THE EXPONENTIAL PROPERTY OF SOLUTIONS BOUNDED FROM BELOW TO DEGENERATE EQUATIONS IN UNBOUNDED DOMAINS*

Lidan WANG (王麗丹)

School of Mathematical Sciences，Shanghai Jiao Tong University，Shanghai 200240，China E-mail:wanglidan@sjtu.edu.cn

Abstract This paper is focused on studying the structure of solutions bounded from below to degenerate elliptic equations with Neumann and Dirichlet boundary conditions in unbounded domains.After establishing the weak maximum principles，the global boundary H?lder estimates and the boundary Harnack inequalities of the equations，we show that all solutions bounded from below are linear combinations of two special solutions (exponential growth at one end and exponential decay at the other) with a bounded solution to the degenerate equations.

Key words degenerate elliptic equations;unbounded domains;boundary Harnack inequalities

1 Introduction

In this paper，we study the behavior of solutions bounded from below to degenerate elliptic equations with mixed boundary conditions in unbounded domains.A series of papers，see for example[4-6，9，12]，have established a systematic theory for degenerate elliptic operators.In addition，degenerate elliptic equations with mixed boundary conditions also have been studied by many authors;we refer to[2，8，14，17].The references above provide us with useful tools for studying degenerate equations further.The main motivations for studying degenerate elliptic equations are twofold.It is well known that Caffarelli and Silvestre[3]obtained an extension theorem through a Dirichlet-Neumann map.The Caffarelli-Silvestre extension，because of its local nature，is very often used to prove qualitative properties of solutions to problems involving the fractional Laplacian.With the same purpose，we plan to study the behavior of solutions to extension formulations for the fractional Laplacian established by Caffarelli and Silvestre[3]，and hope to obtain the corresponding behavior of solutions to a fractional Laplacian.The other motivation comes from the fact that we previously considered the linear uniformly elliptic equations Lu=Di(aij(x) Dju)+bi(x) Diu+c (x) u=f or Lu=aij(x) Diju+bi(x) Diu+cu=f in unbounded cylinders in[15]，followed by the fully nonlinear uniformly elliptic equations F (D2u (x))=0 in unbounded cylinders in[16].In the first paper，we established that all solutions bounded from below are linear combinations of two special solutions with a bounded solution to the term f，and in the later one，we showed results similar to but not exactly like the results in[1]under some conditions.Based on these，it is natural to attempt to develop a degenerate elliptic counterpart of the structure of solutions to uniformly elliptic equations.Therefore，our main objective here is to obtain analogous results for the degenerate elliptic equations.

This paper will follow the lines of[1]and[15].As in[15]，if we obtain three fundamental tools-the weak maximum principle in bounded domains and unbounded domains，the boundary H?lder estimates and the boundary Harnack inequalities，of the degenerate elliptic equations with mixed boundary values-then we will get a similar structure of solutions bounded from above.Therefore，we will address these three problems separately.

More precisely，we will consider the following problem，motivated by the realization of fractional elliptic operators，as Dirichlet-to-Neumann maps of degenerate elliptic equations:

We consider these in an unbounded domain C×(0，R0)?Rn+1，where C=D×R?Rnis an unbounded cylinder，D is a bounded Lipschitz domain in Rn-1(n≥2) and R0∈R is a positive constant.

We would like to say that although the degenerate elliptic equations we studied arise from the Caffarelli-Silvestre extension[3]of the Dirichlet problem for the fractional Laplacian，the results obtained in this paper do not hold in the fractional setting.We cannot simply make the trace u (x)=U (x，0) and conclude that the results of problem (1.1) hold for the following fractional order linear equations:

Actually，it is easy to check that the problem (1.1) and the fractional order linear equations (1.2) are not equivalent.Hence，we cannot analyze problem (1.2) through problem (1.1).We also would like to say that we will study problem (1.2) further in the later paper.

