Dirichlet Problems for Harmonic Functions in Half Spaces
In our paper, we prove that if the positive part $u^{+}(x)$ of a harmonic function $u(x)$ in a half space satisfies the condition of slow growth, then its negative part $u^{-}(x)$ can also be dominated by a similar growth condition. Moreover, we give an integral representation of the function $u(x)$...
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| Дата: | 2014 |
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| Мова: | Українська Англійська |
| Опубліковано: |
Institute of Mathematics, NAS of Ukraine
2014
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| Онлайн доступ: | https://umj.imath.kiev.ua/index.php/umj/article/view/2228 |
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Репозитарії
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508181033123840 |
|---|---|
| author | Qiao, Lei Кіао, Лей |
| author_facet | Qiao, Lei Кіао, Лей |
| author_sort | Qiao, Lei |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T10:26:46Z |
| description | In our paper, we prove that if the positive part $u^{+}(x)$ of a harmonic function $u(x)$ in a half space satisfies the condition of slow growth, then its negative part $u^{-}(x)$ can also be dominated by a similar growth condition. Moreover, we give an integral representation of the function $u(x)$. Further, a solution of the Dirichlet problem in the half space for a rapidly growing continuous boundary function is constructed by using the generalized Poisson integral with this boundary function. |
| first_indexed | 2026-03-24T02:21:07Z |
| format | Article |
| fulltext |
UDC 517.9
Lei Qiao (School Math. and Inform. Sci., Henan Univ. Economics and Law, Zhengzhou, China)
DIRICHLET PROBLEMS FOR HARMONIC FUNCTIONS IN HALF SPACES*
ЗАДАЧI ДIРIХЛЕ ДЛЯ ГАРМОНIЧНИХ ФУНКЦIЙ У НАПIВПРОСТОРАХ
In the paper, we prove that if the positive part u+(x) of a harmonic function u(x) in a half space satisfies a condition of
slow growth, then its negative part u−(x) can also be dominated by a similar condition of growth. Moreover, we give
an integral representation of the function u(x). Further, a solution of the Dirichlet problem in the half space for a rapidly
growing continuous boundary function is constructed by using the generalized Poisson integral with this boundary function.
Доведено, що у випадку, коли додатна частина u+(x) гармонiчної функцiї u(x) у напiвпросторi задовольняє умову
повiльного зростання, її вiд’ємна частина u−(x) також може бути домiнована подiбною умовою зростання. Крiм
того, наведено iнтегральне зображення для функцiї u(x). Бiльш того, розв’язок задачi Дiрiхле в напiвпросторi для
швидко зростаючої неперервної граничної функцiї побудовано за допомогою узагальненого iнтеграла Пуассона з
цiєю граничною функцiєю.
1. Introduction and results. Let R and R+ be the sets of all real numbers and of all positive
real numbers, respectively. Let Rn, n ≥ 3, denote the n-dimensional Euclidean space with points
x = (x′, xn), where x′ = (x1, x2, . . . , xn−1) ∈ Rn−1 and xn ∈ R. The boundary and closure
of an open set D of Rn are denoted by ∂D and D respectively. The upper half space is the set
H = {(x′, xn) ∈ Rn : xn > 0}, whose boundary is ∂H.
For a set E, E ⊂ R+ ∪ {0}, we denote {x ∈ H : |x| ∈ E} and {x ∈ ∂H : |x| ∈ E} by HE and
∂HE, respectively. We identify Rn with Rn−1 ×R and Rn−1 with Rn−1 × {0}, writing typical
points x, y ∈ Rn as x = (x′, xn), y = (y′, yn), where y′ = (y1, y2, . . . , yn−1) ∈ Rn−1 and putting
x · y =
n∑
j=1
xjyj = x′ · y′ + xnyn, |x| =
√
x · x, |x′| =
√
x′ · x′.
Let Bn(r) denote the open ball with center at the origin and radius r(> 0) in Rn. We use the
standard notations u+ = max{u, 0}, u− = −min{u, 0} and [d] is the integer part of the positive
real number d. In the sense of Lebesgue measure dy′ = dy1 . . . dyn−1 and dy = dy′dyn. Let σ
denote (n − 1)-dimensional surface area measure and ∂/∂n denote differentiation along the inward
normal into H. For positive functions h1 and h2, we say that h1 . h2 if h1 ≤ dh2 for some positive
constant d.
