Extinction behaviour for mutually enhancing continuous-state population dynamics

Extinction b eha viour for m utually enhancing con tin uous-state p opulation dynamics Jie Xiong 1 , Xu Y ang 2 and Xiao wen Zhou 3 Abstract. In this pap er, w e study a tw o-dimensional process arising as the unique nonnegativ e solution to a system of tw o sto c hastic diﬀerential equations (SDEs) with m utually enhancing t w o-wa y in teractions driv en b y independent Bro wnian motions and sp ectrally positive α -stable random measures. Such a SDE system can b e iden ti- ﬁed as a con tinuous-state Lotk a-V olterra type p opulation mo del. Extinction prop erties of the p opulations are studied for diﬀerent choices of the co eﬃcien ts inv olv ed in the SDEs. Mathematics Subje ct Classiﬁc ations (2010) : 60J80; 92D25; 60G57; 60G17. Key wor ds and phr ases : Contin uous-state branc hing pro cess, comp etition, nonlinear branc hing, m utually enhancing interaction, sto c hastic Lotk a-V olterra type population, extinction. 1 In tro duction and main results Con tinuous-state branc hing pro cesses (CSBPs for short) are mathematical mo dels that describ e the evolution of rescaled p opulations. They arise as scaling limits of Galton-W atson branching pro cesses and ﬁnd applications in v arious ﬁelds such as biology , p opulation genetics, ph ysics, c hemistry , and so on. W e refer to [14, 22, 23] for reviews and literature on CSBPs. Extinction b eha vior is a k ey topic in the study of p opulation mo dels. Through the Lamp erti random time change, the CSBP is asso ciated with a sp ectrally p ositiv e L ´ evy pro cess stopp ed when ﬁrst reac hing 0, which allows man y semiexplicit expressions. In particular, a suﬃcient and necessary condition, called Grey’s condition, is obtained in [13]. In recent years, man y authors hav e studied the generalized nonlinear CSBP characterized b y the follo wing SDE. X t = X 0 + Z t 0 γ 0 ( X s )d s + Z t 0 Z γ 1 ( X s ) 0 W (d s, d u ) + Z t 0 Z ∞ 0 Z γ 2 ( X s − ) 0 z ˜ M (d s, d z , d u ) , (1.1) where X 0 > 0 and γ 0 and γ 1 , γ 2 ≥ 0 are Borel functions on [0 , ∞ ), W and ˜ N are Gaussian white noise and comp ensated Possion random measure. If γ i ( x ) = γ i x for some constan t γ i , then the solution to (1.1) reduces to the classical CSBP . F or γ 2 ≡ 0 and γ 1 ( x ) = γ 1 x , (1.1) is also called the Da wson-Li SDE with its solution called a CSBP with comp etition in [31], where the function γ 0 mo dels an impact of the current p opulation size on the repro duction dynamics of individuals. 1 Departmen t of Mathematics and SUST ech In ternational center for Mathematics, Southern Universit y of Sci- ence & T echnology , Shenzhen, China. Supp orted by National Key R&D Program of China (No. 2022YF A1006102) and National Natural Science F oundation of China Grant 12471418. Email: xiong j@sustec h.edu.cn 2 Sc ho ol of Mathematics and Information Science, North Minzu Universit y , Yinch uan, China. Supp orted by NSF C (No. 12471135). Email: xuyang@mail.bn u.edu.cn. Corresp onding author. 3 Departmen t of Mathematics and Statistics, Concordia Universit y , Montreal, Canada. Supp orted by NSERC (R GPIN-2021-04100). Email: xiaow en.zhou@concordia.ca. 1 Moreo ver, if γ 0 ( x ) = c 1 x − c 2 x 2 for c 1 , c 2 > 0, then the solution to SDE (1.1), called the branc hing pro cess with logistic growth” or “logistic branching pro cess” in [15], mo dels the den- sit y dependence in p opulation. This density dep endence corresp onds to in trasp eciﬁc comp etition pressure, which is ubiquitous in ecology , and translates mathematically into a quadratic death rate. The terms γ i ( x ) /x