1 Introduction and results

Let R and R + be the set of all real numbers and the set of all positive real numbers respectively. We denote the n-dimensional Euclidean space by R n (n2). A point in R n is denoted by P=(X, x n ), where X=( x 1 , x 2 ,, x n 1 ). The Euclidean distance between two points P and Q in R n is denoted by |PQ|. Also |PO| with the origin O of R n is simply denoted by |P|. The boundary and the closure of a set S in R n are denoted by S and S ¯ respectively.

We introduce a system of spherical coordinates (r,Θ), Θ=( θ 1 , θ 2 ,, θ n 1 ), in R n which are related to Cartesian coordinates ( x 1 , x 2 ,, x n 1 , x n ) by x n =rcos θ 1 .

The unit sphere and the upper half unit sphere in R n are denoted by S n 1 and S + n 1 , respectively. For simplicity, a point (1,Θ) on S n 1 and the set {Θ;(1,Θ)Ω} for a set Ω, Ω S n 1 , are often identified with Θ and Ω, respectively. For two sets Ξ R + and Ω S n 1 , the set {(r,Θ) R n ;rΞ,(1,Θ)Ω} in R n is simply denoted by Ξ×Ω.

For P R n and r>0, let B(P,r) denote an open ball with a center at P and radius r in R n . S r =B(O,r). By C n (Ω), we denote the set R + ×Ω in R n with the domain Ω on S n 1 . We call it a cone. We denote the sets I×Ω and I×Ω with an interval on R by C n (Ω;I) and S n (Ω;I). By S n (Ω;r) we denote C n (Ω) S r . By S n (Ω) we denote S n (Ω;(0,+)) which is C n (Ω){O}. We denote the (n1)-dimensional volume elements induced by the Euclidean metric on S r by d S r .

Let A a denote the class of nonnegative radial potentials a(P), i.e., 0a(P)=a(r), P=(r,Θ) C n (Ω), such that a L loc b ( C n (Ω)) with some b>n/2 if n4 and with b=2 if n=2 or n=3.

This article is devoted to the stationary Schrödinger equation

Sch a u(P)=Δu(P)+a(P)u(P)=0,
(1.1)

where P C n (Ω), Δ is the Laplace operator and a A a . These solutions called a-harmonic functions or generalized harmonic functions are associated with the operator Sch a . Note that they are (classical) harmonic functions in the case a=0. Under these assumptions, the operator Sch a can be extended in the usual way from the space C 0 ( C n (Ω)) to an essentially self-adjoint operator on L 2 ( C n (Ω)) (see [13]). We will denote it Sch a as well. This last one has a Green’s function G(Ω,a)(P,Q). Here G(Ω,a)(P,Q) is positive on C n (Ω) and its inner normal derivative G(Ω,a)(P,Q)/ n Q 0. We denote this derivative by P(Ω,a)(P,Q), which is called the Poisson a-kernel with respect to C n (Ω). We remark that G(Ω,0)(P,Q) and P(Ω,0)(P,Q) are the Green’s function and Poisson kernel of the Laplacian in C n (Ω) respectively.

Given a domain D R n and a continuous function u on (D), we say that h is a solution of the Dirichlet problem for the Schrödinger operator on D with u if Sch a h=0 in D and

lim P D , P Q h(P)=u(Q)

for every Q(D). Note that h is a solution of the classical Dirichlet problem for the Laplacian in the case a=0.

Let Δ be a Laplace-Beltrami operator (the spherical part of the Laplace) on Ω S n 1 and λ j (j=1,2,3,,0< λ 1 < λ 2 λ 3 ) be the eigenvalues of the eigenvalue problem for Δ on Ω (see, e.g., [4], p. 41])

Corresponding eigenfunctions are denoted by φ j v (1v v j ), where v j is the multiplicity of λ j . We set λ 0 =0, norm the eigenfunctions in L 2 (Ω) and φ 1 = φ 11 >0. Then there exist two positive constants d 1 and d 2 such that

d 1 δ(P) φ 1 (Θ) d 2 δ(P)
(1.2)

for P=(1,Θ)Ω (see Courant and Hilbert [5]), where δ(P)= inf Q C n ( Ω ) |PQ|.