We always assume that the coefficient and inhomogeneous term satisfy

Before we state our main results，we give some notations for the reader’s convenience.Let X=(x，t)∈Rn×R，where x=(x′，y)=(x1，···，xn-1，y)∈Rn-1×R，n≥2.For E?R，CE:=D×E={(x′，y)∈Rn|x′∈D，y∈E}，?bCE:=?D×E={(x′，y)∈Rn|x′∈?D，y∈E}.For any y∈R，write Cy:=C{y}，:=C(y，+∞)and:=C(-∞，y).For simplicity，we denote that C+:=，C-:=.

In addition，we use～S to denote the set of solutions bounded from below to problem (1.1).If b=0，we denote S as the set of positive solutions to problem (1.1)(we will see that=S with b=0).

Theorem 1.1Suppose that condition (1.3) holds.Then the boundary problem

has a unique bounded solution U in H (t1-2σ，C×[0，R0])∩C (C×(0，R0)).

The following theorem is about the exponential decay of solutions bounded from above in C+×[0，R0]:

Theorem 1.2Suppose that U is bounded from above and satisfies

Then there exist positive constants α，C0and C1depending only on n and diam (D) such that

Following from Theorem 1.2，we obtain a corollary in C-×[0，R0]:

Corollary 1.3Suppose that U is bounded from above and satisfies

Then there exist positive constants α，C0and C1depending only on n and diam (D) such that

Next，we pursue further the structure of solutions to (1.1):

Theorem 1.4For the problem (1.1)，if b=0，then the positive solution sets S+and Sare well defined.Furthermore，S is a linear combination of S+and S-，that is，for any U∈S+and V∈S-，we have that

S=S++S-={pU+qV|p，q≥0，p+q＞0}.

Theorem 1.5For the problem (1.1)，the set of solutions bounded from below can be presented by，for any U∈S+，V∈S-，

where S0={U0}is the unique bounded solution to (1.4).

Our paper is organized as follows.In Section 2，we collect some auxiliary results.In Section 3，we prove the weak the maximum principle in bounded domains.In Section 4，we mainly study the global boundary H?lder estimate in bounded domains.In Section 5，we prove the weak maximum principle in unbounded domains.In Section 6，we prove the existence and uniqueness of a bounded solution and the exponential decay of bounded solutions.In the last Section，we analyze the structure theorem with an inhomogeneous term.

2 Preliminary Results

In this section，we will collect some basic results which will be used throughout the rest of the paper.First，we present some important inequalities with general A2weights.Then we introduce weighted Sobolev spaces.

Denote that QR=BR×(0，R)?Rn×R+，?′QR=BR，?′′QR=?QR?′QR，where BRis a ball centered at the origin with the radius R.

Recalling the definition of the Muckenhoupt A2class in Rn+1，that is，if there exists a constant Cωsuch that，for any ball B?Rn+1，

we say that ω(X) belongs to the class A2，where ω(X) is a nonnegative measurable function in Rn+1.

Now we quote some inequalities related to A2weights;these results can be found in[14]or[6].

Lemma 2.1(Weighted embedding inequality) Let f (X)∈and ω(X)∈A2.Then there exist positive constants C and δ depending only on n and Cwsuch that，for all 1≤k≤+δ，

where ω(QR)=.

Lemma 2.2(Weighted Poincaré inequality) Let f (X)∈C1(QR) and ω(X)∈A2.Then there exist positive constants C and δ such that，for all 1≤k≤+δ，

Lemma 2.3(Trace embedding inequality) Let f (X)∈and α∈(-1，1).Then there exists a positive constant δ depending only on α such that，for any ε＞0，

Next，we introduce weighted Sobolev spaces.Assume that σ∈(0，1) and that t∈R.According to the definition of A2，we see that|t|1-2σbelongs to the class A2.

Suppose that D is an open domain in Rn+1，and denote L2(|t|1-2σ，D) as the Banach space of all measurable functions U，defined on D，which satisfies

Now we can define

with the norm

Clearly，H (|t|1-2σ，D) is a Hilbert space and C∞(D) is dense in H (|t|1-2σ，D).Moreover，if D is a Lipschitz domain，then there exists a bounded linear extension operator from H (|t|1-2σ，D) to H (|t|1-2σ，Rn+1).