Given a continuous function f on ∂H, we say that h is a solution of the (classical) Dirichlet
problem on H with f, if ∆h = 0 in H and limx∈H,x→z′ h(x) = f(z′) for every z′ ∈ ∂H.
The classical Poisson kernel for H is defined by
P (x, y′) =
2xn
ωn|x− y′|n
,
where ωn = 2πn/2/Γ(n/2) is the area of the unit sphere in Rn.
To solve the Dirichlet problem on H, as in [3, 4, 8, 11], we use the following modified Poisson
kernel of order m defined by
* This work was supported by the National Natural Science Foundation of China (Grant No. 11301140 and U1304102).
c© LEI QIAO, 2014
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10 1367
1368 LEI QIAO
Pm(x, y′) =
P (x, y′) when |y′| ≤ 1,
P (x, y′)−
∑m−1
k=0
2xn|x|k
ωn|y′|n+k
C
n/2
k
(
x · y′
|x||y′|
)
when |y′| > 1,
where m is a nonnegative integer, Cn/2k (t) is the ultraspherical (Gegenbauer) polynomials [10]. The
expression arises from the generating function for Gegenbauer polynomials
(1− 2tr + r2)−n/2 =
∞∑
k=0
C
n/2
k (t)rk,
where |r| < 1 and |t| ≤ 1. The coefficient Cn/2k (t) is called the ultraspherical (Gegenbauer) polyno-
mial of degree k associated with n/2, the function Cn/2k (t) is a polynomial of degree k in t.
Put
U(f)(x) =
∫
∂H
P (x, y′)f(y′)dy′ and Um(f)(x) =
∫
∂H
Pm(x, y′)f(y′)dy′,
where f(y′) is a continuous function on ∂H.
For any positive real number α, we denote by Aα the space of all measurable functions f(y) on
H satisfying ∫
H
yn|f(y)|dy
1 + |y|α+2
<∞ (1.1)
and Bα the set of all measurable functions g(y′) on ∂H such that∫
∂H
|g(y′)|dy′
1 + |y′|α
<∞. (1.2)
We also denote by Cα the set of all continuous functions u(x) onH, harmonic onH with u+(y) ∈ Aα
and u+(y′) ∈ Bα.
Theorem A (see [1, 9]). If u(x) > 0 and u ∈ Cn, then there exists a positive constant d1 such
that u(x) = d1xn + U(u)(x) for all x ∈ H.
Using the modified Poisson kernel Pm(x, y′),H. Yoshida (cf. [11], Theorem 1) and Siegel – Talvila
(cf. [8], Corollary 2.1) obtained classical solutions of the Dirichlet problem on H respectively.
Theorem B. If f ∈ Bn+m, then Um(f)(x) is a solution of the Dirichlet problem on H with f
satisfying
lim
|x|→∞, x∈H
|x|−m−1Um(f)(x) = 0.
Motivated by Theorems A and B, we first prove the following theorem.
Theorem 1. If α ≥ n and u ∈ Cα, then u ∈ Bα.
Then we concerned with the growth property of Um(f)(x) at infinity on H. In the half plane, this
result for α = 2 was obtained by Pan – Deng ([7], Theorem 1.1 and Remark 1.1).
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
DIRICHLET PROBLEMS FOR HARMONIC FUNCTIONS IN HALF SPACES 1369
Theorem 2. If α−n ≤ m < α−n+1 and f ∈ Cα, then Um(f)(x) is a solution of the Dirichlet
problem on H with f satisfying
lim
|x|→∞, x∈H
xn−1n |x|−αUm(f)(x) = 0. (1.3)
Then we are concerned with Dirichlet problems for harmonic functions of infinite order on H
(see [5] (Definition 4.1) for the definition of the order). For this purpose, we define a nondecreasing
and continuously differentiable function ρ(R) ≥ 1 on the interval [0,+∞). We assume further that
ε0 = lim sup
R→∞
ρ′(R)R lnR
ρ(R)
< 1. (1.4)
Let D(ρ, β) be the set of continuous functions f on ∂H such that∫
∂H
|f(y′)|dy′
1 + |y′|ρ(|y′|)+n+β−1
<∞, (1.5)
where β is a positive real number.