for i = 1 , 2 can b e interpreted as p opulation-size-dependent branching rates, and the drift in volving γ 0 can also b e related to contin uous p opulation-size-dependent immigration. P opulation-size-dep enden t CSBPs arise as scaling limits of the corresp onding discrete-state branc hing pro cesses (see, e.g., [20, 21]). If γ i ( x ) = γ i x r for i = 0 , 1 , 2, the solu- tion to SDE (1.1) is called a CSBP with p olynomial branching whose exp ectation of extinction time is discussed via Lamperti type transformations in [16], where the parameter r describ es the degree of in teraction. The CSBPs with polynomial branching arise as time-space scaling limits of discrete-state nonlinear branc hing processes. Intuitiv ely , the functions γ 1 and γ 2 are p opulation-dependent rates that generate a small and large n umber of branching even ts in c hil- dren, resp ectiv ely . Rather sharp conditions on extinction/n on-extinction for SDE (1.1) are giv en in [19] using martingale tec hniques, and critical cases are further studied in [28]. The exp onential ergo dicit y and the strong exponential ergodicity for the solution to SDE (1.1) is di scussed in [17] using coupling techniques. More recen tly , mo dels related to (1.1) ha v e also attracted atten tion. The extinction/non-extinction for generalized nonlinear CSBPs with catastrophes is studied in [29]. [1] establishes necessary and suﬃcient conditions of extinction for the generalized nonlinear CSBPs with Nev eu’s branc hing. The ergo dic prop erty of a CSBP with immigration and com- p etition is giv en in [18]. General nonlinear CSBPs with comp etition in the L´ evy environmen t and CSBPs in v arying environmen ts are studied in [27] and [12], resp ectiv ely . By [23, Theorems 7.1 and 8.1], on an enlarged probabilit y space, on the right of (1.1), the third term can b e transformed into R t 0 p γ 1 ( X s )d B s , and the fourth term can be conv erted in to R t 0 R ∞ 0 z ˜ M γ 2 (d s, d z ), where ( B t ) t ≥ 0 is a Bro wnian motion and ˜ M γ 2 (d s, d z ) is an optional comp en- sated P oisson random measure. Using the Lamp erti transform for p ositiv e self-similar Mark ov pro cesses, in certain cases, [5] studies the necessary and suﬃcient condition for extinction. Compared with the one-dimensional mo del (1.1), the study of tw o-dimensional interaction of generalized nonlinear CSBP c haracterized by the following SDE system is more challenging and the existing literature on this is sparse:                            X t = X 0 + Z t 0 θ 1 ( X s , Y s )d s + Z t 0 γ 10 ( X s )d s + Z t 0 p γ 11 ( X s )d B 1 ( s ) + Z t 0 Z ∞ 0 Z γ 12 ( X s − ) 0 z ˜ N 1 (d s, d z , d u ) , Y t = Y 0 + Z t 0 θ 2 ( Y s , X s )d s + Z t 0 γ 20 ( Y s )d s + Z t 0 p γ 21 ( Y s )d B 2 ( s ) + Z t 0 Z ∞ 0 Z γ 22 ( Y s − ) 0 z ˜ N 2 (d s, d z , d u ) , (1.2) where ( B 1 ( t )) t ≥ 0 and ( B 2 ( t )) t ≥ 0 are Bro wnian motions, and ˜ N 1 (d s, d z , d u ) and ˜ N 2 (d s, d z , d u ) are comp ensated Poisson random measures. The SDE system (1.2) is a solution to the CSBP of t wo-t yp es when γ ij ( x ) /x are nonnegativ e constants for all i, j = 1 , 2. If γ i 0 ( x ) = b i 0 x , γ i 1 ( x ) = b i 1 x with b i 1 > 0 and γ i 2 ≡ 0 for i = 1 , 2, then the SDE system (1.2) is studied in [33] and [6] when θ 1 ( x, y ) = a 1 y , θ 2 ( y , x ) = a 2 x and θ 1 ( x, y ) = a 1 xy , θ 2 ( y , x ) = a 2 xy for a 1 , a 2 > 0, resp ectiv ely . If γ i 0 ( x ) = a i 0 x + η i , γ ij ( x ) = b ij x and θ 1 ( x, y ) = a 1 y , θ 2 ( y , x ) = a 2 x with b ij , a i > 0 for i, j = 1 , 2, then the SDEs system (1.2) is studied in [26] and the more general system with 2 m ultiple SDEs in [2]. Two-t yp e CSBP in v arying en vironments characterized b y the form of (1.2) is studied in [24, 25]. The SDE system (1.2) is also a dynamic sto chastic Lotk a-V olterra-t yp e p opulation. If γ i 0 ( x ) = b i 0 x and γ i 1 ( x ) = b i 1 x 2 for i = 1 , 2, θ 1 ( x, y ) = a 2 xy , θ 2 ( y , x ) = a 2 xy with b 11 , b 21 , a 1 , a 2 > 0, γ 12 ( x ) = γ 22 ( x ) ≡ 0, then SDE (1.2) is also called a competitive Lotk a-V olterra mo del in random environmen ts in [3] and references, and [11] considers the case that (( B 1 ( t ) , B 2 ( t ))) t ≥ 0 is a correlated tw o-dimensional Brownian motion. If γ i 2 ( x ) ≡ 0 and γ i 1 ( x ) = b i 1 x 2 with b i 1 > 0 for i = 1 , 2, and θ 1 ( x, y ) θ 2 ( x, y ) < 0, then SDE (1.2) is called sto c hastic predator-prey mo dels, and we refer to [10] and [4] for related p ersistence/extinction results. If θ 1 ( x, y ) ≡ 0, θ 2 ( x, y ) =: θ ( x ) κ ( y ) < 0 and γ 10 ( x ) , γ 20 ( y ) ≤ 0, then the extinction- extinguishing dic hotomy for SDE system (1.2) is discussed in [32]. If θ i ( x, y ) = θ i ( x ) κ i ( y ) < 0 for i = 1 , 2 and γ ij , θ i , κ i are p o wer functions for i = 1 , 2 and j = 0 , 1 , 2, the SDE (1.2) is m utually comp etitiv e and rather sharp conditions on the extinction-extinguishing dic hotomy are giv en in [34]. Our aim of this pap er is to establish the extinction/non-extinction conditions for SDE (1.2) with m utual enhancing, that is, interacting functions θ 1 , θ 2 > 0. F or simplicity and readabilit y , in this pap er we only consider the sp ecial form of (1.2) with p o w er function co eﬃcien ts and pow er function in tensities for ˜ N i (d s, d z , d u ), that is, the following SDE system:                            X t = X 0 + a 1 Z t 0 X θ 1 s Y κ 1 s d s − b 10 Z t 0 X r 10 s d s + b 11 Z t 0 p 2 X r 11 s d B 1 ( s ) + Z t 0 Z ∞ 0 Z b 12 X r 12 s − 0 z ˜ N 1 (d s, d z , d u ) , Y t = Y 0 + a 2 Z t 0 Y θ 2 s X κ 2 s d s − b 20 Z t 0 Y r 20 s d s + b 21 Z t 0 p 2 Y r 21 s d B 2 ( s ) + Z t 0 Z ∞ 0 Z b 22 Y r 22 s − 0 z ˜ N 2 (d s, d z , d u ) , (1.3) where for i = 1 , 2 , j = 0 , 1 , 2, a i , κ i > 0 and θ i , r ij , b ij ≥ 0. F or i = 1 , 2, ( B i ( t )) t ≥ 0 are t wo Brownian motions and ˜ N i (d s, d z , d u ) are tw o comp ensated P oisson random measures with in tensity d sµ i (d z )d u . Here µ i (d z ) = α i ( α i − 1) Γ( α i )Γ(2 − α i ) z − 1 − α i 1 { z > 0 } d z for α i ∈ (1 , 2), i = 1 , 2, and Γ denotes the Gamma function. W e alwa ys assume that b 11 + b 12 > 0 and b 21 + b 22 > 0. W e also assume that ( B 1 ( t )) t ≥ 0 , ( B 2 ( t )) t ≥ 0 , { ˜ N 1 (d s, d z , d u ) } , and { ˜ N 2 (d s, d z , d u ) } are independent of eac h other. The main metho ds to develop the criteria for extinction/non-extinction in [28, 1] are adapta- tions of the approac h for Chen’s criteria on the uniqueness problem of Mark ov jump pro cesses. These Chen’s criteria are ﬁrst established in [7, 8] and can also b e found in [9, Theorems 2.25 and 2.27]. A similar approac h to studying the b oundary behaviors for Marko v pro cesses can also b e found in [30]. Suc h an approach typ ically inv olv es identifying an appropriate test function that is applied to the inﬁnitesimal generator of the generalized nonlinear CSBP and proving the desired result using a martingale argument. In [32] a sto chastic Lotk a-V olterra-t yp e p opu- lation dynamical system is prop osed as a solution to an SDE system with one-sided in teraction. The tw o-population model (1.3) with mutually comp etitiv e interaction is further studied in [34] where the co eﬃcients a 1 and a 2 are both negativ e. The main methods of [32] and [34] rely on the generalization