# Solving Integral Representations Problems for the Stationary Schrödinger Equation

Authors: Yudong Ren
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

When solutions of the stationary Schrödinger equation in a half-space belong to the weighted Lebesgue classes, we give integral representations of them, which imply known representation theorems of classical harmonic functions in a half-space.


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## Body

## 1. Introduction and Results

LetR and R+ be the sets of all real numbers and of all positive real numbers, respectively. Let Rn(n≥2) denote the n-dimensional Euclidean space with points x=(x′,xn), where x′=(x1,x2,…,xn-1)∈Rn-1 and xn∈R. The unit sphere and the upper half unit sphere in Rn are denoted by Sn-1 and S+n-1, respectively. 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 identifyRn 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)x·y=∑j=1n‍xjyj,|x|=x·x,Θ=x|x|,Φ=y|y|.Forx∈Rn and r>0, let B(x,r) denote the open ball with center at x and radius r(>0) in Rn. Sr=∂B(O,r), where O is the origin of Rn. For a set E, E⊂R+∪{0}, we denote {x∈H;|x|∈E} and {x∈∂H;|x|∈E} by HE and ∂HE, respectively. By Hr we denote H∩Sr. We denote by dSr the (n-1)-dimensional volume elements induced by the Euclidean metric on Sr.Let𝒜b denote the class of nonnegative radial potentials b(x), that is, 0≤b(x)=b(|x|), x∈H, such that b∈Lloca(H) with some a>n/2 if n≥4 and with a=2 if n=2 or n=3.This paper is devoted to the stationary Schrödinger equation:(2)SSEbh(x)=-Δh(x)+b(x)h(x)=0,
where x∈H, Δ is the Laplace operator and b∈𝒜b. Note that solutions of (2) are (classical) harmonic functions in the case b=0. Under these assumptions the operator SSEb can be extended in the usual way from the space C0∞(H) to an essentially self-adjoint operator on L2(H) (see [1]). We will denote it by SSEb as well. This last one has a Green function Gb(x,y). Here Gb(x,y) is positive on H and its inner normal derivative ∂Gb(x,y)/∂n(y′)≥0. We denote this derivative by Pb(x,y′), which is called the Poisson b-kernel with respect to H. If G(x,y) and P(x,y′) are denoted by the Green function and Poisson kernel of the Laplace operator in H, respectively, then
(3)P(x,y′)=-∂G(x,y)∂yn|yn=0=2xnwn1|x-y′|n,
where x=(x′,xn), y=(y′,yn) and wn is the area of the unit sphere in Rn.Letg be a continuous function on ∂H. We say that h is a solution of the Dirichlet problem for the stationary Schrödinger operator SSEb on H with g, if
(4)h∈C2(H)∩C0(H¯),SSEbh=0inH,h=gon∂H.
Note that h is a solution of the classical Dirichlet problem for the Laplace operator Δ on H with g in the case b=0.LetΛ be a Laplace-Beltrami operator (spherical part of the Laplace) on the unit sphere. It is known (see, e.g., [2, page 41]) that the eigenvalue problem
(5)Λφ(Θ)+τφ(Θ)=0Θ∈S+n-1,φ(Θ)=0Θ∈∂S+n-1
has the eigenvalues τj=j(j+n-2), where j=0,1,2…. Corresponding eigenfunctions are denoted by φjv(1≤v≤vj), where vj is the multiplicity of τj. We norm the eigenfunctions in L2(S+n-1) and φ1=φ11>0.LetPj(r) and Qj(r) stand, respectively, for the increasing and nonincreasing, as r→+∞, solutions of the equation:
(6)-T′′(r)-n-1rT′(r)0000+(τjr2+b(r))T(r)=0,0<r<∞,
