Research

# Dirichlet problem for the Schrödinger operator on a cone

Lei Qiao1* and Guan-Tie Deng2

Author Affiliations

1 Department of Mathematics and Information Science, Henan University of Economics and Law, Zhengzhou, 450002, P.R. China

2 School of Mathematical Science, Laboratory of Mathematics and Complex Systems, MOE Beijing Normal University, Beijing, 100875, P.R. China

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Boundary Value Problems 2012, 2012:59  doi:10.1186/1687-2770-2012-59

 Received: 16 February 2012 Accepted: 2 May 2012 Published: 18 June 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this article, a solution of the Dirichlet problem for the Schrödinger operator on a cone is constructed by the generalized Poisson integral with a slowly growing continuous boundary function. A solution of the Poisson integral for any continuous boundary function is also given explicitly by the Poisson integral with the generalized Poisson kernel depending on this boundary function.

MSC: 31B05, 31B10.

##### Keywords:
Dirichlet problem; stationary Schrödinger equation; cone

### 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 ( n 2 ). 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 | P Q | . Also | P O | 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 = r cos 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 ; r Ξ , ( 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 ( n 1 ) -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., 0 a ( P ) = a ( r ) , P = ( r , Θ ) C n ( Ω ) , such that a L loc b ( C n ( Ω ) ) with some b > n / 2 if n 4 and with b = 2 if n = 2 or n = 3 .

Sch a u ( P ) = Δ u ( 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 [1-3]). 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 ( 1 v 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 ) | P Q | .

In order to ensure the existences of λ j ( j = 1 , 2 , 3 , ). We put a rather strong assumption on Ω: if n 3 , 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 , , 1 v 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 , as r , (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 ) r t and χ ( s ) = w ( W 1 ( r ) , V 1 ( r ) ) | r = s , is their Wronskian. The series converges uniformly if either r s t or t s r ( 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 AIfuis 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 ) withgand 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 1Let γ 0 (resp. γ < 0 ), ι [ γ ] , k + + { γ } > ι 1 , k + + 1 (resp. ι [ γ ] , k + { γ } > ι 1 , k + + 1 ) and

Ifuis 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 2Letγand ι m + 1 , k + be as in Theorem 1. Ifuis a continuous function on C n ( Ω ) satisfying (1.6), then U ( Ω , a , m ; u ) ( P ) is a solution of the Dirichlet problem for the Schrödinger operator on C n ( Ω ) withuand (1.7) (resp. (1.8)) holds.

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

CorollaryIfuis 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 ) withuand 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 3Ifuis a continuous function on C n ( Ω ) satisfying (1.9) and h ( r , Θ ) is a solution of the Dirichlet problem for the Schrödinger operator on C n ( Ω ) withusatisfying

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 ) ,

where P = ( 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 any P = ( r , Θ ) C n ( Ω ) and any Q = ( t , Φ ) S n ( Ω ) satisfying 0 < 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 any P = ( r , Θ ) C n ( Ω ) and any Q = ( 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 integerm, 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 any P = ( r , Θ ) C n ( Ω ) and Q = ( t , Φ ) S n ( Ω ) satisfying r s t ( 0 < s < 1 ), where M ( n , m , s ) is a constant dependent ofn, mands.

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

If u ( 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 + ) , as r , (2.5)
then

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

Lemma 4Obviously, 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 + as r

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 γ 0 , 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 ) = M r 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 | P Q | 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 : | P Q | < 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 ) M δ ( 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 ) M ϵ r ι [ γ ] , k + + { γ } n + 1 . (3.9)

We only consider U Ω , a , 6 ( P ) in the case m 1 , 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 ) M ϵ r ι [ γ ] , k + + { γ } n + 1 .

Thus we can conclude that

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

which yields

U Ω , a , 6 ( P ) M ϵ 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 ) , t l } 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 , ; 1 v 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.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.

### Acknowledgements

The authors would like to thank anonymous reviewers for their valuable comments and suggestions about improving the quality of the manuscript. This work is supported by The National Natural Science Foundation of China under Grant 11071020 and Specialized Research Fund for the Doctoral Program of Higher Education under Grant 20100003110004.

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