In order to ensure the existences of λ j (j=1,2,3,). We put a rather strong assumption on Ω: if n3, then Ω is a C 2 , α -domain (0<α<1) on S n 1 surrounded by a finite number of mutually disjoint closed hypersurfaces (e.g., see [6], pp. 88-89] for the definition of C 2 , α -domain). Then φ j v C 2 ( Ω ¯ ) (j=1,2,3,,1v v j ) and φ 1 /n>0 on Ω (here and below, /n denotes differentiation along the interior normal).

Hence well-known estimates (see, e.g., [7], p. 14]) imply the following inequality:

v = 1 v j φ j v (Θ) φ j v ( Φ ) n Φ M(n) j 2 n 1 ,
(1.3)

where the symbol M(n) denotes a constant depending only on n.

Let V j (r) and W j (r) stand, respectively, for the increasing and nonincreasing, as r+, solutions of the equation

Q (r) n 1 r Q (r)+ ( λ j r 2 + a ( r ) ) Q(r)=0,0<r<,
(1.4)

normalized under the condition V j (1)= W j (1)=1.

We shall also consider the class B a , consisting of the potentials a A a such that there exists a finite limit lim r r 2 a(r)=k[0,); moreover, r 1 | r 2 a(r)k|L(1,). If a B a , then the solutions of Equation (1.1) are continuous (see [8]).

In the rest of the article, we assume that a B a and we shall suppress this assumption for simplicity. Further, we use the standard notations u + =max(u,0), u =min(u,0), [d] is the integer part of d and d=[d]+{d}, where d is a positive real number.

Denote

ι j , k ± = 2 n ± ( n 2 ) 2 + 4 ( k + λ j ) 2 (j=0,1,2,3,).

It is known (see [9]) that in the case under consideration the solutions to Equation (1.4) have the asymptotics

V j (r) d 3 r ι j , k + , W j (r) d 4 r ι j , k ,asr,
(1.5)

where d 3 and d 4 are some positive constants.

If a A a , it is known that the following expansion for the Green function G(Ω,a)(P,Q) (see [10], Ch. 11], [1, 11])

G(Ω,a)(P,Q)= j = 0 1 χ ( 1 ) V j ( min ( r , t ) ) W j ( max ( r , t ) ) ( v = 1 v j φ j v ( Θ ) φ j v ( Φ ) ) ,

where P=(r,Θ)Q=(t,Φ)rt and χ (s)=w( W 1 (r), V 1 (r)) | r = s , is their Wronskian. The series converges uniformly if either rst or tsr (0<s<1).

For a nonnegative integer m and two points P=(r,Θ), Q=(t,Φ) C n (Ω), we put

K(Ω,a,m)(P,Q)={ 0 if 0 < t < 1 , K ˜ ( Ω , a , m ) ( P , Q ) if 1 t < ,

where

K ˜ (Ω,a,m)(P,Q)= j = 0 m 1 χ ( 1 ) V j (r) W j (t) ( v = 1 v j φ j v ( Θ ) φ j v ( Φ ) ) .

We introduce another function of P=(r,Θ) C n (Ω) and Q=(t,Φ) C n (Ω)

G(Ω,a,m)(P,Q)=G(Ω,a)(P,Q)K(Ω,a,m)(P,Q).

The generalized Poisson kernel P(Ω,a,m)(P,Q) (P=(r,Θ) C n (Ω), Q=(t,Φ) S n (Ω)) with respect to C n (Ω) is defined by

P(Ω,a,m)(P,Q)= G ( Ω , a , m ) ( P , Q ) n Q .