Now we have the following theorem.
Theorem 3. If f ∈ D(ρ, β), then the integral U[ρ(|y′|)+β](f)(x) is a solution of the Dirichlet
problem on H with f.
Theorem B follows from Theorem 3 (the case [ρ(|y′|) + β] = m), Theorems 1 and 2 (the case
α = n+m).
About integral representations for harmonic functions of finite order on H, we have the following
result.
Corollary 1. Let u ∈ Cα, α ≥ n, and let m be an integer such that n+m < α ≤ n+m+ 1.
(I) If α = n, then U(u)(x) is a harmonic function on H and can be continuously extended to H
such that u(x′) = U(u)(x′) for x′ ∈ ∂H. There exists a constant d2 such that u(x) = d2xn+U(u)(x)
for all x ∈ H.
(II) If α > n, then Um(u)(x) is a harmonic function on H and can be continuously extended
to H such that u(x′) = Um(u)(x′) for x′ ∈ ∂H. There exists a harmonic polynomial Qm(u)(x) of
degree at most m−1 which vanishes on ∂H such that u(x) = Um(u)(x) +Qm(u)(x) for all x ∈ H.
Finally, we have the following theorem.
Theorem 4. Let u be harmonic in H and continuous on H. If u ∈ D(ρ, β), then we have
u(x) = U[ρ(|y′|)+β](u)(x) + Π(x)
for all x ∈ H, where Π(x) is harmonic on H and vanishes continuously on ∂H.
2. Lemmas. The Carleman’s formula refers to holomorphic functions in a half space, which is
due to T. Carleman (see [2]). Recently, Y. H. Zhang, G. T. Deng and K. Kou (see [12], Lemma 1)
generalized it to harmonic functions on H, which plays an important role in our discussions.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1370 LEI QIAO
Lemma 1. If R > 1 and u(y) is a harmonic function on H with continuous boundary on ∂H,
then we have
m−(R) +
∫
∂H(1,R)
u−(y′)
(
1
|y′|n
− 1
Rn
)
dy′ =
= m+(R) +
∫
∂H(1,R)
u+(y′)
(
1
|y′|n
− 1
Rn
)
dy′ − d3 −
d4
Rn
,
where
m±(R) =
∫
{y∈H : |y|=R}
u±(y)
nyn
Rn+1
dσ(y),
d3 =
∫
{y∈H : |y|=1}
(
(n− 1)ynu(y) + yn
∂u(y)
∂n
)
dσ(y),
and
d4 =
∫
{y∈H : |y|=1}
(
ynu(y)− yn
∂u(y)
∂n
)
dσ(y).
Lemma 2 (see [5], Lemma 4.2).
|Pm(x, y′)| . xn|x|m−1|y′|−n−m+1 (2.1)
for any y′ ∈ ∂H[1, |x|/2),
|Pm(x, y′)| . x1−nn (2.2)
for any y′ ∈ ∂H[|x|/2, 2|x|),
|Pm(x, y′)| . xn|x|m|y′|−n−m (2.3)
for any y′ ∈ ∂H[1,∞) ∩ ∂H[2|x|,∞).
Lemma 3 (see [6], Theorem 10). Let h(x) be a harmonic function onH such that h(x) vanishes
continuously on ∂H. If
lim
|x|→∞, x∈H
|x|−m−1h+(x) = 0,
then h(x) = Qm(h)(x) on H, where Qm(h) is a polynomial of (x′, xn) ∈ Rn of degree less than m
and odd with respect to the variable xn.
3. Proof of Theorem 1. We distinguish the following two cases.
Case 1. α = n.
If R > 2, Lemma 1 gives
m−(R) +
(
1− 1
2n
) ∫
∂H(1,R/2)
u−(y′)
|y′|n
dy′ ≤
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
DIRICHLET PROBLEMS FOR HARMONIC FUNCTIONS IN HALF SPACES 1371
≤ m−(R) +
∫
∂H(1,R)
u−(y′)
(
1
|y′|n
− 1
Rn
)
dy′ ≤
≤ m+(R) +
∫
∂H(1,R)
u+(y′)
|y′|n
dy′ + |d3|+ |d4|. (3.1)
Since u ∈ Cn, we obtain
1
n
∞∫
1
m+(R)
R
dR =
∫
H(1,∞)
ynu
+(y)
|y|n+2
dy .