Suppose that Ω is an open domain in Rn.Recall that Hσ(Ω) is the fractional Sobolev space defined as

with the norm

Then C∞(Ω) is dense in Hσ(Ω).What’s more，if Ω is a Lipschitz domain，then there exists a bounded linear extension operator from Hσ(Ω) to Hσ(Rn).If Ω=Rn，Hσ(Rn) can also be expressed by

Hσ(Rn)={u∈L2(Rn):|ξ|σ(Fu)(ξ)∈L2(Rn)}，

where F denotes the Fourier transform operator.By a result in[11]，it is known that the space Hσ(Rn) coincides with the trace on，that is，every U∈H (t1-2σ，) has a well-defined trace u=U (·，0)∈Hσ(Rn).

The following results follow from results in[8]:

Lemma 2.4Suppose that D=Ω×(0，R)?Rn×R+，?′D=Ω，?′′D=?D?′D，where Ω is a Lipschitz domain.Then

(i) If U∈H (t1-2σ，D)∩C (D∪?′D)，then u (x)=U (x，0)∈Hσ(Ω)，and

where C is a positive constant depending only on n，σ，R and Ω.Hence，every U∈H (t1-2σ，D) has a well-defined trace U (·，0)∈Hσ(Ω) on?′D.Furthermore，there exists a constant Cn，σdepending only on n and σ such that

(ii) If u∈Hσ(Ω)，then there exists U∈H (t1-2σ，D) such that the trace of U on Ω equals u and

where C is a positive constant depending only on n，σ，R and Ω.

3 Weak Maximum Principle in Bounded Domains

In this section，we assume that D=Ω×(0，R*)?Rn×R+，?′D=Ω?Rn，?′′D=?D?′D，where D is a bounded domain in Rn+1.We will consider the maximum principle of the following boundary value problem:

Set φ∈H (t1-2σ，D) and H0={U∈H (t1-2σ，D):(U-φ)|?′′D=0 in the trace sense}.

Definition 3.1We say that U∈H0is a weak solution (supersolution，subsolution) of (3.1) in D.If，for every non-negative Φ∈，

Lemma 3.2(Weak maximum principle) Suppose that a (x)，b (x)∈L∞(Ω) and a (x)≤0 in Ω.If U (x，t)∈H (t1-2σ，D) satisfies the equations

then we have that U (x，t)≥0 in D.

ProofSince U is a weak supersolution，we have，for any nonnegative Φ∈，that

By a density argument，we can take U-as a test function.Therefore，we obtain

By using U=U+-U-and U+U-=0，we have that

Therefore，U-=0 in D，and consequently we have that U≥0 in D. □

Theorem 3.3(Weak maximum principle) Suppose that a (x)，b (x)∈L∞(Ω) and a (x)≤0 in Ω and that φ(x，t)∈C (?′′D).If U (x，t)∈H (t1-2σ，D) is a solution of the problem

then we have that

where C depends only on n，σ and R*.

ProofSet Φ=，B=‖b‖L∞(Ω).Since a (x)≤0 on?′D，obviously，we have that

We construct an auxiliary function V (x，t)=Φ+(μ-) B for (x，t)∈，where μ=.By a straightforward calculation，we have that

Therefore，

By Lemma 3.2，we obtain that

U (x，t)≤V (x，t)，(x，t)∈D.

