Martin Boundary of a Degenerate Reflected Brownian Motion in a Wedge

Published Online:https://doi.org/10.1287/stsy.2024.0090

Abstract

We consider an outward degenerate drifted Brownian motion in the quarter plane with oblique reflections on the boundaries. In this article, we explicitly compute the Laplace transforms of the Green’s functions associated with the process. These Laplace transforms are expressed as an infinite sum of products by iterating a functional equation, which is deeply linked to the compensation method. We also derive the asymptotics of the Green’s functions along all possible paths and determine the (minimal) Martin boundary. Finally, we provide explicit formulae for all the corresponding positive harmonic functions.

Funding: This project received funding from ENS Rennes (PhD funding) and from the ANR RESYST [ANR-22-CE40-0002].

1. Introduction and Main Results

1.1. Context

The semimartingale reflecting Brownian motion (SRBM) in two-dimensional convex cones is a classical topic in probability theory. Problems such as existence and uniqueness (Harrison and Reiman 1981, Taylor and Williams 1993), recurrence and transience conditions (Williams 1985, Hobson and Rogers 1993), the study of stationary distribution properties (Harrison and Williams 1987b, Dieker and Moriarty 2009, Dai and Miyazawa 2011, Franceschi and Kourkova 2017), and many others have been studied extensively in the literature, mostly under the assumption of a nondegenerate covariance matrix.

An important problem in transient SRBM is the analysis of Green’s functions, which can be divided into two parts:

  • (P1) Obtaining the Laplace transforms of the Green’s functions;

  • (P2) Computing the asymptotics of the Green’s functions along all trajectories of the SRBM.

Solutions to (P1) in the half-plane can be expressed directly in terms of a rational function of two variables, (x,Y(x)), where Y(x) is a branch of a certain two-valued algebraic function, as detailed in Ernst and Franceschi (2021). However, solving (P1) in a general cone presents a significantly greater challenge. Specifically, for nondegenerate SRBM in the quarter plane with three domains, the Laplace transforms are obtained as singular integral representations via boundary value problems, as shown in Franceschi and Raschel (2019) and Franceschi (2020). Although these expressions are explicit, they are not particularly amenable to in-depth analysis. Fortunately, they are not required to resolve the second issue (P2). In fact, only the locations of the dominant singularities of unknown Laplace transforms are necessary to compute the asymptotics of the Green’s functions. Problem (P2) for nondegenerate SRBM has been solved in the half-plane in Ernst and Franceschi (2021) and in an arbitrary wedge in Franceschi et al. (2024b). The approach followed in these articles has been developed in Malyshev (1973), Kurkova and Malyshev (1998), Dai and Miyazawa (2011), Franceschi and Kourkova (2017), Kurkova and Raschel (2011), and Fayolle et al. (2017) and can be considered as a version of the so-called kernel method. For more information, see the survey by Zhao (2022). The kernel γ(x,y) of the SRBM is given by one half of the quadratic form of the covariance matrix plus the linear form of the drift inside the cone. The interplay between the branches of algebraic functions X(y) and Y(x), defined by the kernel equation γ(x,y)=0, allows us to analytically continue unknown Laplace transforms and to determine their singularities. The inverse Laplace transforms, combined with the saddle point method, then yield asymptotic expansions for the Green’s functions. This procedure provides asymptotic developments of Green’s functions with arbitrarily many terms, but with unknown multiplicative constants. These constants may be derived—albeit somewhat indirectly—from the solutions to (P1).

The degenerate SRBM in two-dimensional cones, that is, with a covariance matrix of rank one, has been studied far less extensively. In Ichiba and Karatzas (2022) and Franceschi et al. (2024a), it arises as the gap process between three particles moving and colliding in R1. The construction of this three-particle process relies on the Skorokhod reflection approach, as developed in Harrison and Reiman (1981), to define pathwise reflected Brownian motion.

In the present article, we consider a class of degenerate transient SRBMs in the quadrant, defined by conditions (1.1)–(1.3), and solve both problems (P1) and (P2). The Laplace transforms of the Green’s functions are expressed in terms of infinite series in product form. This result follows from the compensation method, initially introduced in Adan et al. (1993) to obtain the stationary measure for certain degenerate random walks in a quadrant. This approach has since been successfully applied to queueing systems (Adan et al. 1991, Adan 1994). It has also been used to derive generating functions for random walks with small steps Adan et al. (2013) and, more recently, to determine the harmonic functions of singular walks in the quadrant (Hoang et al. 2023). In Franceschi et al. (2024a), for instance, it was used to derive the explicit form of the stationary distribution and in Franceschi (2024) to determine the Martin boundary of killed degenerate Brownian motion in a two-dimensional cone.

In this article, we compute the asymptotics of the Green’s functions along all trajectories. To achieve this, we adapt the approach described earlier and developed in Kurkova and Malyshev (1998), Dai and Miyazawa (2011), Fayolle et al. (2017), Franceschi and Kourkova (2017), Ernst and Franceschi (2021), and Franceschi et al. (2024b) to this class of degenerate SRBMs. A key difference is that, unlike the nondegenerate case—in which the kernel equation for the process defines an ellipse—the kernel equation for the degenerate case defines a parabola in R2. The multiplicative constants in the asymptotic expressions of the Green’s functions, derived from the solution to (P1), are made explicit in terms of infinite series in product form. The significance of these constants—viewed as functions of the starting point of the process—extends beyond asymptotic precision; they also yield all positive harmonic functions for the DRBM via the Martin boundary theory.

Initiated by Martin (1941) and further developed by Hunt (1957), Doob (1959), and Kunita and Watanabe (1965), this theory is summarized in Doob (1984) and Chung and Walsh (2005). Its aim is to describe the asymptotic behavior of the process and to characterize all nonnegative superharmonic and harmonic functions. The limits of the Martin kernel along the trajectories of the process, when they exist, compactify the state space and form the so-called Martin boundary. This procedure allows every nonnegative harmonic function to be expressed as an integral representation over the Martin boundary. In Ney and Spitzer (1966), Ignatiouk-Robert (2009, 2010), Ignatiouk-Robert and Loree (2010), and Duraj et al. (2022), the Martin boundary is identified via large deviation principles. It has also been obtained from the asymptotics of Green’s functions in Kurkova and Malyshev (1998), Kurkova and Raschel (2011), Ernst and Franceschi (2021), and Franceschi et al. (2024b). In this article, using the solutions to problems (P1) and (P2) described above, we determine the Martin boundary, the minimal one, and provide explicit expressions for all positive harmonic functions.

1.2. Main Results

The Degenerate Reflected Brownian Motion, Assumptions.

We consider a degenerate Brownian motion (Zt)t0 in a quadrant, with oblique reflection at the boundaries. By degenerate, we mean that the covariance matrix is of rank 1. This obliquely reflected process was studied in Ichiba and Karatzas (2022), and its rigorous definition is provided in Section 2. The parameters of the degenerate reflected Brownian motion are given by

Σ=(σ12σ1σ2σ1σ2σ22),μ=(μ1μ2),R=(1r2r11)=(R1R2),(1.1)
where Σ is the degenerate covariance matrix (det(Σ)=0), μ is the drift, and columns of R represent the reflection directions from the axes. The direction v=(v1,v2)T=(σ1,σ2) is antidiagonal, that is, v1v2<0 (see Figure 1). When the process does not hit the boundaries, it behaves like a one-dimensional Brownian motion along the direction v (plus the drift). Our main assumptions in this article are as follows:
μ1>0,μ2>0,(1.2)
r1>σ2σ1,r2>σ1σ2.(1.3)

Figure 1. Reflections R1,R2 on the Edges, the Drift μ, and the Direction v of the Degenerate Brownian Motion
Note. The process starting from z0 never reaches the hatched region.
Figure 2. Example of a Typical Path (Over a Finite Time Horizon) of the Drifted Degenerate Brownian Motion
Note. The initial point is marked in orange.

Assumption (1.2) ensures that the process is transient, whereas (1.3) specifies that the reflection vectors R1=(1,r1)T (on {x=0}) and R2=(1,r2)T (on {y=0}) point outward from the direction v of the Brownian motion (see Figure 1).

In Sections 18, we state and prove our results under the additional assumption

σ1=σ2=1,μ1+μ2=1.(1.4)

Results for the general case, that is, without Assumption (1.4), are stated and proved in Section 9. In fact, they are easily deduced from the results under (1.4) by means of a simple space-time transformation.

Green’s Functions.

We show that for any starting point z0R+2, there exists a density gz0(·) of the Green’s measure G(z0,·) on the quadrant defined as

G(z0,A)0+Pz0(ZtA)dt=Agz0(z)dz.(1.5)

Functions gz0(·) are called the Green’s functions. We also define the Green’s measures on the sides of the wedge

Hi(z0,A)Ez0[0𝟙A(Zt)dLti],i=1,2,(1.6)
where (Lt1)t0 (resp. (Lt2)t0) is the local time of the process on the axis {x=0} (resp. {y=0}). The measure H1 has its support on the vertical axis, and H2 has its support on the horizontal axis. Laplace transforms φ(x,y) of G(z0,·) and φ1(y),φ2(x) of H1(z0,·),H2(z0,·) are related by the functional equation,
γ(x,y)φ(x,y)=γ1(x,y)φ1(y)+γ2(x,y)φ2(x)+e(x,y)·z0,Re(x)<0,Re(y)<0,(1.7)
where
γ(x,y)=12(xy)2+μ1x+μ2y,(1.8)
and
γ1(x,y)=R1·(x,y)=x+r1y,γ2(x,y)=R2·(x,y)=r2x+y.(1.9)

It can be viewed as a balance equation for Green’s measures between the interior and the edges of the quadrant. Let us define

P={(x,y)R2,γ(x,y)=0}.(1.10)

The functional Equation (3.1) is similar to that in Franceschi et al. (2024b); however, an important difference is that P is now a parabola rather than an ellipse. This distinction is what allows the compensation method to be effective, leading to explicit expressions for the Laplace transforms and positive harmonic functions.

Explicit Expressions for Laplace Transforms.

The first results of the article provide explicit expressions for Laplace transforms φ1 and φ2 in terms of infinite series of product forms, given by formulae (4.24) and (4.23), which we do not specify here. Function φ is derived from φ1 and φ2 via the functional Equation (1.7).

Asymptotics of Green’s Functions.

We now focus on the asymptotics of gz0(rcos(α),rsin(α)) as r+ and αα0[0,π/2]. For any direction α, we denote by (x(α),y(α)) a corresponding point on the parabola given by

(x(α),y(α))=arg max(x,y)P(cos(α)x+sin(α)y),(1.11)
see Figure 3(a). It can be computed explicitly as
(x(α),y(α))=((μ2tan(α)μ1)(μ2+tan(α))(1+μ2)2(1+tan(α))2,(μ2tan(α)μ1)(1+μ1(1+tan(α)))2(1+tan(α))2).(1.12)
Figure 3. Geometric Interpretation of (x(α),y(α)), α*, and α**

Let us define two particular directions

α*{0if(r1+1)μ22arctan((1+r1)μ222+(1+r1)μ1)if(r1+1)μ2>2.(1.13)
α**{arctan((1+r2)μ2+2(1+r2)μ12)if(r2+1)μ1>2π/2if(r2+1)μ12,(1.14)
see Figure 3(b) for their geometric interpretation. We always have α*<α**, as will be proved in Section 5.

In the following theorem, we summarize the asymptotics of Green’s functions for directions α0(0,π/2)\{α*,α**}. The ones for α0{0,α*,α**,π/2} are given later in Theorems 6, 7, and 8.