normalized under the condition Pj(1)=Qj(1)=1.We shall also consider the classℬb, consisting of the potentials b∈𝒜b such that there exists a finite limit limr→∞r2b(r)=s∈[0,∞); moreover, r-1|r2b(r)-s|∈L(1,∞). If b∈ℬb, then solutions of (2) are continuous (see [3]).In the rest of paper, we assume thatb∈ℬb and we shall suppress this assumption for simplicity. Further, we use the standard notations u+=max{u,0}, u-=-min{u,0}, and [d] is the integer part of d and d=[d]+{d}, where d is a positive real number.Denote(7)λj,s±=2-n±(n-2)2+4(s+τj)2(j=0,1,2,3,…).Remark 1.
λ
j
,
0
+
=
j in the case b=0, where j=0,1,2,3,….It is known (see [4]) that in the case under consideration the solutions to (6) have the asymptotics
(8)Pj(r)~d1rλj,s+,Qj(r)~d2rλj,s-,asr⟶∞,
where d1 and d2 are some positive constants.Ifb∈𝒜b, it is known that the following expansion for the Green function Gb(x,y) (see [5, Chapter 11], [6]):
(9)Gb(x,y)=∑j=0∞‍1χ′(1)Pj(min(|x|,|y|))00×Qj(max(|x|,|y|))(∑v=1vj‍φjv(Θ)φjv(Φ)),
where |x|≠|y| and χ′(1)=w(Q1(r),P1(r))|r=1 is its Wronskian. The series converges uniformly if either |x|≤k|y| or |y|≤k|x|, where 0<k<1.For a nonnegative integerm and two points x,y∈H, we define a modified Green function:
(10)G(b,m)(x,y)=Gb(x,y)-V(b,m)(x,y),
where
(11)V(b,m)(x,y)00={0if|y|<1,∑j=0m‍1χ′(1)Pj(|x|)Qj(|y|)×(∑v=1vj‍φjv(Θ)φjv(Φ))if1≤|y|<∞.If the generalized Poisson kernelP(b,m)(x,y′) with respect to H is defined by
(12)P(b,m)(x,y′)=∂G(b,m)(x,y)∂n(y′),
then we have P(b,0)(x,y′)=Pb(x,y′) and P(0,m)(x,y′) coincides with ones in Finkelstein and Scheinberg [7], Siegel and Talvila [8], Deng [9], Qiao and Deng [10], and Qiao [11] (see [5, Chapter 11]).Put(13)U(b,m;u)(x)=∫∂H‍P(b,m)(x,y′)u(y′)dy′,
where u(y′) is a continuous function on ∂H.For real numbersβ≥1, we denote 𝒞(β,b) the class of all measurable functions f(y) satisfying the following inequality:
(14)∫H‍|f(y)|φ11+P[β](|y|)|y|n+{β}dy<∞,
and the class 𝒟(β,b) consists of all measurable functions g(y′)(y′∈∂H) satisfying
(15)∫∂H‍|g(y′)|P1(|y′|)Q1(|y′|)1+χ′(|y′|)P[β](|y′|)|y′|n+{β}-1∂φ1∂ndy′<∞.We will also consider the class of all continuous functionsh(y), which is the solution of (2), with h+(y)∈𝒞(β,b) and h+(y′)∈𝒟(β,b), is denoted by ℰ(β,b).Remark 2.
Ifb=0 and β=α+1, then (14) and (15) are equivalent to
(16)∫H‍yn|f(y)|1+|y|n+α+2dy<∞,∫∂H‍|g(y′)|1+|y′|n+αdy′<∞,
respectively, from Remark 1 and (14), which yield that ℰ(α+1,0) is equivalent to (CH)α in the notation of [9].
Let us recall the classical caseb=0. If h(x)≥0 is harmonic in H, continuous on H¯, and h∈ℰ(1,0), then there exists a constant d3≥0 such that (see [12])
(17)h(x)=d3xn+∫∂H‍P(x,y′)u(y′)dy′,
where x=(x′,xn)∈H.Deng (see [9]) has constructed a similar representation to (17) for h∈ℰ(α+1,0)(α≥0), which is the integral with a modified Poisson kernel derived by subtracting some special harmonic polynomials from P(x,y′). By virtue of this modified Poisson kernel, Qiao  (see [11, 13]) and Qiao and Deng (see  [14–18]) have constructed different integral representations for harmonic functions of finite order and infinite order.Especially, Su (see [6, 19]) recently writes solutions to the half-space Dirichlet problem with respect to the stationary Schrödinger operator SSEb. Now we state our main results as follows.Theorem 3.
Ifh∈ℰ(β,b), then h∈𝒟(β,b).Theorem 4.
Ifh∈ℰ(β,b), m is an integer such that Pm(|y|)<P[β](|y|)|y|{β}≤Pm+1(|y|)(|y|≥1), and then the following properties hold.(I)
Ifβ=1, then there exists a constant d4 such that
(18)h(x)=d4P1(|x|)φ1(Θ)+U(b,0;h)(x)forx∈H.(II)
Ifβ>1, then we have h(x)=U(b,m;h)(x)+u(x), where
(19)u(x)=∑j=0m‍(∑v=1vj‍djvφjv(Θ))Pj(|x|)vanishing continuously on∂H, x=(x′,xn)∈H and djv are constants.