In fact,

P(Ω,a,0)(P,Q)=P(Ω,a)(P,Q).

We remark that the kernel function P(Ω,0,m)(P,Q) coincides with the one in Yoshida and Miyamoto [12] (see [10], Ch. 11]).

Put

U(Ω,a,m;u)(P)= S n ( Ω ) P(Ω,a,m)(P,Q)u(Q)d σ Q ,

where u(Q) is a continuous function on C n (Ω) and d σ Q is a surface area element on S n (Ω).

With regard to classical solutions of the Dirichlet problem for the Laplacian, Yoshida and Miyamoto [12], Theorem 1] proved the following result.

Theorem A If u is a continuous function on C n (Ω) satisfying

S n ( Ω ) | u ( t , Φ ) | 1 + t ι m + 1 , 0 + + n 1 d σ Q <,

then U(Ω,0,m;u)(P) is a classical solution of the Dirichlet problem on C n (Ω) with g and satisfies

lim r , P = ( r , Θ ) C n ( Ω ) r ι m + 1 , 0 + U(Ω,0,m;u)(P)=0.

Our first aim is to give growth properties at infinity for U(Ω,a,m;u)(P).

Theorem 1 Letγ0 (resp. γ<0), ι [ γ ] , k + +{γ}> ι 1 , k + +1 (resp. ι [ γ ] , k + {γ}> ι 1 , k + +1) and

If u is a measurable function on C n (Ω) satisfying

S n ( Ω ) | u ( t , Φ ) | 1 + t ι [ γ ] , k + + { γ } d σ Q < ( resp. S n ( Ω ) | u ( t , Φ ) | ( 1 + t ι [ γ ] , k + + { γ } ] ) d σ Q < ) ,
(1.6)

then

(1.7)
(1.8)

Next, we are concerned with solutions of the Dirichlet problem for the Schrödinger operator on C n (Ω).

Theorem 2 Let γ and ι m + 1 , k + be as in Theorem 1. If u is a continuous function on C n (Ω)satisfying (1.6), thenU(Ω,a,m;u)(P)is a solution of the Dirichlet problem for the Schrödinger operator on C n (Ω)with u and (1.7) (resp. (1.8)) holds.

If we take ι [ γ ] , k + +{γ}= ι m + 1 , k + +n1, then we immediately have the following corollary, which is just Theorem A in the case a=0.

Corollary If u is a continuous function on C n (Ω) satisfying

S n ( Ω ) | u ( t , Φ ) | 1 + t ι m + 1 , k + + n 1 d σ Q <,
(1.9)

then U(Ω,a,m;u)(P) is a solution of the Dirichlet problem for the Schrödinger operator on C n (Ω) with u and satisfies

lim r , P = ( r , Θ ) C n ( Ω ) r ι m + 1 , k + U(Ω,a,m;u)(P)=0.
(1.10)

By using Corollary, we can give a solution of the Dirichlet problem for any continuous function on C n (Ω).

Theorem 3 If u is a continuous function on C n (Ω)satisfying (1.9) andh(r,Θ)is a solution of the Dirichlet problem for the Schrödinger operator on C n (Ω)with u satisfying

lim r , P = ( r , Θ ) C n ( Ω ) r ι m + 1 , k + h + (P)=0,
(1.11)

then

h(P)=U(Ω,a,m;u)(P)+ j = 0 m ( v = 1 v j d j v φ j v ( Θ ) ) V j (r),

whereP=(r,Θ) C n (Ω)and d j v are constants.

2 Lemmas

Throughout this article, let M denote various constants independent of the variables in questions, which may be different from line to line.

Lemma 1

(2.1)
(2.2)

for anyP=(r,Θ) C n (Ω)and anyQ=(t,Φ) S n (Ω)satisfying0< t r 4 5 (resp. 0< r t 4 5 );

| P ( Ω , 0 ) ( P , Q ) | M 1 t n 1 +M r | P Q | n
(2.3)

for anyP=(r,Θ) C n (Ω)and anyQ=(t,Φ) S n (Ω;( 4 5 r, 5 4 r)).