Theorem 1

(Asymptotics in the Quadrant, General Case). Assume (1.2) to (1.4). Then, the Green’s density function gz0 of this process has the following asymptotics as αα0 and r.

  • If α*<α0<α**, then

    gz0(rcos(α),rsin(α))rαα0cα0hα0(z0)er(cos(α)x(α)+sin(α)y(α))r.(1.15)

  • If α0<α*, then

    gz0(rcos(α),rsin(α))rαα0c*hα*(z0)er(cos(α)x(α*)+sin(α)y(α*)).(1.16)

  • If α0>α**, then

    gz0(rcos(α),rsin(α))rαα0c**hα**(z0)er(cos(α)x(α**)+sin(α)y(α**)).(1.17)

where cα0=12π(cos(α0)+sin(α0)), c* and c** are positive explicit constants depending only on the parameters of the degenerate reflected Brownian motion (see (6.1)), and hα(z0),hα*(z0),hα**(z0) are harmonic functions given in Theorem 2. Furthermore, hα(z0),hα*(z0),hα**(z0) are nonzero.

Explicit Expressions for Positive Harmonic Functions with the Compensation Method.

Let us recall the following definition: a function h:R+2R is harmonic if and only if for all t0 and z0R+2,

Ez0[h(Zt)]=h(z0).(1.18)

All functions hα,α[α*,α**] are harmonic. These functions are explicitly stated in Theorem 2 below and will be derived in this article using the compensation method. The essence of this method is to construct functions that satisfy the partial differential equation along with boundary conditions:

{(H0)Gh=0on(0,+)2,(H1)R1h(0,y)=0,y0(H2)R2h(x,0)=0,x0(1.19)
where G=12·Σ+μ·. Those function are harmonic, as will be noticed in Section 4.1.

For (a0,b0)P and kZ\{0}, we set

a2k=2k2+2(a0b0μ2)k+a0,a2k+1=a2k(1.20)
b2k=2k2+2(a0b0+μ1)k+b0,b2k+1=b2k+2(1.21)

As illustrated in Figure 4, points (ap,bp)P are constructed by following the “downstairs” path on the parabola, applying successively automorphisms that leave invariant the first or the second coordinate, respectively.

Figure 4. Parabola P and Points (an,bn) on the Parabola
Theorem 2

(Explicit Expressions for Harmonic Functions (hα)α[α*,α**]). Assume (1.2) to (1.4). Then, the functions (hα)α[α*,α**] are harmonic and are given by the following formulae:

  • For α(α*,α**), taking (a0,b0)=(x(α),y(α)), we have

    hα:z0m=+κm(α)ez0·(am,bm)(1.22)

    where κ0(α)=1 and

    κm(α)={(1)m[k=0m21γ1γ2(a2k+1,b2k+1)γ1γ2(a2k+2,b2k+2)]γ2(a0,b0)γ2(am,bm)ifm>0(1)m[k=0m21γ2γ1(a2k1,b2k1)γ2γ1(a2k2,b2k2)]γ1(a0,b0)γ1(am,bm)ifm<0(1.23)

    (with the convention Πk=01=1).

  • For α=α*,

     If 2r2+1<μ2, then α*=0 and h0:z0α[hα(z0)]α=0+.

     If 2r2+1>μ2, then α*>0 and taking (a0,b0)=(x(α*),y(α*)),

    hα*:z0ez0·(a1,b1)+m=2+κ^m(α*)ez0·(am,bm),(1.24)
    where
    κ^m(α*)=(1)m+1γ1(a1,b1)γ1γ2(a2,b2)[k=1m21γ1γ2(a2k+1,b2k+1)γ1γ2(a2k+2,b2k+2]1γ2(am,bm).

     If 2r2+1=μ2, then α*=0 and taking (a0,b0)=(x(0),y(0)),

    h0:z02ez0·(a0,b0)+m=1κm(α*)ez0·(am,bm)+m=2+κ˜m(α*)ez0·(am,bm)(1.25)
    where
    κ˜m(α*)=(1)m+1γ1(a1,b1)γ1γ2(a2,b2)[k=1m21γ1γ2(a2k+1,b2k+1)γ1γ2(a2k+2,b2k+2)]1γ2(am,bm).

  • For α=α**, symmetrical formulae hold replacing r1 by r2, μ1 by μ2, and 0 by π2.

Note that if α<α* or α>α**, expression (1.22) may define a harmonic function that is not necessarily nonnegative everywhere.

The Martin boundary and its minimality are derived from Theorems 1 and 2, together with the further technical results in Theorems 6, 7, and 8 concerning the asymptotics of Green functions along the directions 0,α*,α** and π/2.

Figure 5. Martin Boundary Γ When 0<α* and α**<π/2
Theorem 3

(Martin Boundary). Under (1.2) to (1.4), the Martin boundary Γ of the degenerate reflected Brownian motion is homeomorphic to [α*,α**] via the mapping

α[α*,α**]hα(·)/hα(0)Γ.(1.26)

Furthermore, the Martin boundary is minimal.

Remark 1

(On Assumptions (1.2) and (1.3) and Possible Extensions). Some results may be generalised with extended parameters

  • - Regarding assumption (1.2), similar results could be established under the more general condition μ1σ2+μ2σ1>0. This condition is equivalent to the orientation of the parabola toward x and y. It is also necessary for the convergence of the expressions defining hα—specifically, equation (1.22). Namely, if μ2<0, then the Laplace transform φ2 would have a pole at zero. Because of the technical nature of this paper, we have chosen to restrict our analysis to Assumption (1.2). Investigating how the Martin boundary is affected by the presence of such a pole could be an interesting direction for future work.

  • - If (1.3) is not satisfied, then the arguments that yield the explicit expressions of the harmonic functions fail. In particular, attempts to construct the functions hα without this assumption often lead to signed functions that, although possibly harmonic, are not necessarily nonnegative. For interested readers, the only step in our argument that fails for general reflection vectors is equation (4.18), which may offer a direction for future investigation.

1.3. Plan of the Article

In Section 2, we define the degenerate reflected Brownian motion. We then derive the functional Equation (1.7) in Section 3 and meromorphically extend Laplace transforms on the edges up to their singularities. In Section 4, we obtain the explicit form of the Laplace transforms iterating the functional Equation (1.7). Next, in Section 5, we carry out preparatory work to derive the asymptotics of Green’s functions. These asymptotics are computed in all directions in Sections 6 and 7 by the saddle point method. This enables us to prove Theorems 1 and 2 by employing the explicit expressions from Section 4. In Section 8, we establish the asymptotics of the Martin kernel and identify all the harmonic functions. We also prove the minimality of the Martin boundary and conclude the proof of Theorem 3. Finally, in Section 9, we treat the general case of the model without Assumption (1.4) via a linear transformation of space and time.

2. Definition of the Process

Throughout the following, the filtered space we consider is always the space of continuous functions C(R+,R+2) with the standard σfield and the usual filtration. The following background definition is taken from Taylor and Williams (1993), where the nondegenerate reflected Brownian motion is studied.

Definition 1

(Degenerate Reflecting Brownian Motion). Let Σ,R, and μ be defined as in (1.1). A degenerate reflecting Brownian motion (DRBM) associated with the data (Σ,μ,R) is a process (Zt)t0 and a family of measures (Pz0)z0R+2 such that (Zt)t0 can be written as

Zt=Xt+RLtR+2,t0,(2.1)
where
  • (Xtμt)t0 is an adapted degenerate Brownian motion (with zero drift) of covariance Σ starting from z0 under Pz0.

  • L is an adapted two-dimensional process starting from 0 such that Pz0 almost surely, and its components L1,L2 are continuous and nondecreasing with supp(dLi){t0,Zti=0}; that is, Li increases only when Zti=0.

Note that under Pz0, Z can be written as (Zt)t0=(z0+vBt+μt+RLt)t0, where (Bt)t0 is a one-dimensional Brownian motion and v=(σ1,σ2) (=(1,1) under (1.4)) is the unique eigenvector (up to a scalar multiplication) associated with the positive eigenvalue of the covariance matrix.

Theorem 4

(Existence, Uniqueness, and Strong Markov Property). Suppose that |r1r2|<1. Then, for any starting point z0, there exists a DRBM associated with (Σ,μ,R). The processes Z and (Z,L) are pathwise unique (according to the associated degenerate Brownian motion). Furthermore, Z is a semimartingale, a Feller process (i.e., for any t0, xEx[f(Zt)] is continuous whenever f is bounded and continuous), and a Strong Markov process.

Proof.

Define the matrix Q=IR, whose spectral radius is ρ(Q)=|r1r2|<1. By theorem 1 in Harrison and Reiman (1981), for any continuous path x=(xt)t0R2, there exists a unique solution (zt)t0=ψ(x) of the Skorokod problem

zt=xt+R(lt1,lt2)T,t0
where (zt)t0R+2, and for i{1,2}, (li)t0 is a continuous, increasing function with supp(dli){t0,zti=0}. Moreover, ψ is continuous in the topology of uniform convergence on compact sets. This yields the stated result with Z=ψ(X). □

As in the nondegenerate case, there may be existence and uniqueness in law if R is a general S-matrix Taylor and Williams (1993) (without assuming |r1r2|<1) but not path-wise uniqueness (Bass and Burdzy 2024). To avoid excessive technicality, we work under assumption |r1r2|<1.

Proposition 1

(Transience). Under conditions (1.2) and (1.3), the DRBM is a transient Markov process.

Proof.

Consider w=(σ2,σ1), which is orthogonal to the direction of the Brownian motion. It suffices to note that (Zt·w)t0 is almost surely strictly increasing and tends to + because Zt·wμ·wt by (1.3). □

We recall the definition of Green’s measure G(z0,·) and Hi(z0,·) from (1.5) and (1.6). Assumption (1.2) on the drift is crucial for the following proposition.

Proposition 2

(Densities and Laplace Transforms). Suppose that assumptions (1.2) and (1.3) hold. Then, Green’s measure G(z0,·) has a density gz0(·) with respect to the Lebesgue measure. Functions gz0(·) are called Green’s functions. Furthermore, measures Hi(z0,·) (i=1,2) have densities fiz0(·) with respect to the one-dimensional Lebesgue measure.

Proof.

Let A be a compact set of R+2 at a positive distance of the edges. Define the stopping times:

σ=inf{t0,ZtA},τ=inf{tσ,ZtR+2}.

Considering the back-and-forth trajectories between A and R+2 (see Harrison and Williams 1987a and Lemma 9 in Section 7), we can reduce the proof to showing that

Ez0[στ1A(Zs)ds]=0.

Then, by the Strong Markov property, it suffices to prove the result for a nonreflected degenerate Brownian motion. By Assumption (1.2), rotating the plane so that the x-axis aligns with the drift direction reduces the problem to one-dimensional Brownian motion. The proposition then follows from elementary properties of the latter. □

Definition 2

(Laplace Transforms of Green’s Measures). We denote the Laplace transforms of G(z0,·) by

φ(x,y)Ez0[0e(x,y)·Ztdt]=R+2e(x,y)·zgz0(z)dz
and the Laplace transforms of H1(z0,·),H2(z0,·) by
φ1(y)Ez0[0e(0,y)·ZtdLt1],φ2(x)Ez0[0e(x,0)·ZtdLt2].

For brevity, we omit the dependence on the starting point in the notation for the Laplace transforms. However, when relevant, we will denote this dependence explicitly as φz0(x,y),φ1z0(y) and φ2z0(x).