## 2. Lemmas

The following Lemma plays an important role in our discussions, which is due to Levin and Kheyfits (see [5, page 356]).Lemma 5.
IfR>1 and h(y) is a solution of (2) on a domain containing H(1,R), then
(20)∫HR‍χ′(R)P1(R)hφ1(Φ)dSR+∫∂H(1,R)‍h(y′)∂φ1∂nW(|y′|)dy′0000+d5+d6Q1(R)P1(R)=0,
where
(21)W(|y′|)=Q1(|y′|)-Q1(R)P1(R)P1(|y′|),d5=∫H1‍hφ1(Φ)Q1′(r)-Q1(1)φ1(Φ)∂h∂ndS1,d6=∫H1‍P1(1)φ1(Φ)∂h∂n-hφ1(Φ)P1′(1)dS1.Lemma 6 (see [6, Corollary 1.6]).
Ifu is a continuous function on ∂H satisfying
(22)∫∂H‍|u(y′)|1+|y′|λm+1,s++n-1dy′<∞,
then U(b,m;u)(x) is a solution of the Dirichlet problem for SSEb on H with h and
(23)lim|x|→∞,x∈H|x|-λm+1,s+U(b,m;u)(x)=0.Lemma 7 (see [6, Lemma 2.1] or [20, Theorem 1]).
Ifu(x) is a solution of (2) on H satisfying
(24)lim|x|→∞,x∈H|x|-λm+1,s+u+(x)=0,
then (19) holds.


