# The Dirichlet Problem on the Upper Half-Space

Authors: Jinjin Huang; Lei Qiao
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

A solution of the Dirichlet problem on the upper half-space is constructed by the generalized Dirichlet integral with a fast-growing continuous boundary function.


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## Body

## 1. Introduction and Results

LetRn(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. 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
(1.1)x·y=∑j=1n‍xjyj=x'·y'+xnyn,|x|=x·x,|x′|=x'·x'.LetB(r) denote the open ball with center at the origin and radius r, and let σ denote (n-1)-dimensional surface area measure. Let [d] denote the integer part of the positive real number d. In the sense of Lebesgue measure, dy′=dy1⋯dyn-1 and dy=dy'dyn.Given a continuous functionf on ∂H, we say that h is a solution of the (classical) Dirichlet problem on H with f if Δh=0 in H and limx∈H,x→z′h(x)=f(z') for every z'∈∂H.The classical Poisson kernel forH is defined by P(x,y')=2xnωn-1|x-y'|-n, where ωn=2πn/2/Γ(n/2) is the area of the unit sphere in Rn.To solve the Dirichlet problem onH, as in [1–6], we use the following modified Poisson kernel of order m defined by
(1.2)Pm(x,y′)={P(x,y′)when|y′|≤1,P(x,y′)-∑k=0m-12xn|x|kωn|y′|n+kCkn/2(x·y′|x||y′|)when|y′|>1,
where m is a nonnegative integer, and Ckn/2(t) is the ultraspherical (Gegenbauer) polynomials [7]. The expression arises from the generating function for Gegenbauer polynomials
(1.3)(1-2tr+r2)-n/2=∑k=0∞‍Ckn/2(t)rk,
where |r|<1 and |t|≤1. The coefficient Ckn/2(t) is called the ultraspherical (Gegenbauer) polynomial of degree k associated with n/2, and the function Ckn/2(t) is a polynomial of degree k in t.Put(1.4)Um(f)(x)=∫∂H‍Pm(x,y′)f(y′)dy',
where f(y') is a continuous function on ∂H.Using the modified Poisson kernelPm(x,y'), Yoshida (cf. [6, Theorem 1]) and Siegel and Talvila (cf. [5, Corollary 2.1]) gave classical solutions of the Dirichlet problem on H, respectively. Motivated by their results, we consider the Dirichlet problem for harmonic functions of infinite order (e.g., see [8, Definition 4.1, page 2, Line 12] for the definition of harmonic functions).To do this, we define a nondecreasing and continuously differentiable functionρ(r)≥1 on the interval [0,+∞). We assume further that
(1.5)ε0=limsupr→∞ρ′(r)rlogrρ(r)<1.LetF(ρ,β) be the set of continuous functions f on ∂H such that
(1.6)∫∂H‍|f(y′)|dy'1+|y′|ρ(|y′|)+n+β-1<∞,
where β is a positive real number.Now, we have the following.Theorem 1.1.
Iff∈F(ρ,β), then the integral U[ρ(|y'|)+β](f)(x) is a solution of the Dirichlet problem on H with f.
If one puts[ρ(|y'|)+β]=m in Theorem 1.1, one immediately obtains the following (cf. [6, Theorem 1] and [5, Corollary 2.1]).Corollary 1.2.
Iff is a continuous function on ∂H satisfying ∫∂H‍|f(y')|(1+|y'|)-n-mdy'<∞, then Um(f)(x) is a solution of the Dirichlet problem on H with f.Theorem 1.3.
Letu be harmonic in H and continuous on H¯. If u∈F(ρ,β), then one has
(1.7)u(x)=U[ρ(|y′|)+β](u)(x)+h(x),
for all x∈H¯, where h(x) is harmonic in H and vanishes continuously on ∂H.


## 2. Proof of Theorem1.1

We need to use the following inequality (see [5, page 3]):
(2.1)|Pm(x,y′)|≤Mxn|x|m|y′|-n-m,
for any x∈H and y'∈∂H satisfying |y'|≥max{1,2|x|}, where M is a positive constant.For anyϵ(0<ϵ<1-ϵ0), there exists a sufficiently large positive number R such that r>R, and by (1.5), we have
(2.2)ρ(r)<ρ(e)(lnr)(ϵ0+ϵ),
which yields that there exists a positive constant M(r) dependent only on r such that
(2.3)k-β/2(2r)ρ(k+1)+β+1≤M(r),
for any k>kr=[2r]+1.For anyx∈H and |x|≤r, we have by (1.6), (2.1), (2.3), 1/p+1/q=1, and Hölder’s inequality
(2.4)M∑k=kr∞‍∫{y′∈∂H:k≤|y′|<k+1}‍(2|x|)[ρ(|y′|)+β]+1|y′|[ρ(|y′|)+β]+n|f(y′)|dy'222≤M∑k=kr∞‍(2r)ρ(k+1)+β+1(∫{y′∈∂H:k≤|y′|<k+1}‍|f(y′)|p|y′|ρ(|y′|)+n+pβ/2-1dy')1/p22211×(∫{y′∈∂H:k≤|y′|<k+1}‍|y′|-q{[ρ(|y′|)+β]+n-(ρ(|y′|)+n-1)/p-β/2}dy')1/q222≤M∑k=kr∞‍(2r)ρ(k+1)+β+1kβ/2∫{y′∈∂H:k≤|y′|<k+1}‍|f(y′)||y′|ρ(|y′|)+n+β/2-1dy'222≤2MM(r)∫{y′∈∂H:|y′|≥kr}‍|f(y′)|1+|y′|ρ(|y′|)+n+β/2-1dy'<∞.Thus,U[ρ(|y'|)+β](f)(x) is finite for any x∈H. 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 ofU[ρ(|y'|)+β](f)(x), we fix a boundary point z'∈∂H, choose a large t>|z'|+1, and write
(2.5)U[ρ(|y′|)+β](f)(x)=X(x)-Y(x)+Z(x),
where
(2.6)X(x)=∫{y′∈∂H:|y′|≤t}‍P(x,y′)f(y′)dy′,Y(x)=∑k=0[ρ(|y′|+β)]-1‍2xn|x|kωn∫{y′∈∂H:1<|y′|≤t}‍1|y′|n+kCkn/2(x′·y′|x||y′|)f(y′)dy',Z(x)=∫{y′∈∂H:|y′|>t}‍P[ρ(|y′|+β)](x,y′)f(y′)dy'.Notice thatX(x) is the Poisson integral of f(y')χB(t)(y'), where χB(t) is the characteristic function of the ball B(t). So it tends to f(z') as x→z'. Since Y(x) are polynomial times xn and Z(x)=O(xn), both of them tend to zero as x→z'. Thus, 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 1.1 is proved.


## 3. Proof of Theorem1.3

Consider that the functionu(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 [9, page 68] applied to u(x)-U[ρ(|y'|)+β](u)(x) shows that there exists a harmonic function h(x) in H such that h(x*)=-h(x)=-(u(x)-U[ρ(|y'|)+β](u)(x)) for x∈H¯, where * denotes reflection in ∂H just as x*=(x',-xn).Thus,u(x)=h(x)+U[ρ(|y'|)+β](u)(x) for all x∈H¯, where h(x) is a harmonic function on H vanishing continuously on ∂H. We complete the proof of Theorem 1.3.



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*Source: 203096-2012-10-17.xml*