3. Functional Equation, Kernel, and Analytic Continuation

From now on, we assume (1.2) to (1.4). As mentioned in the introduction, Laplace transforms φ,φ1,φ2 are linked by a functional equation.

Proposition 3

(Functional Equation). If Re(x)<0 and Re(y)<0, then φ1(y), φ2(x), and φ(x,y) converge, and the following equation holds:

γ(x,y)φ(x,y)=γ1(x,y)φ1(y)+γ2(x,y)φ2(x)+e(x,y)·z0,(3.1)
where γ,γ1, and γ2 are defined in (1.8) and (1.9).

Proof.

We apply Itô’s formula to the semimartingale (Zt)t0 and the function (u,v)exu+yv. Then,

e(x,y)·Zte(x,y)·z0=0te(x,y)·Zt(x,y)T·dBs+γ(x,y)0te(x,y)·Zsds+i=12γi(x,y)0te(x,y)·ZsdLsi.,(3.2)
where (Bt)t0=(Xtμt)t0 is the non-reflected degenerate Brownian motion associated with the process (see Definition 1). Next, taking the expectation and letting t to +, we derive (3.1). See (Franceschi et al. 2024b, proposition 2.7) for a detailed version of the proof in the nondegenerate case. □

Considering γ(x,y) as a polynomial in x (resp. y) with coefficients depending on y (resp. x), we obtain two complex branches Y+(x), Y(x) (resp. X+(y), X(y)) satisfying γ(x,Y±(x))=γ(X±(y),y)=0:

Y±(x)=xμ2±2x+μ22,X±(y)=yμ1±2y+μ12.(3.3)

We have one branching point xmax=μ222>0 (resp. ymax=μ122>0) for Y± (resp. X±). The square roots are chosen to be defined as holomorphic functions on C\(,0) and take nonnegative values on the nonnegative reals.

Lemma 1.

Let u,vR such that u+iv[xmax,+[. Then, we have

Re(Y±(u+iv))=uμ2±12μ222u+(μ222u)2+4v2(3.4)

If u,vR satisfy u+iv[ymax,+[, then

Re(X±(u+iv))=uμ1±12μ122u+(μ122u)2+4v2.(3.5)

Let δ=min(μ1,μ2)>0. Then, Re(Y(x))<0 for all x such that Re(x)<xmax+δ,x[xmax,+[. Similarly, Re(X(y))<0 for all y such that Re(y)<ymax+δ,y[ymax,+[.

Proof.

Equations (3.4) and (3.5) follow directly from the expression (1.8) of γ. The last statements come from the inequalities xmax=μ222<μ2 and ymax=μ122<μ1. □

Corollary 1

(Continuation of Laplace Transforms). The Laplace transforms φ1 and φ2 can be extended as meromorphic functions on {yC,Re(y)<ymax+δ}\[ymax,ymax+δ] and {xC,Re(x)<xmax+δ}\[xmax,xmax+δ], respectively via the formulae

φ1(y)=γ2(X(y),y)φ2(X(y))exp(a0X(y)+b0y)γ1(X(y),y)(3.6)
φ2(x)=γ1(x,Y(x))φ1(Y(x))exp(a0x+b0Y(x))γ2(x,Y(x)).(3.7)

Proof.

This follows directly from Lemma 1 and the functional Equation (3.1). □

From now on, φ1 and φ2 will be considered over their extended domains. Let us define

x*=2μ2r2μ1(1+r2)2,y**=2μ1r1μ2(r1+1)2.(3.8)

If equation γ2(x,Y(x))=0 (resp. γ1(X(y),y)=0) has a solution in the complex plane, then it is unique and is given by x=x* (resp. y=y**). We also define

y*=Y+(x*),x**=X+(y**),(3.9)
see Figure 6.

Figure 6. In the Case of the Figure, Both φ1 and φ2 Have a Pole
Proposition 4

(Poles of Laplace Transform). The Laplace transforms φ1 and φ2 satisfy the following properties:

  • (i) x=0 (resp. y=0) is not a pole of φ2(x) (resp. φ1(y)).

  • (ii) If x (resp. y) is a pole of φ2(x) (resp. φ1(y)), then x=x* (resp. y=y**) and γ1(x*,Y(x*))=0 (resp. γ2(X(y**),y**)=0). Furthermore, x* is a pole of φ2 (resp. y** is a pole of φ1) if and only if (r2+1)μ2>2(resp.(r1+1)μ1>2).

Proof.

The first point follows from the continuation Equation (3.7) because γ2(0,Y(0))=2μ20.

For (ii), if x is a pole of φ2, then it follows from (3.7) that γ2(x,Y(x))=0, which implies that x=x*. Moreover, the Laplace transform φ2 is holomorphic in Re(x)<0. Thus x*, being a pole of φ2, must be positive. Note that equation γ2(x,Y(x))=0 has a positive solution if and only if γ2(xmax,Y±(xmax))>0. This last condition is equivalent to (r2+1)μ2>2.

Let us assume that (r2+1)μ2>2. Then, x*>0. Because γ2(x*,Y(x*))=0, it follows from (3.7) that x* is a pole of φ2 if the numerator of the right-hand side of (3.6) does not vanish at x*. The last fact holds true and is actually equivalent to the nonnullity of the function hα*(z0) defined in (3.24); this equivalence and the non-nullity are postponed till the end of Section 6.2. □

The following proposition provides some estimates for the Laplace transforms. These estimates will be useful in Section 5.

Proposition 5

(Decay of Laplace Transforms on Re=ϵ). Let z0=(a0,b0)R+2 be an initial condition with a00,b00 and ϵ>0. Then, there exist constants c,C>0 such that for l=1,2,

vR,  |φlz0(ϵ+iv)|Cec|v|.(3.10)

Proof.

The expressions γ1(ϵ+iv,Y(ϵ+iv)) and γ2(ϵ+iv,Y(ϵ+iv)) grow linearly with respect to v as v tends to ±. Furthermore, expression (3.4) provides inequality exp(b0Y(ϵ+iv))C1eb0c1|v| for some constants c1,C1>0 and

φ1z0(Y(ϵ+iv))φ1z0(0)e(a0+b0)Re(Y(ϵ+iv))C2ec2|v|
for some constants c2,C2>0. Finally, Equation (3.7) implies the conclusion for φ2. The proof for φ1 is analogous. □

We also give some further estimates for Laplace transforms that will be useful in Section 4.3.

Lemma 2

(Decay of Laplace Transforms). Assume that (1.2) to (1.4) hold. For any initial point z0=(a0,b0) and p0,

φ2z0(p)ep(a0+b0)φ2z0(0).(3.11)

The symmetric result holds for φ1z0.

Proof.

By (1.3), note that the support of the measure H2((a0,b0),·) is [a0+b0,+). Then,

φ2z0(p)=a0+b0+epxf2z0(x)dxep(a0+b0)a0+b0+f2z0(x)dx=ep(a0+b0)φ2z0(0).

4. The Compensation Method and the Explicit Expressions of the Laplace Transforms

4.1. Heuristic of the Compensation Method

Let h be a smooth function satisfying the following partial differential equation with boundary conditions,

{(H0)Gh=0on(0,+)2(H1)R1h(0,y)=0,y0(H2)R2h(x,0)=0,x0(4.1)
with G=12·Σ+μ·, and then h is harmonic (see Ernst and Franceschi 2021, section 6). To demonstrate this, one may apply Itô’s formula to the process (Zt)t0 and hC2(R+2,R):
h(Zt)=h(Z0)+0th(Zs)dBs+0tGh(s)ds+i=120tRi·h(Zs)dLsi,
where (Bt)t0=(Xtμt)t0 is the non-reflected degenerate Brownian motion associated with the process (see Definition 1). If h satisfies (4.1), then h(Zt)=h(Z0)+0th(Zs)dBs, and thus E[h(Zt)]=E[h(Z0)] (at least formally), which implies that h is harmonic (cf. (1.18)).

The principle of the compensation method is to find functions of the form h(x,y)=nZcneanx+bny such that each exponential term satisfies condition (H0): Geanx+bny=0 (i.e., (an,bn)P, see Figure 4) and to “compensate” the constants (cn)nZ so as to ensure that conditions (H1) and (H2) are satisfied. We require that

h(x,y)=+c2ea2x+b2y+c1ea1x+b1y(H2)+c0ea0x+b0y+c1ea1x+b1y(H2)(H1)+c2ea2x+b2y(H1)+(4.2)

Given that conditions (H0),(H1),(H2) are linear, it follows that h is a harmonic function. By a direct computation, we find that conditions (H2) on the right-hand side of (4.2) are satisfied if and only if a2k=a2k+1 and c2n+1=γ2(a2n,b2n)γ2(a2n+1,b2n+1)c2n for any integer k. Similarly, conditions (H1) in the right-hand side of (4.2) are satisfied if and only if b2n+1=b2n+2 and c2n+2=γ1(a2n+1,b2n+1)γ1(a2n+2,b2n+2)c2n+1 for any integer n.

We will see in Section 6.1 that the harmonic functions we obtain can be written as

(x,y)γ1(a0,b0)φ1(x,y)(b0)+γ2(a0,b0)φ2(x,y)(a0)+ea0x+b0y.

The explicit expressions of φ1 and φ2 in Section 4.3 then provide the exact Equation (4.2) suggested by the compensation method. Moreover, the approach of Section 6.1 justifies why the harmonic functions given by (4.2) are nonnegative when (a0,b0) is well chosen.

4.2. Parabola and Automorphisms

Let us recall that P is the parabola defined by P={(x,y)R2,γ(x,y)=0} (see (1.10)). Before defining the sequence ((an,bn))nZ motivated by Section 4.1 (see Figure 4), we first give a parametrization of P.

Proposition 6

(Parameterization of P). The parabola P (see (1.10)) admits the following parameterization:

{x(s)=12s(s2μ2)y(s)=12s(s+2μ1),sR.(4.3)

This means that {(x,y)R2,γ(x,y)=0}={(x(s),y(s)),sR}.

Proof.

The relation γ(x(s),y(s))=0 is easily verified by substituting x(s),y(s) into the expression (1.8) of γ(x,y). Furthermore, the parameterization is injective. To show this, assume that

{s(s2μ2)=s(s2μ2)s(s+2μ1)=s(s+2μ1).

Subtracting the second equation from the first gives 2s(μ1+μ2)=2s(μ1+μ2), which implies s=s. Similarly, surjectivity can be verified by elementary considerations. □

To define the “downstairs” as in Figure 4, we introduce two transformations on the parabola that leave the first (resp. second) coordinate invariant. This is the aim of the following proposition (which also serves as a definition). This proposition is illustrated by Figure 7.

Figure 7. Parabola P and Automorphisms η and ζ
Proposition 7

(Automorphisms η,ζ). For sR, we define

{ζs=s+2μ2ηs=s2μ1.(4.4)

Then, x(ζs)=x(s) and y(ηs)=y(s) for all sR. Therefore, φ2(x(ζs))=φ2(x(s)) and φ1(y(ηs))=φ1(y(s)) in their respective domains of definition. Furthermore, for all nZ and sR, we have

(ηζ)ns=s2n.(4.5)

Proof.

The formulae x(s+2μ2)=x(s) and y(s2μ1)=y(s) are easily verified. The expression of (ηζ)n is a consequence of expressions of η,ζ and of the equation μ1+μ2=1 (see Assumption (1.4)). □

Note that ζ2=Id, η2=Id. By the parameterization (4.3), ζ and η can be regarded as reflections (see (4.4)) and their composition as a translation (see (4.5)).