## 3. Proof of Theorem3

We apply the formula (20) to h=h+-h- in H(1,R):
(25)m+(R)+∫∂H(1,R)‍h+(y′)W(|y′|)∂φ1∂ndy′+d5+Q1(R)P1(R)d600=m-(R)+∫∂H(1,R)‍h-(y′)W(|y′|)∂φ1∂ndy′,
where(26)m±(R)=∫HR‍χ′(R)P1(R)h±φ1dSR.Sinceu∈ℰ(β,b), we obtain by (8)
(27)∫1∞‍m+(R)P1(R)χ′(R)P[β](R)Rn+{β}dR00=∫H(1,∞)‍h+φ1P[β](|y|)|y|n+{β}dy00≤2∫H‍h+φ11+P[β](|y|)|y|n+{β}dy<∞.From (15), we conclude that
(28)∫1∞P1(R)χ′(R)P[β](R)Rn+{β}‍0000×∫∂H(1,R)h+(y′)W(|y′|)∂φ1∂ndy′dR‍00=∫∂H(1,∞)h+(y′)P1(|y′|)‍00000×∫|y′|∞P1(R)χ′(R)P[β](R)Rn+{β}‍00000×(Q1(|y′|)P1(|y′|)-W1(R)P1(R))dR∂φ1∂ndy′00≤M∫∂H(1,∞)h+(y′)P1(|y′|)Q1(|y′|)χ′(|y′|)P[β](|y′|)|y′|n+{β}-1∂φ1∂ndy′‍00≤M∫∂Hh+(y′)P1(|y′|)Q1(|y′|)1+χ′(|y′|)P[β](|y′|)|y′|n+{β}-1∂φ1∂ndy′‍00<∞.Combining (25), (27), and (28), we obtain
(29)∫1∞‍P1(R)χ′(R)P[β](R)Rn+({β}/2)0000×∫∂H(1,R)‍h-(y′)W(|y′|)∂φ1∂ndy′dR00≤∫1∞‍m+(R)P1(R)χ′(R)P[β](R)Rn+({β}/2)dR0000+∫1∞‍P1(R)χ′(R)P[β](R)Rn+({β}/2)0000×∫∂H(1,R)‍h+(y′)W(|y′|)∂φ1∂ndy′dR0000+∫1∞‍1χ′(R)P[β](R)Rn+({β}/2)0000000×(P1(R)d5+Q1(R)d6)dR00<∞.Set(30)ℋ(β)=lim|y′|→∞χ′(|y′|)P[β](|y′|)|y′|n+{β}-1Q1(|y′|)×∫|y′|∞‍P1(R)χ′(R)P[β](R)Rn+({β}/2)00000×(Q1(|y′|)P1(|y′|)-W1(R)P1(R))dR=lim|y′|→∞|y′|λ[β],s++λ1,s++n+{β}-2×∫|y′|∞‍1Rλ[β],s+-λ1,s++({β}/2)+1(1|y′|χ1,s-1Rχ1,s)dR,
where χ1,s=λ1,s+-λ1,s-.By the L’hospital’s rule, we have(31)ℋ(β)={χ1,s(λ[β],s+-λ1,s+)(λ[β],s++λ1,s++n-2)if{β}=0,+∞if{β}≠0,
which yields that there exists a positive constant M such that, for any |y′|≥1,
(32)∫|y′|∞‍P1(R)χ′(R)P[β](R)Rn+({β}/2)W(|y′|)dR00≥MP1(|y′|)Q1(|y′|)χ′(|y′|)P[β](|y′|)|y′|n+{β}-1.Then(33)M∫∂H(1,∞)‍h-(y′)P1(|y′|)Q1(|y′|)χ′(|y′|)P[β](|y′|)|y′|n+{β}-1∂φ1∂ndy′00≤∫∂H(1,∞)‍h-∫|y′|∞‍P1(R)χ′(R)P[β](R)Rn+({β}/2)0000000000000000×W(|y′|)dR∂φ1∂ndy′00<∞,
which shows that h∈𝒟(β,b) from |h|=h++h-. Then Theorem 3 is proved.


## 4. Proof of Theorem4

To prove (II), notice thatPm(|y|)<P[β](|y|)|y|{β}≤Pm+1(|y|)(|y|≥1) and condition (15) is stronger than condition (22) from Theorem 3.Consider the functionh(x)-U(a,m;h)(x). Then it follows from Lemma 6 and Theorem 3 that this is a solution of (2) in H and vanishes continuously on ∂H.Then(34)0≤(h(x)-U(b,m;h)(x))+≤h+(x)+(U(b,m;h))-(x)
for any P∈H. Further, (8) gives that
(35)lim|x|→∞,x∈H|x|-λm+1,s+h+(x)=0,
which together with (34) and Lemmas 6 and 7 give the result of (II).Ifu∈ℰ(1,b), then h∈ℰ(β,b) for each β>1 and there exists a constant d7 such that
(36)h(x)=d7P1(|x|)φ1(Θ)+U(b,1;h)(x)
for all x∈H. So if we take
(37)d4=d7-∫∂H[1,∞)‍P(b,1)(0,y′)h(y′)dy′,
we see that (18) holds for all x∈H.Thus we complete the proof of Theorem4.



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