∫
H
ynu
+(y)
1 + |y|n+2
dy <∞
from (1.1) and hence
lim inf
R→∞
m+(R) = 0, (3.2)
where m+(R) is defined in Lemma 1.
Then from (1.2), (3.1) and (3.2) we have
lim inf
R→∞
∫
∂H(1,R/2)
u−(y′)
|y′|n
dy′ <∞,
which gives ∫
∂H
u−(y′)
1 + |y′|n
dy′ <∞.
Thus u ∈ Bn from |u| = u+ + u−.
Case 2. α > n.
Since u ∈ Cα, we see from (1.1) that
1
n
∞∫
1
m+(R)
Rα−n+1
dR =
∫
H(1,∞)
ynu
+(y)
|y|α+2
dy .
∫
H
ynu
+(y)
1 + |y|α+2
dy <∞ (3.3)
and see from (1.2) that
∞∫
1
1
Rα−n+1
∫
∂H(1,R)
u+(y′)
(
1
|y′|n
− 1
Rn
)
dy′dR =
=
∫
∂H(1,∞)
u+(y′)
∞∫
|y′|
1
Rα−n+1
(
1
|y′|n
− 1
Rn
)
dRdy′ .
.
n
(α− n)α
∫
∂H(1,∞)
u+(y′)
|y′|α
dy′ <∞. (3.4)
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1372 LEI QIAO
We have from (3.3), (3.4) and Lemma 1
∫
∂H(1,∞)
u−(y′)
∞∫
|y′|
1
Rα−n+1
(
1
|y′|n
− 1
Rn
)
dRdy′ ≤
≤
∞∫
1
m+(R)
Rα−n+1
dR−
∞∫
1
1
Rα−n+1
(
d3 +
d4
Rn
)
dR+
+
∞∫
1
1
Rα−n+1
∫
∂H(1,R)
u+(y′)
(
1
|y′|n
− 1
Rn
)
dy′dR <∞.
Set
I(α) = lim
|y′|→∞
|y′|α
∞∫
|y′|
1
Rα−n+1
(
1
|y′|n
− 1
Rn
)
dR.
We get
I(α) =
n
α(α− n)
from the L’hospital’s rule and hence we have
∞∫
|y′|
1
Rα−n+1
(
1
|y′|n
− 1
Rn
)
dR & |y′|−α.
So ∫
∂H[1,∞)
u−(y′)
|y′|α
dx′ .
∫
∂H[1,∞)
u−(y′)
∞∫
|y′|
1
Rα−n+1
(
1
|y′|n
− 1
Rn
)
dRdy′ <∞.
Then u ∈ Bα from |u| = u+ + u−.
Theorem 1 is proved.
4. Proof of Theorem 2. For any fixed x ∈ H, take a number R1 satisfying R1 > max{1, 2|x|},
we have ∫
∂H(R1,∞)
|Pm(x, y′)||f(y′)|dy′ .
. xn|x|m
∫
∂H(R1,∞)
|y′|−n−m|f(y′)|dy′ .
. xn|x|α−n
∫
∂H(R1,∞)
|y′|−α|f(y′)|dy′ <∞
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
DIRICHLET PROBLEMS FOR HARMONIC FUNCTIONS IN HALF SPACES 1373
from (2.3) and Theorem 1. Thus Um(f)(x) is finite for any x ∈ H. Since Pm(x, y′) is a harmonic
function of x ∈ H for any fixed y′ ∈ ∂H. Um(f)(x) is also a harmonic function of x ∈ H.
To verify the boundary behavior of Um(f)(x). For any fixed boundary point z′ ∈ ∂H, we can
choose a number R2 such that R2 > |z′|+ 1. Now we write
Um(f)(x) = I1(x)− I2(x) + I3(x),
where
I1(x) =
∫
∂H[0,R2]
P (x, y′)f(y′)dy′,
I2(x) =
m−1∑
k=0
2xn|x|k
ωn
∫
∂H(1,R2]
1
|y′|n+k
C
n/2
k
(
x′ · y′
|x||y′|
)
f(y′)dy′,
and
I3(x) =
∫
∂H(R2,∞)
Pm(x, y′)f(y′)dy′.