Lemma 3

(Explicit Form of (an,bn)). Let sR and (a0,b0)=(x(s),y(s)). For any integer nZ, we define

(a2n,b2n)=(x((ηζ)ns),y((ηζ)ns)),(a2n+1,b2n+1)=(x(ζ(ηζ)ns),y(ζ(ηζ)ns))

(see Figure 4). Then, for any nZ, the following expressions hold:

a2n=2n2+2(a0b0μ2)n+a0,a2n+1=a2n(4.6)
b2n=2n2+2(a0b0+μ1)n+b0,b2n+1=b2n+2.(4.7)

Proof.

The invariance of the first and the second coordinate of ζ and η, respectively, implies equalities a2n+1=a2n and b2n+1=b2n+2. The explicit expressions of a2n and b2n are obtained from the explicit expression (4.5). □

Notations 1.

For sR, we define

z(s)=(x(s),y(s))(4.8)
as the point of the parabola corresponding to the parameter sR. We also define
smax=μ2,smin=μ1(4.9)
and write γi(s) instead of γi(x(s),y(s)) for i=1,2. Finally, let s* and s** be defined as
s*=2r2+1,s**=2r1+1.(4.10)

With these notations,

z(smax)=(xmax,Y±(xmax))andz(smin)=(X±(ymax),ymax).

Note that the curve (z(s))s[smin,smax] is the portion of the parabola from (xmax,Y+(xmax)) to (X+(ymax),ymax) going counterclockwise (see Figure 6). Furthermore, x(s*)=x* and y(s**)=y** with definition (3.8). We can now provide explicit expressions for the Laplace transforms φ1 and φ2.

4.3. Explicit Expression of Laplace Transforms via the Compensation Approach

Theorem 5

(Explicit Expressions for Laplace Transforms). Let z0R+2{(0,0)} be the initial condition. Then, for any s(max(smin,s**),min(smax,s*)),

φ2z0(x(s))=1γ2(ζs)ez0·z(ζs)+n=1+[k=0n1G(s2k)][ez0·z(s2n)γ2(s2n)ez0·z(ζ(s2n))γ2(ζ(s2n))](4.11)
where
G(s)=γ1γ2(ζs)γ1γ2(s2).(4.12)

Similarly, for all s(max(smin,s**),min(smax,s*))

φ1z0(y(s))=1γ1(ηs)ez0·z(ηs)+n=1+[k=0n1G˜(s+2k)][ez0·z(s+2n)γ1(s+2n)ez0·z(η(s+2n))γ1(η(s+2n))](4.13)
where
G˜(s)=γ2γ1(ηs)γ2γ1(s+2).

Before proving Theorem 5, we establish a technical lemma.

Lemma 4.

For all n1 and s(max(smin,s**),min(smax,s*)), we have γ1(s2n)0 and γ2(ζ(s2n))0. Furthermore, γ2(s),γ2(ζs),γ1(s),γ1(ηs) are also nonzero.

Proof.

We define two portions of the parabola E+ and E given by

E+={(x,Y+(x)),xX+(0)}andE={(X(y)),y),yY(0)}.

By Assumption (1.3), the line {γ2=0} (resp. {γ1=0}) cannot pass through E (resp. E+). Additionally, note that η(E)E+ and ζ(E+)E. Becauses(smin,smax), z((ηζ)ns)=z(s2n) belongs to E+ for all n1. Thus, γ2(ζ(s2n))0 for any n0. By similar reasoning, γ1(s2n)0 for any n0. The last statement comes from the fact that s(s**,s*). □

Proof of Theorem 5.

The main idea of the proof is to get a recursive formula for Laplace transforms. To do this, we rewrite the functional Equation (3.1) in z(ζs) and z(ηζs)=z(s2), which holds because x(ζs),y(ζs),x(s2) and y(s2) are negative:

{0=γ1(ζs)φ1(y(ζs))+γ2(ζs)φ2(x(ζs))+ez0·z(ζs)0=γ1(s2)φ1(y(s2))+γ2(s2)φ2(x(s2))+ez0·z(s2).

By the invariance of φ2 (resp. φ1) under ζ (resp. η), we have φ2(x(ζs))=φ2(x(s)) and φ1(y(s2))=φ1(y(ζs)). Then, by eliminating φ1(y(ζs)) from the equations (which is possible by Lemma 4), we obtain

φ2(x(s))=γ1γ2(ζs)γ1γ2(s2)φ2(x(s2))+[γ1γ2(ζs)ez(s2)·z0γ1(s2)ez(ζs)·z0γ2(ζs)](4.14)
=G(s)φ2(x(s2))+[G(s)γ2(s2)ez0·z(s2)ez0·z(ζs)γ2(ζs)].(4.15)

Similarly, we get

φ2(x(s2))=G(s2)φ2(x(s4))+[G(s2)γ2(s4)ez0·z(s4)ez0·z(ζ(s2))γ2(ζ(s2))].(4.16)

Substituting this into (4.15), we get

φ2(x(s))=G(s)G(s2)φ2(x(s4))+G(s)G(s2)ez0·z(s4)γ2(s4)G(s)ez0·z(ζ(s2))γ2(ζ(s2))+G(s)γ2(s2)ez0·z(s2)ez0·z(ζs)γ2(ζs).

Then, by induction on N, we obtain the following equality for all N1:

φ2(x(s))=[k=0NG(s2k)]φ2(x(s2(N+1)))ez0·z(ζs)γ2(ζs)+[k=0NG(s2k)]ez0·z(s2(N+1))γ2(s2(N+1))+n=1N[k=0n1G(s2k)][ez0·z(s2n)γ2(s2n)ez0·z(ζ(s2n))γ2(ζ(s2n))](4.17)

The proof is then reduced to proving the following limit:

[k=0nG(s2k)]φ2(x(s2(n+1)))n+0.(4.18)

To justify this, note using Equation (4.12) and Lemma 3 that

G(s2k)=(k+a)(k+b)(k+c)(k+d).(4.19)
for some constants a, b, c, and d defined by
a=s2+r11+r1,b=1s2+μ2r2μ11+r2c=s2+11+r2,d=1s2+μ2μ1r11+r1.

By elementary considerations, the following asymptotic behavior holds:

[k=0nG(s+2k)]nCnac+bd,(4.20)
where C is a real constant. Moreover,
a+bcd=22(11+r1+11+r2)(4.21)
because μ1+μ2=1. Then, the exponential decay in (3.11) for φ2, together with the polynomial rate of expression (4.20), yields (4.11). Note that inequality (3.11) is the only (and crucial) reason why we work under Assumption (1.3). Equation (4.13) is obtained with symmetric arguments. □

Remark 2.

The exponent given by (4.20) is exactly the parameter 2γ introduced in Dreyfus et al. (2025), which determines the algebraic nature of the Laplace transforms for the same degenerate particle model in the recurrent case. Furthermore, the constants κm=κm(α) in (1.22) satisfy

κmm±C±m2γ2.(4.22)
for some constant C±>0, where 2γ2<0 by (1.3).

In (4.11) (resp. (4.13)), φ2 (resp. φ1) is not given as a function of x (resp. y) but of s. Therefore, we establish the following corollary.

Corollary 2.

The following expressions hold in the domains Re(x)<xmax and Re(y)<ymax, respectively:

φ2(x)=1γ2(x,Y(x))ez0·(x,Y(x))+n=1+[k=1nγ1γ2(ψ2k1(x,Y+(x))γ1γ2(ψ2k(x,Y+(x)))][ez0·ψ2n(x,Y+(x))γ2(ψ2n(x,Y+(x))ez0·ψ2n+1(x,Y+(x))γ2(ψ2n+1(x,Y+(x))](4.23)
φ1(y)=1γ1(X+(y),y)ez0·(X+(y),y)+n=1+[k=1nγ2γ1(ψ2k+1(X+(y),y))γ2γ1(ψ2k(X+(y),y))][ez0·ψ2n(X+(y),y))γ1(ψ2n(X+(y),y))ez0·ψ2n+1(X+(y),y)γ1(ψ2n+1(X+(y),y))],(4.24)
where
ψ2n(a,b)=(2n2+2(abμ2)n+a,2n2+2(ab+μ1)n+b)
and
ψ2n+1(a,b)=(2n2+2(abμ2)n+a,2(n+1)2+2(ab+μ1)(n+1)+b).

Proof.

By Lemma 3 and equalities z(s)=(x(s),Y+(x(s))=(X+(y(s)),y(s)) for s(smin,smax), Equations (4.23) and (4.24) hold on the curve {(x,y)=(x(s),y(s)):s((max(smin,s**),min(smax,s*)))}. By Corollary 1, Laplace transforms φ2(x) and φ1(y) are meromorphic on Re(x)<xmax and Re(y)<ymax, respectively. Consequently, the explicit expressions (4.23) and (4.24) remain valid in these domains. □

5. Laplace Inverse and Saddle Point Method

To avoid certain technical complications, we first derive the asymptotic behavior of the Green functions gz0 for z00 and later address the case z0=0 with additional arguments.

5.1. Inverse Laplace Theorem and Saddle Point

Let z0(0,0) be a starting point of the process. The inverse Laplace transform formula (see (Doetsch 1974 (theorems 24.3 and 24.4) and Brychkov 1992) yields the following representation for gz0(a,b): for ϵ>0 sufficiently small,

gz0(a,b)=1(2πi)2ϵiϵ+iϵiϵ+iφz0(x,y)exp(axby)dxdy,(5.1)
where the convergence is in the sense of principal value. This can be justified by the functional Equation (1) and the decay properties of the Laplace transforms established in Proposition 5.

Lemma 5

(From Double to Simple Integrals). Denote by z0=(a0,b0) the starting point of the process. Then, for any (a,b)R+2 satisfying a>0 or b>0,

g(a,b)=I1(a,b)+I2(a,b)+I3(a,b),
where
I1(a,b)=12πiϵiϵ+iφ2(x)γ2(x,Y+(x))exp(axbY+(x))dxyγ(x,Y+(x)),I2(a,b)=12πiϵiϵ+iφ1(y)γ1(X+(y),y)exp(aX+(y)by)dyxγ(X+(y),y),I3(a,b)=12πiϵiϵ+iexp(a0x+b0Y+(x))exp(axbY+(x))dxyγ(x,Y+(x)) if b>b0,I3(a,b)=12πiϵiϵ+iexp(a0X+(y)+b0y)exp(aX+(y)by)dyxγ(X+(y),y) if a>a0.

Proof.

By the functional Equation (3.1), φ(x,y) can be decomposed as

φ(x,y)=γ1(x,y)φ1(y)γ(x,y)γ2(x,y)φ2(x)γ(x,y)e(x,y)·z0γ(x,y).(5.2)

Substituting this expression into the double integral (5.1), gz0(a,b) is written as the sum of three double integrals. Let us consider the first term, given by

1(2πi)2ϵiϵ+iφ2(x)ϵiϵ+iγ2(x,y)γ(x,y)eaxbydydx.(5.3)

Let CR be the closed oriented contour defined by

CR={ϵ+it|t[R,R]}{ϵ+Reiθ|θ[π/2,π/2]}.