Proof (2.1) and (2.2) are obtained by Kheyfits (see [10], Ch. 11]). (2.3) follows from Azarin (see [13], Lemma 4 and Remark]). □

Lemma 2 (see [1])

For a nonnegative integer m, we have

| P ( Ω , a , m ) ( P , Q ) | M(n,m,s) V m + 1 (r) W m + 1 ( t ) t φ 1 (Θ) φ 1 ( Φ ) n Φ
(2.4)

for anyP=(r,Θ) C n (Ω)andQ=(t,Φ) S n (Ω)satisfyingrst (0<s<1), whereM(n,m,s)is a constant dependent of n, m and s.

Lemma 3 (see [2], Theorem 1])

Ifu(r,Θ)is a solution of Equation (1.1) on C n (Ω)satisfying

Ω u + (r,Θ)d S 1 =O ( r ι m , k + ) ,asr,
(2.5)

then

u(r,Θ)= j = 0 m ( v = 1 v j d j v φ j v ( Θ ) ) V j (r).

Lemma 4 Obviously, the conclusion of Lemma 3 holds true if (2.5) is replaced by

lim r , ( r , Θ ) C n ( Ω ) r ι m + 1 , k + u + (r,Θ)=0.
(2.6)

Proof Since

V m + 1 (r) r ι m + 1 , k + asr

from (1.5) and

ι m + 1 , k + ι m , k + ,

(2.6) gives that (2.5) holds, from which the conclusion immediately follows. □

3 Proof of Theorem 1

We only prove the case γ0, the remaining case γ<0 can be proved similarly.

For any ϵ>0, there exists R ϵ >1 such that

S n ( Ω ; ( R ϵ , ) ) | u ( Q ) | 1 + t ι [ γ ] , k + + { γ } d σ Q <ϵ.
(3.1)

The relation G(Ω,a)(P,Q)G(Ω,0)(P,Q) implies this inequality (see [14])

P(Ω,a)(P,Q)P(Ω,0)(P,Q).
(3.2)

For 0<s< 4 5 and any fixed point P=(r,Θ) C n (Ω) satisfying r> 5 4 R ϵ , let I 1 = S n (Ω;(0,1)), I 2 = S n (Ω;[1, R ϵ ]), I 3 = S n (Ω;( R ϵ , 4 5 r]), I 4 = S n (Ω;( 4 5 r, 5 4 r)), I 5 = S n (Ω;[ 5 4 r, r s )), I 6 = S n (Ω;[1, r s )) and I 7 = S n (Ω;[ r s ,)), we write

U(Ω,a,m;u)(P) i = 1 7 U Ω , a , i (P),

where

By ι [ γ ] , k + +{γ}> ι 1 , k + +1, (1.6), (2.1) and (3.1), we have the following growth estimates

(3.3)
(3.4)
(3.5)

We obtain by ι m + 1 , k + ι [ γ ] , k + +{γ}n+1, (2.2) and (3.1)

U Ω , a , 5 ( P ) M r ι 1 , k + S n ( Ω ; [ ( 5 / 4 ) r , ) ) t ι 1 , k 1 | u ( Q ) | d σ Q M r ι 1 , k + S n ( Ω ; [ ( 5 / 4 ) r , ) ) t ι [ γ ] , k + + { γ } + ι 1 , k 1 | u ( Q ) | t ι [ γ ] , k + + { γ } d σ Q M ϵ r ι [ γ ] , k + + { γ } n + 1 .
(3.6)

By (2.3) and (3.2), we consider the inequality

U Ω , a , 4 (P) U Ω , 0 , 4 (P) U Ω , 0 , 4 (P)+ U Ω , 0 , 4 (P),

where

U Ω , 0 , 4 (P)=M I 4 t 1 n | u ( Q ) | d σ Q , U Ω , 0 , 4 (P)=Mr I 4 | u ( Q ) | | P Q | n d σ Q .