Notice that I1(x) is the Poisson integral of f(y′)χBn−1(R2)(y
′), where χBn−1(R2) is the charac-
teristic function of the ball Bn−1(R2). So it tends to f(z′) as x → z′. Since I2(x) is a polynomial
times xn and I3(x) = O(xn), both of them tend to zero as x → z′. So the function Um(f)(x) can
be continuously extended to H such that Um(f)(z′) = f(z′) for any z′ ∈ ∂H. Then Um(f)(x) is a
solution of the Dirichlet problem on H with f.
For any ε > 0, there exists Rε > 2 such that∫
∂H[Rε,∞)
|f(y′)|
1 + |y′|α
dy′ < ε (4.1)
from Theorem 1. For any fixed x ∈ H[2Rε,∞), we write
Um(f)(x) =
4∑
i=1
∫
Gi
Pm(x, y′)f(y′)dy′ =
4∑
i=1
Vi(x),
where G1 = ∂H[0, 1), G2 = ∂H[1, |x|/2), G3 = ∂H[|x|/2, 2|x|), and G4 = ∂H[2|x|,∞).
First note that
|V1(x)| . xn
(
|x|
2
)−n ∫
G1
|f(y′)|dy′ . xn|x|−n. (4.2)
By (2.1) we have
|V2(x)| . xn|x|m−1
∫
G2
|y′|−n−m+1|f(y′)|dy′ . xn|x|α−n
∫
G2
|y′|−α|f(y′)|dy′. (4.3)
Write
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1374 LEI QIAO
V2(x) = V21(x) + V22(x),
where
V21(x) =
∫
G2
⋂
Bn−1(Rε)
Pm(x, y′)f(y′)dy′
and
U22(x) =
∫
G2−Bn−1(Rε)
Pm(x, y′)f(y′)dy′.
If |x| > 2Rε1 , then
|V21(x)| . Rα−n−m+1
ε xn|x|m−1.
Moreover, by (4.1) and (4.3) we obtain
|V22(x)| . εxn|x|α−n.
That is
|V2(x)| . εx1−nn |x|α. (4.4)
We also obtain the following estimates:
|V3(x)| . εx1−nn |x|α, (4.5)
|V4(x)| . xn|x|m
∫
G4
|y′|−n−m|f(y′)|dy′ . εxn|x|α−n (4.6)
from (2.2), (2.3), and (4.1).
Combining (4.2) and (4.4) – (4.6), (1.3) holds.
Theorem 2 is proved.
5. Proof of Theorem 3. Take a number r satisfying r > R3, where R3 is a sufficiently large
positive number. For any ε (0 < ε < 1− ε0), by (1.4) we have
ρ(r) < ρ(e)(ln r)(ε0+ε), (5.1)
which yields that there exists a positive constant M(r) dependentbonly on r such that
k−
β
2 (2r)ρ(k+1)+β+1 ≤M(r) (5.2)
for any k ≥ kr = [2r] + 1.
For any x ∈ H and |x| ≤ r, we obtain by (1.5), (2.3) and (5.2)
∞∑
k=kr
∫
∂H[k,k+1)
(2|x|)[ρ(|y′|)+β]+1
|y′|[ρ(|y′|)+β]+n
|f(y′)|dy′ .
.
∞∑
k=kr
∫
∂H[k,k+1)
(2r)ρ(k+1)+β+1
|y′|[ρ(|y′|)+β]−ρ(|y′|)−
β
2
+1
|f(y′)|
|y′|ρ(|y′|)+n+
β
2
−1
dy′ .
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
DIRICHLET PROBLEMS FOR HARMONIC FUNCTIONS IN HALF SPACES 1375
.
∞∑
k=kr
k−
β
2 (2r)ρ(k+1)+β+1
∫
∂H[k,k+1)
2|f(y′)|
1 + |y′|ρ(|y′|)+n+
β
2
−1
dy′ .
.