This yields the desired result:

Notice that the above weak maximum principle holds under the assumptions of a (x)，b (x)∈L∞(Ω) and a (x)≤0 in Ω.If we reduce the integrability of a (x)，b (x)∈L∞(Ω) to a (x)∈Lp(Ω)，b (x)∈for p＞，we have the following weak maximum principle:

Theorem 3.4(Weak maximum principle) Suppose that a (x)∈Lp(Ω)，b (x)∈and a (x)≤0 in Ω.If U (x，t)∈H (t1-2σ，D) satisfies the equations

ProofLet L=，and assume furthermore that＞L.For any K＞L，choosing a test function Φ=(U-K)+with support in D∪?′D，by the definition of a weak subsolution，we have that

It follows that

Since a∈Lp(Ω) for some p＞，by H?lder’s inequality，we have that

By Lemma 2.4，there exists a constant C＞0 depending only on n，σ such that

By Lemma 2.3，there exist η，C＞0 both depending on n and σ such that

Substituting the above inequalities into (3.5)，we obtain

By using H?lder’s inequality and Lemma 2.4 again，we have that

where A (K)={(x，t)∈D|U (x，t)＞K}.Combining (3.4)，(3.6) with (3.7)，and taking，we have that

For the second term on the right hand side in (3.8)，we use the ε-Cauchy inequality to get that

If there exists K0≥L such that

then，for any K≥K0，we have，from (3.9)，that

By using Lemma 2.1 again，we have that

Noting that Φ=(U-K)+，when H＞K，we have that

Combining this with (3.11)，when H＞K≥K0，we have that

By Lemma 4.1 in[13]，we obtain

Next we estimate K0.We divide things into two steps to estimate K0.

Step 1Since

we can choose K0≥and.Thus (3.10) holds.In combination with (3.12)，this gives that

Step 2U has an upper bound from (3.13);the difficulty is to eliminate the second term of the right hand side in (3.13).Therefore，for any ε＞0，we consider the function

Noticing that Φ∈H (t1-σ，D) with compact support in D∪?′D，we put Φ into (3.3).For the left hand side of (3.3)，we have that

For the right hand side of (3.3)，we have that

Combined with (3.3)，(3.14) and (3.15)，this gives us that

Therefore，by Lemma 2.1，we have that

For any K＞L，from (3.16)，we have that

Taking K0-L=(1-ξ)(M+ε+B0)，where ξ＞0 is small and to be determined later，we get that

It is easy to see that there exists ξ＞0 such that (3.10) holds.Therefore，from (3.12)，we have that

4 Global Boundary H?lder Estimates

In this section，we will prove the global boundary H?lder estimate of solutions to (3.1).For this purpose，we assume further that Ω?Rnis a bounded Lipschitz domain and that Ω satisfies a uniform exterior cone condition.Let a，b∈Lp(Ω) for some p＞，and let φ(x，t)∈.

We denote that QR(x)=BR(x)×(0，R) for R＜R*，where?′QR(x)=BR(x)，?′′QR(x)=?QR(x)?′QR(x) and BR(x) is a ball centered at x∈Rnwith the radius R.What’s more，we denote that Q1(0)=Q1.

Lemma 4.1Let U∈H0be a weak subsolution of (3.1) in D.Then，for any x0∈?′D∩?′′D，R＞0 and q≥1，we have that

where

and C depends only on n，σ，p，q and.Here we have extended a and b to zero outside?′D=Ω.

ProofWithout loss of generality，we can assume that x0is the origin and that R=1.The general case can be recovered by means of the coordinate transform (x，t)→.We consider q=2 first.

as the nonnegative test function，where β＞0 and η∈is a non-negative function.A direct calculation yields that

Since U is a weak subsolution of (3.1) in D，it is easy to knowsatisfies

where we have used the ε-Cauchy inequality and the fact thatand K≤.Hence，we have that

Now we can rewrite (4.2) as

Since|?(ηW)|2≤2(η2|?W|2+|?η|2W2)，(4.3) can be rewritten as

Due to a，b∈Lp(B1) for some p＞，it follows from H?lder’s inequality that

By Lemma 2.4，there exists a C＞0 depending only on n and σ such that

By Lemma 2.3，there exist ξ，C＞0 both depending on n and σ such that

By choosing ε small and substituting the above inequalities into (4.4)，we obtain

By Lemma 2.1 and (4.5)，we have that

By the definition of W，we obtain

Set γ=β+1.Then γ＞1，and we get that

Therefore，we obtain

By Moser’s iteration，we then obtain that

This finishes the proof of q=2.This also holds for any q≥1，by standard arguments.Finally，through a simple coordinate transformation (x，t)→，we obtain the desired the result:

Lemma 4.2Let U∈H0be a nonnegative weak supersolution of (3.1) in D.Then，for any x0∈?′D∩?′′D，R＞0，0＜q≤and 0＜θ＜?＜1，we have that

ProofWithout loss of generality，we assume that x0is the origin and that R=1.We set+K with K＞0.Then.We also choose

as the nonnegative test function，where β＜0 and where η∈is a nonnegative function.A direct calculation yields that

Since U is a weak supersolution of (3.1) in D，we know thatsatisfies

Applying the above test function Φ to (4.6)，we obtain that

where we have used the ε-Cauchy inequality (0＜ε≤1) and the fact thatand K≤.By choosing ε=min (1，)，we obtain that

where C (β) is bounded if|β|is bounded away from zero.Now we define W as

Letting γ=β+1，we rewrite (4.7) as

Since|?(ηW)|2≤2(η2|?W|2+|?η|2W2)，we can rewrite the first inequality (when β-1) of (4.8) as

The next proof is similar to that of Lemma 4.1.Hence，for any 0＜r＜τ≤1，we obtain that

If we can show that there exists some q0＞0 such that，for any 0＜μ＜1，

where C depends on n，σ and μ，then the desired result will be obtained from (4.10) after finitely many iterations (Moser’s iteration).

In order to establish (4.11)，we turn to the second inequality of (4.8).It follows from the H?lder’s inequality and the Sobolev inequality that

Therefore，by Lemma 2.2，the H?lder’s inequality and (4.13)，we get that

where Wx，rdenotes the weighted averageof W in Qr(x).Hence，we see that W∈BMO (t1-2σ，Q1).Then，by similar arguments to those of the John-Nirenberg type lemma in[10]，we can show that≤C for any μ＜1.Recalling the definition of W，we obtain the estimate (4.11)，and consequently establish the desired result. □

Lemma 4.3Let U∈H0be a weak solution of (3.1) in D.Then，for any x0∈?′D∩?′′D and any 0＜R≤R0，there exist positive constants C and α depending on n，σ，p，R0andsuch that

ProofLet x0∈?′D∩?′′D.By the uniform cone condition，we know that|Q2R(x0)D|≥ξ|Q2R(x0)|for some R1＞0，some ξ＞0，and any Q2R(x0) with R≤R1.

We may assume，without loss of generality，that R≤.Writing

Then M4-U is a nonnegative solution of the equations

while U-m4is a nonnegative solution of the equations

Since Ω satisfies a uniform exterior cone condition，we know that Q2R(x0)D contains a uniform exterior cone∩Q2R(x0).Then we can apply Lemma 4.2 to the functions M4-U and U-m4in Q4R(x0) by taking q=1，and obtain that

By addition，we obtain that

where γ=1-1/C＜1 and where C depends on n，σ，p and R0.

Then the desired result follows from Lemma 8.23 in[7]. □

Theorem 4.4Let U∈H0be a weak solution of (3.1) in D.Then there exist constants κ，α0such that，for any x0∈?′D and R＞0，we have that

where C，α both depend on n，σ，p and α0.

ProofIt is easy to obtain (4.15) from Lemma 4.3.From the H?lder continuity at the boundary?′′D，one can con firm Theorem 2.4.6(Dirichlet boundary problem) in[6].Combining the interior H?lder estimate of Proposition 2.6 in[8]with the boundary H?lder estimate (4.15) on?′D，we can obtain our result. □

5 Weak Maximum Principle in Unbounded Domains

In this section，we mainly prove the weak maximum principle in unbounded domains.That is，we will consider the domain C×[0，R0]?Rn+1，where C=D×R?Rnis an unbounded cylinder，and assume that C×[0，R0]satisfies a uniform exterior cone condition.We note that the diameter of C(k，k+2)only depends on n and diam (D)，i.e.diam (C(k，k+2)) is independent of k and can be denoted by diam (C(k，k+2))=C (n，diam (D)).Therefore，the diameter of the domain C×[0，R0]satisfies diam (C×[0，R0])=C (n，R0，diam (D)).