We first have

U Ω , 0 , 4 ( P ) = M I 4 t ι 1 , k + + ι 1 , k 1 | u ( Q ) | d σ Q M r ι 1 , k + S n ( Ω ; ( ( 4 / 5 ) r , ) ) t ι 1 , k 1 | u ( Q ) | d σ Q M ϵ r ι [ γ ] , k + + { γ } n + 1 ,
(3.7)

which is similar to the estimate of U Ω , a , 5 (P).

Next, we shall estimate U Ω , 0 , 4 (P). Take a sufficiently small positive number d 5 such that I 4 B(P, 1 2 r) for any P=(r,Θ)Π( d 5 ), where

Π( d 5 )= { P = ( r , Θ ) C n ( Ω ) ; inf z Ω | ( 1 , Θ ) ( 1 , z ) | < d 5 , 0 < r < }

and divide C n (Ω) into two sets Π( d 5 ) and C n (Ω)Π( d 5 ).

If P=(r,Θ) C n (Ω)Π( d 5 ), then there exists a positive d 5 such that |PQ| d 5 r for any Q S n (Ω), and hence

U Ω , 0 , 4 ( P ) M I 4 t 1 n | u ( Q ) | d σ Q M ϵ r ι [ γ ] , k + + { γ } n + 1 ,
(3.8)

which is similar to the estimate of U Ω , 0 , 4 (P).

We shall consider the case P=(r,Θ)Π( d 5 ). Now put

H i (P)= { Q I 4 ; 2 i 1 δ ( P ) | P Q | < 2 i δ ( P ) } .

Since S n (Ω){Q R n :|PQ|<δ(P)}=, we have

U Ω , 0 , 4 (P)=M i = 1 i ( P ) H i ( P ) r | u ( Q ) | | P Q | n d σ Q ,

where i(P) is a positive integer satisfying 2 i ( P ) 1 δ(P) r 2 < 2 i ( P ) δ(P).

Since we see from (1.2)

r φ 1 (Θ)Mδ(P)

for P=(r,Θ) C n (Ω). Similar to the estimate of U Ω , 0 , 4 (P), we obtain

for i=0,1,2,,i(P).

So

U Ω , 0 , 4 (P)Mϵ r ι [ γ ] , k + + { γ } n + 1 .
(3.9)

We only consider U Ω , a , 6 (P) in the case m1, since U Ω , a , 6 (P)0 for m=0. By the definition of K ˜ (Ω,a,m), (1.3) and Lemma 2, we see

U Ω , a , 6 (P) M χ ( 1 ) j = 0 m j 2 n 1 q j (r),

where

q j (r)= V j (r) I 6 W j ( t ) | u ( Q ) | t d σ Q .

To estimate q j (r), we write

q j (r) q j (r)+ q j (r),

where

q j (r)= V j (r) I 2 W j ( t ) | u ( Q ) | t d σ Q , q j (r)= V j (r) S n ( Ω ; ( R ϵ , r / s ) ) W j ( t ) | u ( Q ) | t d σ Q .

Notice that

V j (r) V m + 1 ( t ) V j ( t ) t M V m + 1 ( r ) r M r ι m + 1 , k + 1 ( t 1 , R ϵ < r s ) .

Thus, by ι m + 1 , k + < ι [ γ ] , k + +{γ}n+2, (1.5) and (1.6) we conclude

q j ( r ) = V j ( r ) I 2 | u ( Q ) | V j ( t ) t n 1 d σ Q M V j ( r ) I 2 V m + 1 ( t ) t ι m + 1 , k + | u ( Q ) | V j ( t ) t n 1 d σ Q M r ι m + 1 , k + 1 R ϵ ι [ γ ] , k + + { γ } ι m + 1 , k + n + 2 .