∞∑
k=kr
M(r)
∫
∂H[k,k+1)
|f(y′)|
1 + |y′|ρ(|y′|)+n+
β
2
−1
dy′ .
.M(r)
∫
∂H[kr,∞)
|f(y′)|
1 + |y′|ρ(|y′|)+n+
β
2
−1
dy′ <∞. (5.3)
Notice that
|U[ρ(|y′|)+β](f)(x)| ≤W1(x) +W2(x) +W3(x),
where
W1(x) =
∫
∂H[0,1]
2
ωnx
n−1
n
|f(y′)|dy′,
W2(x) =
∫
∂H(1,kr]
2[ρ(|y
′|)+β]
ωn
(
2n + [ρ(|y′|) + β]C
n/2
[ρ(|y′|)+β]−1(1)
)
×
× |x|[ρ(|y′|)+β]+n+1
xn−1n |y′|[ρ(|y′|)+β]+n−1
|f(y′)|dy′,
and
W3(x) =
∫
∂H(kr,∞)
2[ρ(|y
′|)+β]+n+1xnx
[ρ(|y′|)+β]
ωn|y′|[ρ(|y′|)+β]+n
|f(y′)|dy′.
By (1.5) we have
W1(x) . x1−nn
∫
∂H[0,1]
|f(y′)|
1 + |y′|ρ(|y′|)+n+β−1
dy′ <∞ (5.4)
and
W2(x) ≤ 2mr
ωn
(
2n +mrC
n/2
mr−1(1)
) rmr+n−1
xn−1n
∫
∂H(1,kr]
2|y′|ρ(|y′|)+β−[ρ(|y′|)+β]|f(y′)|
1 + |y′|ρ(|y′|)+n+β−1
dy′ ≤
≤ 2mr
ωn
(
2n +mrC
n/2
mr−1(1)
) rmr+n−1
xn−1n
2kr
∫
∂H
|f(y′)|
1 + |y′|ρ(|y′|)+n+β−1
dy′ <∞, (5.5)
where mr = [ρ(kr) + β].
On the other hand we have, by (5.3), that
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1376 LEI QIAO
W3(x) ≤
∞∑
k=kr
∫
∂H[k,k+1)
2n
ωn
(2|x|)[ρ(|y′|)+β]+1
|y′|n+[ρ(|y′|)+β] |f(y′)|dy′ .
.M(r)
∫
∂H
|f(y′)|
1 + |y′|ρ(|y′|)+n+
β
2
−1
dy′ <∞. (5.6)
Thus U[ρ(|y′|)+β](f)(x) is finite for any x ∈ H from (5.4), (5.5) and (5.6). Since P[ρ(|y′|)+β](x, y
′)
is a harmonic function of x ∈ H for any fixed y′ ∈ ∂H. U[ρ(|y′|)+β](f)(x) is also a harmonic function
of x ∈ H.
To verify the boundary behavior of U[ρ(|y′|)+β](f)(x). For any fixed boundary point z′ ∈ ∂H,
we can choose a number R4 such that R4 > 2. Let DR4 = H ∩ Bn−1(R4) and χBn−1(R4) be the
characteristic function of Bn−1(R4).
Set
S[ρ(|y′|+β)](x, y
′) =
[ρ(|y′|+β)]−1∑
k=0
2xn|x|k
ωn
1
|y′|n+k
C
n/2
k
(
x′ · y′
|x||y′|
)
.
Write
U[ρ(|y′|)+β](f)(x) = X(x)− Y (x) + Z(x),
where
X(x) =
∫
∂H[0,2R4]
P (x, y′)f(y′)dy′,
Y (x) =
∫
∂H(1,2R4]
S[ρ(|y′|+β)](x, y
′)f(y′)dy′, and Z(x) =
∫
∂H(2R4,∞)
P[ρ(|y′|+β)](x, y
′)f(y′)dy′.
Note that X(x) =
∫
∂H
P (x, y′)f(y′)χBn−1(R4)(y
′)dy′, which tends to f(z′) as x→ z′. Further,
X(x) is harmonic on DR4 and can be continuously extended to DR4 . Since S[ρ(|y′|+β)](x, y
′) is a
harmonic polynomial of x and tends to zero as x→ z′, Y (x) is also a harmonic polynomial of x and
tends to zero as x→ z′.