By applying the residue theorem along the contour CR and considering the asymptotics as R+ (see Franceschi et al. 2024b (lemma 4.1) for more details), we obtain the identity

ϵiϵ+iγ2(x,y)γ(x,y)eaxbydy=γ2(x,Y+(x))yγ(x,Y+(x))eaxbY+(x),
so that expression (5.3) equals I1(a,b). The remaining terms are handled analogously. □

To find the asymptotics of these integrals as a,b+, we use the saddle point method. For any α[0,π/2], let (x(α),y(α)) be defined as

(x(α),y(α))=arg max(x,y)P(cos(α)x+sin(α)y),(5.4)
see Figure 3(a). For α[0,π/2], we define the real number s(α)R by
s(α)=arg maxsR(cos(α)x(s)+sin(α)y(s)).

Note that (x(α),y(α))=(x(s(α)),y(s(α))), using notation (4.3). By studying the variations of the function sx(s)cos(α)+y(s)sin(α), we prove that

s:{[0,π/2][smin,smax]αμ2tan(α)μ11+tan(α)(withs(π/2)=μ1=smin)(5.5)
is a C diffeomorphism, and
s1:{[smin,smax][0,π/2]sarctan(μ2ss+μ1)(withs1(μ1)=π/2).(5.6)

Using the definitions of α*,α**,x*, and y** given by (1.13), (1.14), and (3.8), if x* (resp. y*) is a pole of φ2 (resp. φ1), then x(α*)=x* (resp. y(α**)=y**). Because s**<0<s* (see Notation 1), then the monotonicity of (5.6) implies that 0α*<αμ<α**π/2, where αμ=arctan(μ2/μ1)(0,π/2) is the angle of the drift. We follow the notation of Franceschi et al. (2024b) and define

F(x,α)=cos(α)xsin(α)Y+(x)+cos(α)x(α)+sin(α)y(α)(5.7)
G(y,α)=cos(α)X+(y)sin(α)y+cos(α)x(α)+sin(α)y(α).(5.8)

By construction, the equations xF(x(α),α)=0 and yG(y(α),α)=0 hold. Then, by differentiating Equation (5.7) and γ(x,Y+(x))=0, we get for any α(0,π/2]:

(Y+(x))|x=x(α)=xγ(x(α),y(α))yγ(x(α),y(α))=cos(α)sin(α),(Y+(x))|x=x(α)=(1+ctan(α))2yγ(x(α),y(α)).(5.9)

Therefore,

xx2F(x(α),α)=(sin(α)+cos(α))2yγ(x(α),y(α))sin(α)>0,α(0,π/2].(5.10)

Similarly,

yy2G(y(α),α)=(sin(α)+cos(α))2xγ(x(α),y(α))cos(α)>0,α[0,π/2).

5.2. Contour of Steepest Descent

Let α0(0,π/2]. The key idea of the saddle point method is to use the parameterized Morse lemma. Because xx2F(x(α),α)>0, lemma A.1 from Franceschi et al. (2024b) yields some ϵ>0,η>0 and a family of smooth paths Γx,α={x(it,α)|t[ϵ,ϵ]},|αα0|<η such that

t[ϵ,ϵ],F(x(it,α),α)=t2.(5.11)

For further details on the construction, please refer to appendix A in Franceschi et al. 2024b. Define

xα+=x(iϵ,α),xα=x(iϵ,α).(5.12)

In particular,

F(xα+,α)=ϵ2,F(xα,α)=ϵ2.(5.13)

Furthermore, Im(xα+)>0 and Im(xα)<0 (see Figure 8 and construction in Franceschi et al. 2024b). The same construction holds for Γy,α={y(it,α)|t[ϵ,ϵ]} for G and α0[0,π/2). These paths satisfy

Γx,α=X+(Γy,α)andΓy,α=Y+(Γx,α),0<α<π/2.

Figure 8. Changing Path for I2
Note. Here, x(α)<x*.

The arrows above and below the paths indicate reversed orientations; this notation is taken from (Fayolle et al. 2017, chapter 5.3).

5.3. Shift of the Integration Contours and Contribution of the Poles

We now apply the saddle point method. To do this, we shift the integration contours of I1, I2, and I3 to contours passing through the saddle point and following the steepest descent contours Γx,α and Γy,α. We define Tx,α=Sx,α+Γx,α+Sx,α+ and Ty,α=Sy,α+Γy,α+Sy,α+ for α[0,π/2], where

Sx,α+={xα++it|t0},Sx,α={xαit|t0},Sy,α+={yα++it|t0},Sy,α={yαit|t0}.

Lemma 6

(Contour Deformation and Contribution of the Pole).

Let α[0,π/2]\{α*,α**} and z0(0,0) be the initial condition of the process. Then, for any a,b>0,

I1(a,b)=(resx=x*φ2(x))γ2(x*,y*)yγ(x*,y*)exp(ax*by*)1α<α*+12πiTx,αφ2(x)γ2(x,Y+(x))yγ(x,Y+(x))exp(axbY+(x))dx,(5.14)
I2(a,b)=(resy=y**φ1(y))γ1(x**,y**)xγ(x**,y**)exp(ax**by**)1α>α**+12πiTy,αφ1(y)γ1(X+(y),y)xγ(X+(y),y)exp(aX+(y)by)dy,(5.15)
I3(a,b)=12πiTx,αexp((a0a)x+(b0b)Y+(x))dxyγ(x,Y+(x)) if b>b0,(5.16)
I3(a,b)=12πiTy,αexp((a0a)X+(y)+(b0b)y)dyxγ(X+(y),y) if a>a0.(5.17)

Proof.

The shift of the path is illustrated in Figure 8 and is the same as in (Franceschi et al. 2024b, lemma 6.1). The proof of (5.14) is a direct consequence of the residue theorem, provided that the integrals over the horizontal contours Rv+ and Rv tend to 0 as v tends to +. Then, it remains to be proven that for any sufficiently small η>0,

supu[X+(ymax)η,xmax+η]|φ2(u+iv)γ2(u+iv,Y+(u+iv))γy(u+iv,Y+(u+iv))exp(a(u+iv)bY+(u+iv))|0, as v.

By the functional Equation (3.1) and continuation Equation (3.7), the term inside the supremum is equal to

|(γ1(u+iv,Y(u+iv))φ1(Y(u+iv))+ea0(u+iv)+b0Y(u+iv))γ2(u+iv,Y+(u+iv))γ2(u+iv,Y(u+iv))γy(u+iv,Y+(u+iv))|×|exp(a(u+iv)bY+(u+iv))|.

By (3.3), Re(Y±(u+iv)) grows like ±|v| uniformly in u[X+(ymax)η,xmax+η] as |v|+. Furthermore, γ2(u+iv,Y±(u+iv)) grows linearly in v uniformly in u[X+(ymax)η,xmax+η] as v+ by Assumption (1.3). The same asymptotics hold for γ1(u+iv,Y(u+iv)). Moreover, yγ(u+iv,Y+(u+iv))=2(u+iv)+μ22, so this expression grows with rate v, uniformly in u[X+(ymax)η,xmax+η]. Considering the exponential decay of φ1 (see Lemma 2) we get the conclusion for I1. Formulae for I2 and I3 are obtained similarly. □

5.4. Negligibility of Some Integrals

For any pair (a,b)R+2 let α(a,b) be the angle in [0,π/2] such that cos(α)=aa2+b2 and sin(α)=ba2+b2. We now aim to evaluate the asymptotics of the integrals over Tx,α± and Ty,α± in Lemma 6 as a2+b2+ and α(a,b)α0 for some α0[0,π/2]. In the next lemma, we establish exponential bounds for the integrals over the vertical contours Sx,α±, Sy,α±. These bounds imply that the main contribution to the above asymptotics comes from the integrals over the steepest descent contours Γx,α, Γy,α, whereas those over Sx,α± and Sy,α± turn out to be negligible.

Lemma 7

(Negligibility of the Integrals Over Sx,α± and Sy,α±). Suppose z0(0,0). Let K be a compact neighborhood of z0 in the quadrant satisfying d((0,0),K)>0. Let α0[0,π/2]. Then, for sufficiently small η>0, there exist constants r0>0 and Dα0>0 such that for any zK and any pair (a,b) satisfying a2+b2>r0 and |α(a,b)α0|<η, the following inequalities hold:

|Sx,α±φ2z(x)γ2(x,Y+(x))yγ(x,Y+(x))exp(axbY+(x))dx|Dα0eax(α)by(α)ϵa2+b2,(5.18)
|Sy,α±φ1z(y)γ1(X+(y),y)xγ(X+(y),y)exp(aX+(y)by)dy|Dα0eax(α)by(α)ϵa2+b2.(5.19)

If b>b0, then

|Sx,α±exp((a0a)x+(b0b)Y+(x))dxyγ(x,Y+(x))|Dα0bb0eax(α)by(α)ϵ2a2+(bb0)2.(5.20)

If a>a0, then

|Sy,α±exp((a0a)X+(y)+(b0b)y)dyxγ(X+(y),y)|Dα0aa0eax(α)by(α)ϵ2(aa0)2+b2.(5.21)

Proof.

We start by showing (5.18). Using notations (5.7) and (5.12), this inequality can be rewritten as

|v>0φ2z(xα++iv)γ2(xα++iv,Y+(xα++iv))yγ(xα++iv,Y+(xα++iv))eaivb(Y+(xα++iv)Y+(xα+))dx|Dα0(5.22)
where α=α(a,b).

Suppose first that α0>0. Let α>0 and 0<η<α0/2. Because yγ(xα++iv,Y+(xα++iv))=2(xα++iv)+μ22, this expression does not vanish and grows at rate |v| as v+, uniformly in α, with |αα0|<η. Similarly, γ2(xα++iv,Y+(xα++iv)) grows with speed |v| as v+, uniformly in α, |αα0|<η. Then, we have, for all v0,

sup|αα0|<ηγ2(xα++iv,Y+(xα++iv))yγ(xα++iv,Y+(xα++iv))Cη(1+v)
for some constant Cη>0. If |αα0|<η, then by (3.4), there exists a constant Cη>0 such that
Re(a2+b2(F(xα++iv,α)F(xα+,α)))=b(Re(Y+(xα++iv))Re(Y+(xα+)))Cηbv(5.23)
for any v1. Furthermore, using the continuation Equation (3.7), the estimates (3.11), and the continuity of φ2z0(0) in z0 (see (4.23)), there exists a constant D such that |φ2z(xα++iv)|D for all v0, zK and |αα0|<η. Then, the left-hand side of (5.22) is bounded by
DCηCη(2+v>1(1+v)ebvdv)=DCηCη(2+1b2+4b3)Dα0
for some constant Dα0>0 because b+ (because α0>0). This inequality implies (5.18).