Lemma 5.1There are constants 0＜ε0，δ＜1 such that，if U (x，t) satisfies

ProofLet U+(x，t) be the solution of

By the weak maximum principle (Lemma 3.2 and Theorem 3.3)，we have that

where C depends only on n，σ，R0and diam (D).

By Theorem 4.4，there exists C0depending only on n，σ and D such that

We apply the weak Harnack inequality (Proposition 2.6) in[8]to (1+Cε0-U+) in×(0，R0)，and we obtain，for some η＞0，that

Remark 5.2To prove Lemma 5.1，one key is the weak maximum principle.If we use the weak maximum principle (Theorem 3.4)，the result in Lemma 5.1 also holds.

Now we use Lemma 5.1 to prove the weak maximum principle in unbounded domains C×[0，R0].

Theorem 5.3(Weak maximum principle) Suppose that U (x，t) is bounded from above，and that U (x，t)∈H (t1-2σ，C×(0，R0)) satisfies

Then we have that

where C depends only on n，σ，R0and diam (D).

Therefore，we only need to prove that

Suppose that U (x，t)≤M since U is bounded from above.Also，let B=.In order to apply Lemma 5.1，we consider the function

Then，by iteration，we have，for any k∈Z and t∈[0，R0]，that

By applying Theorem 3.3，for any y∈[k-1，k+1]and t∈[0，R0]with k∈Z，we have that

Corollary 5.4Suppose that U (x，t) is bounded from above，and that U (x，t)∈H (t1-2σ，C+×(0，R0)) satisfies

Then we have that

where C depends only on n，σ，R0and diam (D).

6 Boundary Harnack Inequality

In this section，we mainly prove the boundary Harnack inequalities in bounded domains.The method of proofs mainly follows the idea of Lemma 4.9 in[2].The boundary Harnack inequality and a comparison theorem are crucial for the proof of our theorems.

Let ψ(x′) be a Lipschitz function in Rn-1(n≥2) with a Lipschitz constant，such that

For r＞0，denote

Fix R0≥1，Qr×(-R0，R0)?Rn×R and Qr×(0，R0)?Rn×R+.Assume，for any X=(x，t)∈?(Qr×(-R0，R0))，that there exists X0=(x0，t0)∈Qr×(-R0，R0) such that|X-X0|＜Cr and Br/C(X0)?Qr×(-R0，R0)，where C＞1，0＜r＜r0are constants.

Lemma 6.1(Boundary Harnack inequality) Suppose that U∈H (t1-2σ，Q4r×(0，R0)) satisfies

Then we have that

where C depends only on n，σ and.

ProofBy scaling，we assume that r=1.For some A＞0 to be determined later，set

The function VA(x，t) satisfies

We consider the even extension of VAacross{t=0}on Q4，defined by

Denote by LAthe operator

We introduce the solution h±Aof

These solutions are obtained from the solutions of the Dirichlet problem (Theorem 2.2) in[6].Hence，we have that

By the weak maximum principle (for the Dirichlet problem also holds;see of Remark 4.2 in[2])，we have that

then according to Remark 1 in[5]，we have that

Therefore，we obtain that

Using (6.2) and (6.3)，we have that

Therefore，(6.4) and (6.5) lead immediately to the desired result:

Lemma 6.2(Comparison theorem) Suppose that Ui∈H (t1-2σ，Q4r×(0，R0))，i=1，2，satisfy

Then we have that

ProofSimilarly，we use the same method as to that of Lemma 6.1 and combine with the boundary Harnarck inequality in[5]to obtain our conclusion.Here，we omit the proof. □

7 The Exponential Decay of Bounded Solutions

In this section，we first prove the existence and uniqueness of a bounded solution in unbounded domains;this plays an important role in proving the structure theorem.Then we show that the exponential decay of bounded solutions.