From (1.5), we have Z(x) is a harmonic function on H. P[ρ(|y′|+β)](x, y
′) = 0 implies that
Z(z′) = 0, where |z′| ≤ R4.
Moreover, (5.1) also implies that there exists a positive constant M(R4) dependent only on R4
such that
2n+1
ωn
(2R4)
ρ(k+1)+β
k
β
2
≤M(R4) (5.7)
for any k ≥ [2R4].
Hence it follows from (1.5) and (5.2) that
|Z(x)− Z(z′)| = |Z(x)| ≤
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
DIRICHLET PROBLEMS FOR HARMONIC FUNCTIONS IN HALF SPACES 1377
≤
∫
∂H(2R4,∞)
|P[ρ(|y′|+β)](x, y
′)||f(y′)|dy′ ≤
≤
∫
∂H[[2R4],∞)
2n+1
ωn
xn(2R4)
[ρ(|y′|)+β]
|y′|[ρ(|y′|)+β]−ρ(|y′|)−
β
2
+1
|f(y′)|
|y′|ρ(|y′|)+
β
2
+n−1
dy′ ≤
≤
∞∑
k=[2R4]
∫
∂H[k,k+1)
2n+1
ωn
xn(2R4)
ρ(k+1)+β
k
β
2
2|f(y′)|
1 + |y′|ρ(|y′|)+
β
2
+n−1
dy′ .
. xn
∞∑
k=[2R4]
∫
∂H[k,k+1)
M(R4)
|f(y′)|
1 + |y′|ρ(|y′|)+
β
2
+n−1
dy′ .
. xnM(R4)
∫
∂H
|f(y′)|
1 + |y′|ρ(|y′|)+
β
2
+n−1
dy′ . xnM(R4),
which tends to zero as x → z′. Thus Z(x) is harmonic on DR4 and can be continuously extended
to DR4 .
From the arbitrariness of R4, we have that the function U[ρ(|y′|)+β](f)(x) can be continuously
extended to H such that U[ρ(|y′|)+β](f)(z′) = f(z′) for any z′ ∈ ∂H.
Theorem 3 is proved.
6. Proof of Corollary 1. To prove (II). Consider the function u(x)−Um(u)(x). Then it follows
from Theorem 2 that this is harmonic in H and vanishes continuously on ∂H. Since
0 ≤ (u(x)− Um(u)(x))+ ≤ u+(x) + Um(u)−(x) (6.1)
for any x ∈ H and
lim inf
|x|→∞
|x|−m−1u+(x) = 0 (6.2)
from (1.1), for every x ∈ H we have
u(x) = Um(u)(x) +Qm(u)(x)
from (1.3), (6.1), (6.2) and Lemma 3, where Qm(u) is a polynomial in Rn of degree at most m− 1
and odd with respect to the variable xn. From these we evidently obtain Corollary (II).
If u ∈ Cn, then u ∈ Cα for α > n. Corollary (II) gives that there exists a constant d5 such that
u(x) = d5xn + U1(u)(x).
Put
d2 = d5 −
1
wn
∫
∂H[1,∞)
f(y′)
|y′|n
dy′.
It immediately follows that u(x) = d2xn +U(u)(x) for every x ∈ H, which is the conclusion of
Corollary (I).
Corollary 1 is proved.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1378 LEI QIAO
7. Proof of Theorem 4. Consider the function u(x)− U[ρ(|y′|)+β](u)(x), which is harmonic in
H, can be continuously extended to H and vanishes on ∂H.
The Schwarz Reflection Principle [1, p.68] applied to u(x)− U[ρ(|y′|)+β](u)(x) shows that there
exists a harmonic function Π(x) on H such that Π(x∗) = −Π(x) = U[ρ(|y′|)+β](u)(x) − u(x) for
x ∈ H, where ∗ denotes reflection on ∂H just as x∗ = (x′,−xn).
Thus u(x) = U[ρ(|y′|)+β](u)(x) + Π(x) for all x ∈ H, where Π(x) is a harmonic function on H
vanishing continuously on ∂H.
Theorem 4 is proved.