Now suppose that α0=0. We no longer use estimate (5.23) because it would produce terms of order 1b, and here b may be close to zero. Let z=(a1,b1)K. We write continuation Equation (3.7) for φ2z(xα++iv), which splits into two terms,

φ2z(xα++iv)=γ1(xα++iv,Y(xα++iv))φ1z(Y(xα++iv))γ2(xα++iv,Y(u+iv))ea1(xα++iv)+b1Y(xα++iv)γ2(xα++iv,Y(u+iv)),(5.24)
and we substitute into the right-hand side of (5.22). Then, the integral (5.22) can be written as the sum of two terms. For the first term, note that there are some constants, c,C0>0 independent on α[0,η] and zK, such that
|γ2(xα++iv,Y+(xα++iv))γ1(xα++iv,Y(xα++iv))φ1z(Y(xα++iv))γ2(xα++iv,Y(u+iv))yγ(xα++iv,Y+(xα++iv))|(5.25)
×eaivb(Y+(xα++iv)Y+(xα+))C0(v+1)e(a1+b1)Re(Y(xα+iv))φ1z(0)C0(v+1)e(a1+b1)cvφ1z(0).
for any v0. We recall that function zφ1z(0) is continuous and therefore locally bounded. The integral of (5.25) over v>0 can then be bounded by a positive constant that is (locally) independent of z and of 0αη. The second term is given by
0+γ2(xα++iv,Y+(xα++iv))ea1(xα++iv)+b1Y(xα++iv)γ2(xα++iv,Y(u+iv))yγ(xα++iv,Y+(xα++iv))eaivb(Y+(xα++iv)Y+(xα+))dv.(5.26)

Note that if b1=0, then the quotient in the integrand is of order O(1/v) as v+. Moreover, it suffices to bound the integral over (v0,+) for some v0>0 because the integrand is uniformly bounded with respect to α[0,η] and zK. By integration by parts, the integral over (v0,+) equals

γ2(xα++iv0,Y+(xα++iv0))ea1(xα++iv0)+b1Y(xα++iv0)eaiv0b(Y+(xα++iv0)Y+(xα))γ2(xα++iv,Y(xα++iv0))yγ(xα++iv0,Y+(xα++iv0))(aibddv(Y+(xα++iv))v=v0)v0ddv(γ2(xα++iv,Y+(xα++iv))ea1(xα++iv)+b1Y(xα++iv)γ2(xα++iv,Y(xα++iv))yγ(xα++iv,Y+(xα++iv))(aibddv(Y+(xα++iv)))×exp(aivb(Y+(xα++iv)Y+(xα+)))dv.(5.27)

Furthermore, ddv(Y+(xα++iv))=i(11μ222(xα+iv)) and Re(11μ222(xα+iv))1/2 for all vv0 with v0 large enough and 0<α<η. With some calculations, the integrand of (5.27) is of order O(1/v3/2) as v+. Hence, the integral in (5.27) is bounded by a positive constant independent of α and of zK. This establishes the bound in (5.18). Inequalities (5.19), (5.20), and (5.21) are obtained similarly. □

6. Proof of Theorem 1

In Section 6.1, we establish the asymptotics stated in Theorem 1. In Section 6.2, we show that all the constants hα(z0) appearing in the asymptotics of Theorem 1 are nonzero, which completes the proof of the theorem.

6.1. Asymptotics in Theorem 1

We now have the tools to derive the asymptotics stated in Theorem 1, where hα0(z0) is given by (1..22), hα*(z0) by (1..24) (with the symmetric formula for hα**(z0)), and

c*=γ2(x*,y*)yγ(x*,y*)x(s*)γ2(x(ζs*),y(ζs*)),c**=γ1(x**,y**)xγ(x**,y**)y(s**)γ1(x(ηs**),y(ηs**)),(6.1)
where x(s) and y(s) are the derivatives of x(s) and y(s) (see (4.3) for the definition of x(s) and y(s), (4.10) for s* and s**, (3.8) and (3.9) for x*,x**,y*, and y**, and (4.4) for ηs and ζs).

Proof of the Asymptotics in Theorem 1 When z0(0,0).

We use the identity g(a,b)=I1(a,b)+I2(a,b)+I3(a,b), using the expressions provided in Lemma 6. By the classical saddle point method (see details in (Franceschi et al. 2024b, lemma 8.1), the sum of the integrals of Lemma 6 along Γα,x and Γy,α=Y+(Γx,α) has the following asymptotic expansion:

12πiΓx,αφ2(x)γ2(x,Y+(x))yγ(x,Y+(x))eaxbY+(x)dx+12πiΓy,αφ1(y)γ1(X+(y),y)xγ(X+(y),y)eaX+(y)bydy+12πiΓy,αe(a0a)X+(y)+(b0b)ydyxγ(X+(y),y)=rαα0er(cos(α)x(α)+sin(α)y(α))(1rk=0nckz0(α)rk+o(1rnr)).(6.2)
where (a,b)=(rcos(α),rsin(α)), and
c0z0(α)=12π(cos(α)+sin(α))2sin(α)yγ(x(α),y(α))(6.3)
×(γ1(x(α),y(α))φ1(y(α))+γ2(x(α),y(α))φ2(x(α))+e(x(α),y(α))·z0)(6.4)
=12π(cos(α)+sin(α))hα(z0)(6.5)
by the explicit expressions of φ1(y(s)) and φ2(x(s)) given in (4.1) and (4.13), evaluated at s=s(α) (see (5.5)).

Lemma 7 shows that, when z0(0,0), integrals over Sx±,Sy± are negligible compared with those over paths of steepest descent. Finally, Theorem 5 gives the explicit form of residues of Lemma 6 providing hα*(z0), hα**(z0). □

For the case z0=(0,0), we establish two preliminary lemmas. The first one is a consequence of the general Martin boundary theory.

Lemma 8.

For α[α*,α**], zhα(z) is harmonic on R+2\{(0,0)}.

Proof.

For z0=(a0,b0)(0,0), we may consider the process evolving in R+2{(x,y),x+ya0+b0}. Because hα is the limit of the quotient of Green’s kernels, Kunita and Watanabe (1965) implies its harmonicity over all these domains and thus over R+2\{(0,0)}. □

Lemma 9.

Let Θ be the contour defined by Θ{zR+2:|z|=1} and TΘinf{t0,ZtΘ} the stopping time at Θ. Then, for all z0R+2 satisfying |z0|<1,

hα0(z0)=Θhα0(z)Pz0(ZTΘ=dz).(6.6)

Proof.

Suppose first that z0(0,0). The process (hα(Zt))t0 is a martingale; indeed, for t,s0,

Ez0[hα(Zt+s)|Ft]=EZt[hα(Zs)]=hα(Zt)
by the strong Markov property and the harmonicity of hα (see Lemma 8). Furthermore, under Pz0, the process (hα(ZtTΘ))t0 is bounded above by sup|z|1hα(z)< because hα is continuous. Then, by the optional stopping theorem for bounded martingales, we obtain hα0(z0)=Ez0[h(ZTΘ)], which is precisely the desired equality.

Now suppose z0=(0,0), and consider a sequence (zn)n1 in the quarter plane converging to (0,0) such that 0<|zn|<1. By continuity of hα, hα(zn) converges to hα(z0) as n goes to +. Because Equation (6.6) holds for all nonzero initial conditions, it suffices to show that

Θhα0(z)Pzn(ZTΘ=dz)n+Θhα0(z)P(0,0)(ZTΘ=dz).

By continuity and boundedness of hα on {zR+2,|z|1}), it is enough to show that LznZTΘn+Lz0ZTΘ weakly, where LzZTΘ denotes the law of ZTΘ with initial condition Z0=z. This follows from Assumption (1.2), combined with Harrison and Reiman (1981, theorem 1), which ensures the continuity of the mapping from the nonreflected to the reflected path under the topology of uniform convergence on compacts. □

We can now prove Theorem 1 in the case of z0=(0,0).

Proof of the Asymptotics in Theorem 1 for z0=(0,0).

By continuity of the process and by the Strong Markov property, if (a,b) lies at a distance >1 from (0,0), then

g(0,0)(a,b)=Θgz(a,b)P(ZTΘ=dz).(6.7)

Because the constant C from the saddle point method (Franceschi et al. 2024b, lemma 8.1) depends continuously on z0, and because the constants Dα0 in Lemma 7 are locally uniform in z0, then for any compact set K in the quadrant R+2 with d((0,0),K)>0, we have

supzK|gz(rcos(α),rsin(α))er(cos(α)x(α)+sin(α)y(α))1rk=0nckz(α)rk|(6.8)
=rαα0o(er(cos(α)x(α)+sin(α)y(α))rnr).

By this expansion,the asymptotics of (6.7) yield

g(0,0)(a,b)rαα0er(cos(α)x(α)+sin(α)y(α))rΘhα0(z)P(ZTΘ=dz).(6.9)

Lemma 9 combined with (6.9) gives the result. □

6.2. Positivity of hα(z0)

To make our asymptotics consistent, we prove here the positivity of the constants hα(z0).

Lemma 10.

Let α(α*,α**). Then, for every R>0, there exists z0 such that |z0|R and hα(z0)>0. If α*>0 (resp. α**<π/2), then the same result holds for hα* (resp. hα**).

Proof.

By the explicit Equation (1.22) for hα, Equation (1.24) for hα*, and its equivalent for hα**, the following asymptotics hold as r+ for α(α*,α**), and for α=α* (resp. α=α**), if α*>0 (resp. α**<π/2), then

hα(rcos(α),rsin(α))rer(x(α)cos(α)+y(α)sin(α)).(6.10)

The conclusion follows with z0=(rcos(α),rsin(α)) for r large enough. □

The following Lemma is inspired by (Franceschi et al. 2024b, lemma 8.3) and establishes the positivity of constants hα(z0) in the framework of Theorem 1.

Lemma 11

(Positivity of hα(z)). Let α(α*,α**) or α=α* (resp. α=α**) if α*>0 (resp. if α**<π/2). Let zR+2. Then, hα(z)0.

Proof.

Let α(α*,α**), and let z0 such that both coordinates are larger than those of z and such that hα(z0)>0; see Lemma 10. Let V be a compact neighborhood of z0, and denote by TVinf{t0:ZtV} the hitting time of V. By the hypothesis on z, Pz(TV<+)>0. By the strong Markov property,

gz(rcos(α),rsin(α))Pz(TV<+)infz0Vgz0(rcos(α),rsin(α))(6.11)
Pz(TV<+)infz0V(hα(z0)+εz0,α(r))er(cos(α)x(α)+sin(α)y(α))r.(6.12)
for r large enough where (6.8) provides the asymptotics
supz0V|εz0,α(r)|r0.

Furthermore, by continuity of hα, the set V can be chosen to satisfy infz0Vhα(z0)>0. On the other hand, we also have

gz(rcos(α),rsin(α))=er(cos(α)x(α)+sin(α)y(α))r(hα(z)+εz0,α(r))
where εz0,α(r)0 as r+. Therefore, comparing the two expressions, we conclude that hα(z)>0. If α=α* or α=α**, then the proof is analogous. □

This is the end of the proof of Proposition 4.

The remaining part of the proof is equivalent to showing that hα*(z0)>0; indeed, hα*(z0) is equal to resx=x*φ2(x) up to a nonnull multiplicative constant (see (1.24) and (4.11)). The positivity is established in the previous lemma. □

7. Asymptotics of Green’s Kernel in the Particular Directions 0,α*,α**, and π/2

In Section 7.1, we study the asymptotics of Green’s functions in the direction α=0 under the assumption that γ2(xmax,Y±(xmax))0. In Section 7.2, we provide these asymptotics in the direction α* if α*>0. Then, in Section 7.3, we analyze the limiting case where α*=0 and γ2(xmax,Y±(xmax))=0. The analysis of the directions α=π/2, α=α** if α*<π/2 and α**=π/2,γ1(X±(ymax,ymax))=0 is symmetrical. We then derive the proof of Theorem 2 in Section 7.4.

7.1. Case α0 If γ2(xmax,Y±(xmax))0

Before deriving the asymptotics, let us relate Green’s densities gz0 to f1z0,f2z0.

Proposition 8

(Link Between Densities). Let a,b0. Suppose z0(a,0) and z0(0,b). Then, we have

f1z0(b)=12gz0(0,b)andf2z0(a)=12gz0(a,0).

Proof.