Proof of Theorem 1.1First，we consider the equations in the bounded domain:

Here N∈Z+.By Proposition 2.4 in[8]，for this problem (7.1)，there exists a unique solution UN(x，t)∈H (t1-2σ，C(-N，N)×(0，R0)).By Theorem 3.3，we have that

where CNdepends only on n，σ，R0and N.

We will prove that there exists a constant C0＞0 not depending on N such that

For convenience，we denote M=.

For any ξ satisfying-N+1≤ξ≤N-1，it is clear that C(ξ-1，ξ+1)×(0，R0)?C(-N，N)×(0，R0).By using arguments similar to that in the proof of Theorem 5.3 in C(ξ-1，ξ+1)×(0，R0)，we get that

Then we obtain

We have，furthermore，that

For any (x，t)∈C(-N，-N+1)×(0，R0)，by Theorem 3.3 again，we have that

where C2=+C1depending only on n，σ，R0and diam (D).

For any (x，t)∈C(N-1，N)×(0，R0)，similarly，we have that

Therefore，for any (x，t)∈C(-N，N)×(0，R0)，we have that

where C3=max{C2，}，which depends only on n，σ，R0and diam (D).That is，from the definition of M，we have that M≤(1-δ) M+.Thus we obtain

where C0=，which depends only on n，σ，R0and diam (D).

Then，by Theorem 4.4，there exists a constant C*＞0 depending only on n，σ，D such that

Thus，for any bounded domain C[-l，l]×[0，R0]with l＞0，there exists a subsequence of{UN(x，t)}which uniformly converges in C[-l，l]×[0，R0]by the Arzela-Ascoli theorem.Without loss of generality，we assume that there exists a function U (x，t) such that UN(x，t) uniformly converges to U (x，t) in C0(C×(0，R0)).Therefore，U (x，t) is bounded in C×(0，R0) and satisfies (1.4) in the weak sense.By Theorem 5.3，we know that U is the desired unique bounded solution. □

Next，we give a proof of the exponential decay of bounded solutions (Theorem 1.2) with general b.

Proof of Theorem 1.2Assume that=B.Since U is bounded from above，Corollary 5.4 leads to that

Applying Lemma 5.1，there exists a constant δ∈(0，1) such that

Similarly，we have that

By induction，we obtain

Therefore，we have the following estimate for (x，t)=(x′，y，t)∈C+×[0，R0]:

where α=-ln (1-δ)＞0.Since B=，we have，for any (x，t)∈C+×[0，R0]，that

8 The Structure Theorems of Solutions

In this section，we show that the structure theorems of solutions with inhomogeneous term b.In the case of b=0，since we have proved the weak maximum principle in unbounded domains (Theorem 5.3) and the boundary Harnack inequalities for the mixed boundary problem (Lemma 6.1 and Lemma 6.2)，we can classify all positive solutions and consider the asymptotic behaviors of the solutions to get the proof of Theorem 1.4.The details of the proofs for these results are very similar to those of[1，15，16]，so we are not going to give them.Here，we only give the proof of the structure Theorem 1.5.

Proof of Theorem 1.5Applying Theorem 1.1，we take V such that V is the unique bounded solution of the following problem:

Then we know that there exists a constant C＞0 such that U-V＞-C and U-V satisfies

Since U-V is bounded below，by applying Theorem 5.3，we obtain U-V≥0.

Thus，either U≡V or U-V＞0.If U=V，then our conclusion clearly holds (taking p=q=0).If U-V＞0，by applying Theorem 1.4，we derive that there exist W∈S+，Z∈Ssuch that U-V=pW+qZ;that is，U=V+pW+qZ，where V is the bounded solution.Therefore，we obtain our conclusion:

Here S0={V}is the unique bounded solution of the problem (1.4). □

AcknowledgementsThe author would like to take this opportunity to express gratitude to her advisors，Professor Chunqin Zhou and Lihe Wang，for their constant encouragements，inspiring discussions and helpful suggestions，without which this work would be impossible to be carried out.