1. Axler S., Bourdon P., Ramey W. Harmonic function theory. – Second Ed. – New York: Springer-Verlag, 1992.
2. Carleman T. Über die Approximation analytischer Funktionen durch lineare Aggregate von vorgegebenen Potenzen //
Ark. mat., astron. och fys. – 1923. – 17. – S. 1 – 30.
3. Finkelstein M., Scheinberg S. Kernels for solving problems of Dirichlet typer in a half-plane // Adv. Math. – 1975. –
18, № 1. – P. 108 – 113.
4. Gardiner S. J. The Dirichlet and Neumann problems for harmonic functions in half-spaces // J. London Math. Soc. –
1981. – 24. – P. 502 – 512.
5. Hayman W. K., Kennedy P. B. Subharmonic functions. – London: Acad. Press, 1976. – Vol. 1.
6. Kuran Ü. Study of superharmonic functions in Rn × (0,∞) by a passage to Rn+3 // Proc. London Math. Soc. –
1970. – 20. – P. 276 – 302.
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Ser. A. Chin. Ed. – 2011. – 31. – P. 892 – 901.
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15. – P. 333 – 360.
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Received 12.10.12,
after revision — 19.09.13
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
|
| id | umjimathkievua-article-2228 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | Ukrainian English |
| last_indexed | 2026-03-24T02:21:07Z |
| publishDate | 2014 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/a0/48f9487dc80f77ba14644c416ff659a0.pdf |
| spelling | umjimathkievua-article-22282019-12-05T10:26:46Z Dirichlet Problems for Harmonic Functions in Half Spaces Задачі Діріхле для гармонічних функцій у напівпросторах Qiao, Lei Кіао, Лей In our paper, we prove that if the positive part $u^{+}(x)$ of a harmonic function $u(x)$ in a half space satisfies the condition of slow growth, then its negative part $u^{-}(x)$ can also be dominated by a similar growth condition. Moreover, we give an integral representation of the function $u(x)$. Further, a solution of the Dirichlet problem in the half space for a rapidly growing continuous boundary function is constructed by using the generalized Poisson integral with this boundary function. Доведено, що у випадку, коли додатна частина $u^{+}(x)$гармонічної функції $u(x)$ у напiвпросторi задовольняє умову повільного зростання, її від'ємна частина $u^{-}(x)$ також може бути домінована подібною умовою зростання. Крім того, наведено інтегральне зображення для функції $u(x)$. Більш того, розв'язок задачі Діріхле в напівпросторі для швидко зростаючої неперервної граничної функції побудовано за допомогою узагальненого інтеграла Пуассона з цією граничною функцією. Institute of Mathematics, NAS of Ukraine 2014-10-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2228 Ukrains’kyi Matematychnyi Zhurnal; Vol. 66 No. 10 (2014); 1367–1378 Український математичний журнал; Том 66 № 10 (2014); 1367–1378 1027-3190 uk en https://umj.imath.kiev.ua/index.php/umj/article/view/2228/1457 https://umj.imath.kiev.ua/index.php/umj/article/view/2228/1458 Copyright (c) 2014 Qiao Lei |
| spellingShingle | Qiao, Lei Кіао, Лей Dirichlet Problems for Harmonic Functions in Half Spaces |
| title | Dirichlet Problems for Harmonic Functions in Half Spaces |
| title_alt | Задачі Діріхле для гармонічних функцій у напівпросторах |
| title_full | Dirichlet Problems for Harmonic Functions in Half Spaces |
| title_fullStr | Dirichlet Problems for Harmonic Functions in Half Spaces |
| title_full_unstemmed | Dirichlet Problems for Harmonic Functions in Half Spaces |
| title_short | Dirichlet Problems for Harmonic Functions in Half Spaces |
| title_sort | dirichlet problems for harmonic functions in half spaces |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2228 |
| work_keys_str_mv | AT qiaolei dirichletproblemsforharmonicfunctionsinhalfspaces AT kíaolej dirichletproblemsforharmonicfunctionsinhalfspaces AT qiaolei zadačídíríhledlâgarmoníčnihfunkcíjunapívprostorah AT kíaolej zadačídíríhledlâgarmoníčnihfunkcíjunapívprostorah |