By the functional Equation (3.1), if x,y<0, then

γ(x,y)xφ(x,y)=γ1(x,y)xφ1(y)+γ2(x,y)xφ2(x)+e(x,y)·z0x.(7.1)

Furthermore, by elementary properties of Laplace transforms,

xφ(x,y)x0+ebyg(0,b)db.

Then, letting x in (7.1), we get 0+ebyg(0,b)db=φ1(y)=0+ebyf1(b)db. The injectivity of the Laplace transform concludes the proof. The case of f2 is symmetrical. □

Lemma 12

(Asymptotics at α=0). Suppose that φ2 does not have a pole. Let κ=(1+μ22Γ(1/2))1, where Γ denotes the usual Gamma function. Then,

f2z0(x)x+κα[hα(z0)]α=0x3/2exmaxx.

Proof.

Note that Y(x)=Y(xmax)2(xmaxx)+o(xxmax) by (3.3). Then, using the continuation Equation (3.7), φ2 is continuous at xmax and

φ2(x)φ2(xmax)2(xmaxx)=r1φ1(Y(xmax))+γ1(xmax,Y±(xmax))φ1(Y±(xmax))+b0ez0·(xmax,Y±(xmax))+γ1(xmax,Y±(xmax))φ1(Y(xmax))+ez0·(xmax,Y±(xmax))γ2(xmax,Y±(xmax))+oxxmax(1)=r1φ1(Y(xmax))+γ1(xmax,Y±(xmax))φ1(Y±(xmax))+b0ez0·(xmax,Y±(xmax))+φ2(xmax)+oxxmax(1)A+oxxmax(1),
where z0=(a0,b0). Then, by the Tauberian theorem given by Dai and Miyazawa (2011, lemma C.2), we obtain
f2z0(x)x+AΓ(1/2)x3/2exmaxx.

It then remains to be shown that AΓ(1/2)=κα[hα(z0)]α=0. From Equation (5.6), we have

α[hα(z0)]α=0+=(s1)(smax)s[γ1(x(s),y(s))φ1(y(s))+γ2(x(s),y(s))φ2(x(s))+ea0x(s)+b0y(s)]s=smax=((μ1+μ2)(μ1+μ2)2)(r1y(smax)φ1(y(smax))+γ1(x(smax),y(smax))y(smax)φ1(y(s))+y(smax)φ2(x(smax))+y(smax)b0ea0x(smax)+b0y(smax))=1+μ22A
because y(smax)=y(μ2)=1+μ22 and x(smax)=0. The conclusion follows. □

We can now establish the asymptotics of Green’s functions as α0. We use the notation c0(α) and c1(α) for the constants in the first and second terms, respectively, in the asymptotic expansion of Green’s functions (cf. (6.8)).

Theorem 6

(Asymptotics with α0). Suppose γ2(xmax,Y±(xmax))0. Then,

hα(z0)α0αα[hα(z0)]α=0(7.2)
and thus c0z0(α)α012παα[hα(z0)]α=0. Furthermore, if α*=0, then
c1z0(α)α02κα[hα(z0)]α=0.(7.3)

Moreover,

  • If α*=0 (i.e. φ2 has no pole), then

    gz0(rcos(α),rsin(α))rα0α[hα(z0)]α=0er(cos(α)x(α)+sin(α)y(α))r(α2π+2κr).

  • If α*>0 (i.e. φ2 has a pole), then

    gz0(rcos(α),rsin(α))rα0c*hα*(z0)er(cos(α)x*+sin(α)y*),

where hα*(z0) and c* are given by (1.24) and (6.1), respectively.

Moreover, the constants α[hα(z0)]α=0 and hα*(z0) are nonzero in the corresponding asymptotics.

Proof.

First, (7.2) follows from the regularity of hα(z0) in α and from the convergence hα(z0)0 as α0 (see (1.22)). Now we analyze the asymptotics of the sum of the three integrals in (2) along the saddle point curves as α0. The integrands in the second and third terms are holomorphic in a neighborhood of the saddle point Y+(xmax). The integrand of the first term, namely φ2, has a branching point at xmax. For this reason, we perform the change of variables Γx,α=X+(Γy,α):

Γx,α(a,b)φ2(x)γ2(x,Y+(x))yγ(x,Y+(x))eaxbY+(x)dx=Γy,α(a,b)φ2(X+(y))γ2(X+(y),y)xγ(X+(y),y)eaX+(y)bydy.(7.4)

Additionally, from (3.7),

φ2(X+(y))=γ1(X+(y),Y(X+(y)))φ1(Y(X+(y)))ea0X+(y)+b0Y(X+(y))γ2(X+(y),Y(X+(y))).(7.5)

Note that X+(y) is holomorphic in a neighborhood of Y±(xmax). The crucial point is that Y(X+(y)) is also holomorphic there. Indeed, it can be expressed as

Y(X+(y))=X+(y)2+2μ1X+(y)y.(7.6)

To see this, note that Y(x) and Y+(x) are the two roots of y12(xy)2+μ1x+μ2y. Then, by Vieta’s equations and because Y+(X+(y))=y, (7.6) follows immediately. Because γ2(xmax,Y±(xmax)0, it follows from (7.5) that φ2(X+(y)) is holomorphic at Y±(xmax), so the saddle point method applies to the right-hand side of (7.4). Then, asymptotics of Green’s functions become

g(rcos(α),rsin(α))rαα0c*hα*(z0)er(cos(α)x*+sin(α)y*)𝟙α*>0+er(cos(α)x(α)+sin(α)y(α))1r(c0(α)+c1(α)r)(7.7)
for all α0[0,ϵ], where ϵ>0 is sufficiently small. It remains to show that if α*=0, then (3) holds. With α=0, Equation (7.7) becomes
g(r,0)rerxmaxr3/2c1(0).(7.8)

By Proposition 8, g(r,0)=2f2(r). Finally, Lemma 12 applies and completes the asymptotic analysis. The nonvanishing of α[hα(z0)]α=0+ is analogous to the case α(α*,α**); see Section 6.2. The nonvanishing of hα*(z0) if α*>0 is already proved in Lemma 11. □

7.2. Case αα* When α*>0

Theorem 7

(Asymptotics with αα*). Suppose α*>0 (i.e., φ2 has a pole). Then,

  • If r(αα*)20, then

    gz0(rcos(α),rsin(α))rαα*12hα*(z0)er(cos(α)x*+sin(α)y*),(7.9)
    where hα*(z0) is given by (1.24).

  • If r(αα*)2K>0 for some constant K, then for α<α* (resp. α>α*),

    gz0(rcos(α),rsin(α))rαα*cKhα*(z0)er(cos(α)x*+sin(α)y*)(resp.gz0(rcos(α),rsin(α))rαα*c˜Khα*(z0)er(cos(α)x*+sin(α)y*)),(7.10)
    where constants cK>0, c˜K>0 are independent of initial condition z0.

  • If r(αα*)2, then

  • If α<α*, then

    gz0(rcos(α),rsin(α))rαα*hα*(z0)er(cos(α)x*+sin(α)y*).(7.11)

  • If α>α*, then

    gz0(rcos(α),rsin(α))rαα*hα*(z0)er(cos(α)x(α)+sin(α)y(α)1rCαα*,(7.12)

where C is a positive constant independent of initial condition z0.

Furthermore, hα*(z0)>0. Constants cK,c˜K, and C are made explicit in Franceschi et al. (2024b, section 10).

The proof is analogous to Franceschi et al. (2024b, section 10), which compares the asymptotic contribution of the pole term and the saddle point term in the expressions of Lemma 6 for gz0=I1+I2+I3. The nonvanishing of hα*(z0) was already proved in Lemma 11.

7.3. Last Particular Case γ2(xmax,Y±(xmax))=0

Theorem 8.

Suppose γ2(xmax,Y±(xmax))=0 (so α*=0). Then,

gz0(rcos(α),rsin(α))rα0er(cos(α)x(α)+sin(α)y(α)h0(z0)r,(7.13)
where h0(z0) is given by (1.25). Furthermore, h0(z0)0.

Proof.

The proof follows the same approach as that of Theorem 6, but here c0(α)h0(z0)0. Thus, we consider only the first term in (6.2) with the representation (7.4) for I1(a,b). First, note from (3.6) that ddy[X+(y)]y=Y±(xmax)=0. Then, by (7.6),

ddy[γ2(X+(y),Y(X+(y)))]y=Y±(xmax)=((μ2/2)2+2μ1μ22/2(μ22/2μ2)2)=1.(7.14)

Hence, γ2(X+(y),Y(X+(y)))=(yY±(xmax))(1+oyY±(xmax)(1)). Furthermore, γ2(X+(y),y)=(yY±(xmax))(1+oyY±(xmax)(1)) by similar calculations.

Using the same arguments as in the proof of Theorem 6, the function

γ2(X+(y),y)φ2(X+(y))xγ(X+(y),y)(7.15)
is holomorphic in a neighborhood of Y±(xmax), expect possibly in Y±(xmax), where φ2(X+(y)) has a simple pole by (7.5) and (7.14). Because γ2(X+(y),y) has a zero of the same order at this point, then the quantity (7.15) turns out to be holomorphic at Y±(xmax) as well. Moreover, by (4.23), the following asymptotic expansion holds as yY±(xmax):
γ2(X+(y),y)φ2(X+(y))xγ(X+(y),y)=(1+oyY±(xmax)(1))(ez0·(X+(y),Y(X+(y)))+γ1(ψ1(X+(y),y))γ1γ2(ψ2(X+(y),y))n=1+[k=2nγ1γ2(ψ2k1(X+(y),y)γ1γ2(ψ2k(X+(y),y))][ez0·ψ2n(X+(y),y))γ2(ψ2n(X+(y),y))ez0·ψ2n+1(X+(y),y)γ2(ψ2n+1(X+(y),y))]).(7.16)

This implies (1.25). The proof of the nonvanishing of h0(z0) is analogous to Lemma 11. □

7.4. Proof of Theorem 2

This is a direct consequence of Theorems 1, 6, 7, and 8.

8. Harmonic Functions and Martin Boundary

In this section, we prove Theorem 3. In particular, we show in Section 8.1 that the Martin boundary is homeomorphic to [α*,α**] and in Section 8.2 that the Martin boundary is minimal.

8.1. Context of Martin the Boundary

In this section, we consider the construction of the Martin boundary as presented in (Pinsky 1995, section 7.1) for elliptic processes, and we adapt this approach to reflected degenerate processes. This method allows us to consistently link the harmonic functions (hα)α[α*,α**] found in Theorem 2 and the Martin boundary. Note that another general construction of the Martin compactification is presented in Kunita and Watanabe (1965).

Definition 3

(Martin Kernel). For z0,z1R+2, we define the Martin kernel,

k(z0,z1)={gz0(z1)g(0,0)(z1)ifz1(0,0)0ifz1=(0,0)orz0=z1,(8.1)
and the Martin metric,
ρ(z1,z2)=R+2|k(x,z1)k(x,z2)|1+|k(x,z1)k(x,z2)|e|x|2dx.(8.2)

By usual considerations (see Pinsky 1995), ρ is a metric equivalent to the Euclidean one on R+2. A sequence (yn)n0 of R+2 is called a Martin sequence if (k(·,yn))n0 converges pointwise. Two Martin sequences are said to be equivalent if their limit functions are equal. We then define M as the quotient of the set of all Martin sequences by this equivalence relation. Each ξM is then naturally associated with function denoted by k(·,ξ). The metric ρ extends naturally to M with the same formula so that the map

ι:{R+2(M,ρ)zk(·,z)(8.3)
is injective and continuous. We define the Martin boundary Γ as Γ=M\ι(R+2).