Analogous to the estimate of q j (r), we have

q j (r)Mϵ r ι [ γ ] , k + + { γ } n + 1 .

Thus we can conclude that

q j (r)Mϵ r ι [ γ ] , k + + { γ } n + 1 ,

which yields

U Ω , a , 6 (P)Mϵ r ι [ γ ] , k + + { γ } n + 1 .
(3.10)

By ι m + 1 , k + ι [ γ ] , k + +{γ}n+1, (1.5), (2.4) and (3.1) we have

U Ω , 0 , 7 ( P ) M V m + 1 ( r ) I 7 | u ( Q ) | V m + 1 ( t ) t n 1 d σ Q M ϵ r ι [ γ ] , k + + { γ } n + 1 .
(3.11)

Combining (3.3)–(3.11), we obtain that if R ϵ is sufficiently large and ϵ is sufficiently small, then U(Ω,a,m;u)(P)=o( r ι [ γ ] , k + + { γ } n + 1 ) as r, where P=(r,Θ) C n (Ω). Then we complete the proof of Theorem 1.

4 Proof of Theorem 2

For any fixed P=(r,Θ) C n (Ω), take a number satisfying R>max(1, r s ) (0<s< 4 5 ). By ι m + 1 , k + ι [ γ ] , k + +{γ}n+1, (1.4), (1.6) and (2.4), we have

Thus U(Ω,a,m;u)(P) is finite for any P C n (Ω). Since P(Ω,a,m)(P,Q) is a generalized harmonic function of P C n (Ω) for any fixed Q S n (Ω), U(Ω,a,m;u)(P) is also a generalized harmonic function of P C n (Ω). That is to say, U(Ω,a,m;u)(P) is a solution of Equation (1.1) on C n (Ω).

Now we study the boundary behavior of U(Ω,a,m;u)(P). Let Q =( t , Φ ) C n (Ω) be any fixed point and l be any positive number satisfying l>max( t +1, 4 5 R).

Set χ S ( l ) is a characteristic function of S(l)={Q=(t,Φ) C n (Ω),tl} and write

U(Ω,a,m;u)(P)= U (P) U (P)+ U (P),

where

Notice that U (P) is the Poisson a-integral of u(Q) χ S ( ( 5 / 4 ) l ) , we have lim P Q , P C n ( Ω ) U (P)=u( Q ). Since lim Θ Φ φ j v (Θ)=0 (j=1,2,3,; 1v v j ) as P=(r,Θ) Q =( t , Φ ) S n (Ω), we have lim P Q , P C n ( Ω ) U (P)=0 from the definition of the kernel function K(Ω,a,m)(P,Q). U (P)=O( r ι [ γ ] , k + + { γ } n + 1 φ 1 (Θ)), and therefore tends to zero.

So the function U(Ω,a,m;u)(P) can be continuously extended to C n ( Ω ) ¯ such that

lim P Q , P C n ( Ω ) U(Ω,a,m;u)(P)=u ( Q )

for any Q =( t , Φ ) C n (Ω) from the arbitrariness of l. Thus we complete the proof of Theorem 2 from Theorem 1.

5 Proof of Theorem 3

From Corollary, we have the solution U(Ω,a,m;u)(P) of the Dirichlet problem on C n (Ω) with u satisfying (1.9). Consider the function h(P)U(Ω,a,m;u)(P). Then it follows that this is the solution of Equation (1.1) in C n (Ω) and vanishes continuously on C n (Ω).

Since

0 ( h U ( Ω , a , m ; u ) ) + (P) h + (P)+ ( U ( Ω , a , m ; u ) ) (P)

for any P C n (Ω), we have

lim r , P = ( r , Θ ) C n ( Ω ) r ι m + 1 , k + ( h U ( Ω , a , m ; u ) ) + (P)=0

from (1.10) and (1.11). Then the conclusions of Theorem 3 follow immediately from Lemma 4.