime \\Л2(-, h)\\p let
(12) m„ (x) = /(/; x, 1/2n+1) —/(/; x, 1/2"), n = 1, 2, ...
From the Fubini’s theorem it follows that
/(/(/; *, 1/2"); X, l/2"+1) = /(/(/; X, l/2"+1); x, 1/2").
Hence, by the properties of the convolution and by (12) we get (13) u„(x)
= / ((/ (• )-/ (/ ; -, 1/2")); x, l/2"+ * )- / ((/ ( )- / ( / ; -, l/2"+1)); x, 1/2").
Further, we find
h/2 h/2
u„ (x + h) - 2 un (x) + un( x - h ) = A 2
dx 2un(x + t1+ t 2)dtl dt2,
-h/2 -h/2hence, by (6) and (13) m- 1
И2(%
h)\\P
^ Z \\Uj( '+ h) - 2uj { ' ) + uJ(--h)\\l j= 1
m-l ( Д2
h2 I \ g p / ((/ (• )-/ (/ ; % i/2J)); x, i/2J+1) +
+ A 2 _
dx2
/ ( ( / ( • ) - / ( / ; • . i / 2 J + l ));x , 1/21)m— 1
< M 3h2 I {! ! / ( • ) - / ( / ; -, s-1
+11 / (• )- / (/ ; -, i/2J+')llp22j! . By assumption and by properties of the function to we have
m— 1
(14) \\A2(-, h)U, ^ M 3h2 £ (<u( l/2j) 22° + u + w ( 1/2-7 + *)22J)
j = i m - 1
^ 5 M 3/i2 X w ( 1/2') 22
jj = i m— 1
= 10M3 h2 £ ûi(l/20(l/20-3(l/ ÿ +1)
i = i
1/2
^ C 8 h 2
J
w ( z ) z ~ 3 d z , 2 ~ mwhere C8 is a positive constant.
Therefore it follows from (10), (11) and (14) that
1/2
\\f(- + h ) - 2 f ( - ) + f ( - - h ) \ \ p ^ C 1h2 + 4co(l/2m) + Cgh2 J ( o( z) z~4z
2~m
for every integer m ^ 2 and real / 1 .
If we assume 0 < t < 1/2 and choose m ^ 2 such that the inequality 2~m
< t < 2~m+i holds, then by properties of the function o> we get
1/2
ll/ (‘ + ^) —2/(•)+/(• —/i)||p ^ C7 h2 + 4(o(t) + C8 h2 J m (z)z-3rfz.
t/2
Consequently
1
(o2(f, 0 < C7 f2 + 4a>(f) + C9f2 Jcu(z)z~3dz t
and by (1) we have (9).
In the case where co(h) = Khx, 0 < a < 1 from Theorem 2 it follows
Co r o l l a r y.
I f f e L p(R),
1 <p
< 00and
ll/(/;% y)-f(-)\\P = o ( f ) , y > 0 ,
then \\f(- + h ) - 2 f ( - ) + f ( - - h ) \ \ p = 0(\h\*) as Л - 0 .
We notice that all the results of the present paper remain true if p = оc .
The proofs are similar, but more elementary.
186 E . W a c h n i c k i
References
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