Lemma 13.

Let (hα)α[α*,α**] be defined in Theorem 2. Then, the map

Φ:α[α*,α**]hα(·)/hα(0)Γ
is a homeomorphism.

Before proving this lemma, we recall some properties of the family (hα)α[α*,α**].

Remark 3.

Note that for zR+2:

  • If α*>0, then by (1.22),

    hα(z)αα*α>α*+.

  • If α*=0 and γ2(xmax,Y±(xmax)=0, then by (1.25),

    hα(z)α0α>0h0(z)>0.

  • If α*=0 and γ2(xmax,Y±(xmax)0, then by (1.24),

    hα(z)α0α>00.

Proof of Lemma 13.

By Theorems 1, 2, 6, 7, and 8, Φ is surjective. To prove the continuity of Φ, note that a sequence (ξn)n0 converges to some ξM if k(·,ξn) converges pointwise toward k(·,ξ) almost everywhere. Therefore, the proof of the continuity of Φ is reduced to showing that, for any zR+2, the map αΦ(α)(z) is continuous. Let zR+2.

  • By (1.22), the map αΦ(α)(z) is continuous on (α*,α**).

  • If α*=0 and γ2(xmax,Y±(xmax))0, then we have

    Φ(α)(z)=hα(z)hα(0)=hα(z)ααhα(0)α0[αhα(z)]α=0[αhα(0)]α=0=Φ(0)(z),
    so αΦ(α)(z) is continuous at α*=0.

  • If α*>0, then hα(z)hα(0) can be written as

    m=0κm(α)ez·(am(s(α)),bm(s(α)))+1γ2(ζs(α))m=1+κm(α)ez·(am(s(α)),bm(s(α)))m=0κm(α)+1γ2(ζs(α))m=1+κm(α),
    where (am(s(α)),bm(s(α))), are defined by (1.20) and (1.21), with (a0(s(α)),b0(s(α)))=(x(α),y(α)), and κm(α) is given by (1.23). Because γ2(ζ(s(α)))αα*0 (see Notation 1 and Proposition 4), the expected continuity in α* follows from standard continuity theorems on series.

  • The remaining case α*=0 and γ2(xmax,Y±(xmax))=0 is analogous.

  • The proof of the continuity of Φ at α** is symmetric.

Next, let us show that Φ is injective. By the explicit expressions in Theorem 2, the following asymptotics hold as r+. For α*<α<α** and 0θπ/2, we have

hα(rcos(θ),rsin(θ))rer(x(α)cos(θ)+y(α)sin(θ)).(8.4)

If α*=0, then for any 0<θπ/2,

h0(rcos(θ),rsin(θ))rrsin(θ)er(x(0)cos(θ)+y(0)sin(θ)).(8.5)

If α*>0 and 0θπ/2, then

hα*(rcos(θ),rsin(θ))rer(x(α*)cos(θ)+y(α*)sin(θ)).(8.6)

The corresponding symmetric asymptotic behavior holds for hα**. If α,α[α*,α**] are distinct, then by (5.4) and the preceding formulae,

hα(rcos(α),rsin(α))hα(rcos(α),rsin(α))r+.

Hence, hαChα for any constant C, and Φ is injective.

Because Φ is continuous, and because [α*,α**] is compact, Φ(F) is closed in Γ for any closed subset F of [α*,α**]. Therefore, Φ is a homeomorphism. □

Corollary 2.

The following properties hold:

  • (i) If η,ξM satisfies k(·,ξ)=k(·,η), then η=ξ.

  • (ii) The metric space (M,ρ) is compact.

  • (iii) ι(R+2) is dense in M with respect to ρ.

  • (iv) If a sequence (yn)n0R+2 converges to ηΓ with respect to ρ, then k(·,yn) converges pointwise to k(·,η).

Proof.

Properties (i), (iii), and (iv) follow directly from our construction. We now prove (ii). Let (yn)n0 be a sequence in M. Then:

  • Either (yn)n0 has infinitely many points in Γ, in which case it has a convergent subsequence because Γ is compact (see Lemma 13);

  • or (yn)n0 has a bounded subsequence, in which case the conclusion follows because ρ|R+2×R+2 is equivalent to the Euclidean metric;

  • or (yn)n0 has a subsequence that tends to infinity. Because [0,π/2] is compact, (yn)n0 has a subsequence to infinity in some direction α[0,π/2]. By Theorems 1, 6, 7, and 8, this subsequence converges (with respect to the metric ρ) to 𝟙α<α*hα*+𝟙α*αα**hα+𝟙α<α**hα**. □

Remark 4.

By Corollary 2, M is the Martin compactification in the sense of Kunita and Watanabe (1965) and Pinsky (1995), and the Martin boundary Γ is homeomorphic to [α*,α**].

In particular, by Kunita and Watanabe (1965, theorem 4), the following representation theorem holds.

Theorem 9

(Integral Representation). If h is a nonnegative harmonic function, then there exists a Radon measure μh on [α*,α**] satisfying

zR+2,h(z)=[α*,α**]hα(z)dμh(α).(8.7)

Furthermore, every function defined by (8.7) is harmonic.

8.2. Minimality of Functions (hα)α[α*,α**] and Martin Boundary

In this section, we prove that the Martin boundary is minimal.

Definition 4

(Minimal Harmonic Function). A nonnegative harmonic function h is said to be minimal if, for every pair of nonnegative harmonic functions f1 and f2 satisfying f1+f2=h, both f1 and f2 are proportional to h.

Proposition 9

(Γ is Minimal). The Martin boundary is minimal in the sense that if ηΓ, then k(·,η) is minimal. In particular, the measure μh in representation (8.7) is unique.

To prove this, we state the following lemma.

Lemma 14.

Let α*<α1<α** and ϵ>0. Then, there exist constants η>0 and r0>0 such that

hα(rcos(α1),rsin(α1))12er(x(α1)cos(α1)+y(α1)sin(α1)ϵ)(8.8)
for all rr0 and α[α1η,α1+η].

Proof.

This follows from Theorem 2, where explicit formulae of hα are given. □

Proof of Proposition 9.

Let α0[α*,α**]. We aim to prove that if hα0=[α*,α**]hαdμ(α) for some Radon measure μ, then μ is the Dirac measure at α0. This directly implies the minimality of Γ using Definition 4 and Theorem 9. It suffices to show that the support of μ is exactly {α0}. Suppose first that α*<α0<α**. Let us prove that μ((α*,α**)\{α0})=0. Let α1(α*,α**)\{α0}. First, by (5.4), we can choose ϵ>0 such that

x(α1)cos(α1)+y(α1)sin(α1)ϵ>x(α0)cos(α1)+y(α0)sin(α1).(8.9)

Second, by Lemma 14, η>0 such that

hα(rcos(α),rsin(α))12er(x(α1)cos(α1)+y(α1)sin(α1)ϵ)
for α[α1η,α1+η] and rr0 large enough. Then,
hα0(rcos(α1),rsin(α1))α[α1η,α1+η]hα(reα1)μ(dα)rr0μ([α1η,α1+η])2er(x(α1)cos(α1)+y(α1)sin(α1)ϵ).

Considering θ=α1 and α=α0 in (5.4), we obtain, for rr1 large enough,

er(x(α1)cos(α0)+y(α1)sin(α0)ϵ)μ([α1η,α1+η])2er(x(α1)cos(α1)+y(α1)sin(α1)ϵ).

By (8.9), the asymptotics of the previous inequality as r+ yield μ([α1η,α1+η])=0. Therefore, μ can be written as μ=Aδα0+Bδα*+Cδα** for some nonnegative constants A, B, and C, that is, (1A)hα0=Bhα*+Chα**. Now, considering the asymptotics (8.4), (8.5), and (8.6), we immediately get B=C=0 and A=1. Hence, μ is the Dirac measure at α0, and hα0 is minimal. The cases α0=α* and α0=α** are treated similarly. □

Proof of Theorem 3.

This is a direct consequence of Lemma 13, Remark 4, and Proposition 9. □

9. From Assumption (1.4) to the General Case

We stated and proved Theorems 1, 2, and 3 under Assumption (1.4). In this section, we generalize these theorems without assuming (1.4). To achieve this, we apply transformations to the xaxis, yaxis, and time t in order to reduce the problem to a process Z˜ that satisfies (1.4).

Proposition 10

(Space-Time Dilatation). Let (Zt)t0 be a degenerate reflected Brownian motion with parameters

Σ=(σ12σ1σ2σ1σ2σ22),μ=(μ1μ2),R=(1r2r11).

Then, the process

(Z˜t)t0((μ1σ1+μ2σ2)(1σ1001σ2)Z(t(μ1σ1+μ2σ2)2))t0
is a degenerate reflected Brownian motion with parameters
(1111),μ˜1(μ1σ1+μ2σ2)(μ1σ1μ2σ2),R˜(1r2σ2σ1r1σ1σ21).

Furthermore, Z˜ satisfies (1.2) to (1.4).

Proof.

This is a direct consequence of Definition 1 applying the corresponding transformation to (2.1). □

Theorem 10

(Harmonic Functions and Martin Boundary: General Case). Suppose that (1.2) and (1.3). Let α*,α** and (h˜α)α[α*,α**] be the angles and the harmonic functions in Theorem 2 for the degenerate reflected Brownian motion of parameters ((1111),μ˜,R˜) (using the notation of Proposition 10). Then, the family of minimal harmonic functions for the initial process is given by

(x0,y0)h˜α((μ1σ1+μ2σ2)(x0σ1,y0σ2)),α[α*,α**].(9.1)

Furthermore, the Martin boundary remains homeomorphic to [α*,α**] and is minimal.

Proof.

Let ψ:R2R2 be the map defined by ψ(z)=(μ1σ1+μ2σ2)(1σ1001σ2)z. We denote by G(z0,·) (resp. G˜(z˜0,·)) the Green’s measure associated with (Zt)t0 (resp. with (Z˜t)t0) and gz0(z) (resp. g˜z˜0(z˜)) the corresponding Green’s functions, where z˜0=ψ(z0). Note that

G(z0,A)=0+Pz0(ZtA)dt=0+Pz0(Z(u(μ1σ1+μ2σ2)2)A)du(μ1σ1+μ2σ2)2=0+Pz˜0(Z˜uψ(A))du(μ1σ1+μ2σ2)2.

Furthermore,

Pz0(Z˜uψ(A))=ψ(A)Pz˜0(Z˜u=u)du=APz˜0(Z˜u=ψ(v))|Jac(ψ)|dv.

Therefore, the following holds for all z0,aR+2:

gz0(a)=1σ1σ2g˜z˜0(ψ(a)).(9.2)

Then,

gz0(rcos(α),rsin(α))=1σ1σ2g˜ψ(z0)(r˜cos(α˜),r˜sin(α˜))(9.3)
with α˜=arctan(σ2σ1tan(α)). The conclusion follows from relation (9.3). □

Acknowledgments

The author thanks Sandro Franceschi and Irina Kourkova for their invaluable insights and discussions about this article. First and foremost, the author sincerely thanks Sandro Franceschi for introducing the author to the compensation method and for the many fruitful discussions they have had. Sandro Franceschi’s guidance has been crucial in shaping this research, and the conversations between him and the author have greatly deepened the author’s understanding of the subject. The author is also deeply grateful to Irina Kourkova for her numerous valuable suggestions regarding the writing of this article as well as for their exchanges on a specific case related to asymptotic analysis.

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