of Chen’s criteria tec hnique from one-dimensional pro cesses to tw o-dimensional pro cesses. In the pap er, we consider the mo del of mutually enhancing p opulations, namely a 1 , a 2 > 0, and lea ve the case of mixing interaction of a 1 a 2 < 0 as a challenging op en problem. The main 3 metho ds of this pap er are to develop the criteria of extinction/non-extinction again for a tw o- dimensional pro cess which is an adaption of the approach for Chen’s criteria again. The key to applying this criterion is to ﬁnd appropriate test functions for whic h our approac h is mostly ad ho c, guessing bit by bit without an obvious intuition. F or the extinction b eha vior, the criteria and the test functions in [32, 34] among other references are no longer applicable in this pap er, and w e need to develop new criteria and new testing functions that are more demanding. F or an y generic sto c hastic pro cess V := ( V ( t )) t ≥ 0 and constan t w > 0, let τ 0 ( V ) := inf { t ≥ 0 : V ( t ) = 0 } , τ − w ( V ) := inf { t ≥ 0 : V ( t ) ≤ w } and τ + w ( V ) := inf { t ≥ 0 : V ( t ) ≥ w } with the conv en tion inf ∅ = ∞ . Let τ 0 := τ 0 ( X ) ∧ τ 0 ( Y ), τ − w := τ − w ( X ) ∧ τ − w ( Y ) and τ + w := τ + w ( X ) ∧ τ + w ( Y ) for w > 0. In the following, we state the deﬁnition of a solution to the SDE system (1.3), whic h is deﬁned b efore the minimum of the ﬁrst hitting time of 0 or the explosion time for the t wo pro cesses X and Y . Deﬁnition 1.1 By a solution to SDE (1.3) w e mean a tw o-dimensional c` adl` ag pro cess ( X , Y ) := (( X t , Y t )) t ≥ 0 satisfying SDE (1.3) up to γ n := τ − 1 /n ∧ τ + n for eac h n ≥ 1 and X t = lim sup n →∞ X γ n − and Y t = lim sup n →∞ Y γ n − for t ≥ lim n →∞ γ n . By Deﬁnition 1.1, 0 and ∞ are absorbing states, and the solution is nonnegative. Deﬁnition 1.1 of the solution allows for weak er conditions for the uniqueness of the solution. In particular, the existence and pathwise uniqueness of SDE (1.3) can b e obtained b y the same argumen ts as in [32, Lemma A.1]. Throughout this pap er, we alw ays assume that the c` adl` ag pro cess ( X , Y ) is the unique solution to (1.3), and consequently , the pro cess ( X , Y ) has the strong Marko v prop ert y . W e also assume that X 0 , Y 0 > 0 are deterministic and that all sto c hastic pro cesses are deﬁned on the same ﬁltered probability space (Ω , F , F t , P ). Let E denote the corresp onding exp ectation. F or i = 1 , 2, let r i := min  r ij − ϱ ij , j ∈ { b ij  = 0 , j = 0 , 1 , 2 }  , b i := X j =0 , 1 , 2 b ij 1 { r i = r ij − ϱ ij } , where ϱ ij = j + 1 and ϱ i 2 = α i for i = 1 , 2 and j = 0 , 1. The ab ov e parameters determine the extinction b eha viors for the pro cesses ( X , Y ). W e ﬁrst consider the non-extinction b ehaviors. Note that the nonlinear one-dimensional CSBP (with a 1 X θ 1 s Y κ 1 s replaced by a 1 X θ 1 s in (1.3)) cannot reac h 0 if r 1 ≥ ( θ 1 − 1) ∧ 0 (see [19, p.2535]). The follo wing result can also b e sho wn similarly to [19, p.2535]. Prop osition 1.2 If r 1 , r 2 ≥ 0 , then P { τ 0 < ∞} = 0 . The next theorem concerns the extinction conditions for the t w o-dimensional mo del (1.3) under whic h Y κ 1 s in the interaction term a 1 X θ 1 s Y κ 1 s do es not aﬀect the extinction of X when r 2 ≥ 0 and θ 1 − 1 < r 1 < 0. Theorem 1.3 Supp ose that one of the follo wing holds: 4 (i) r 1 ≥ 0 and θ 2 − 1 < r 2 < 0 ; (ii) r 2 ≥ 0 and θ 1 − 1 < r 1 < 0 . Then P { τ 0 < ∞} = 0 . The following theorem presents a non-extinction condition, when b oth θ 1 − 1 < r 1 < 0 and θ 2 − 1 < r 2 < 0 hold, whic h dep ends on the in teraction term a 1 X θ 1 s Y κ 1 s and a 2 Y θ 2 s X κ 2 s . Theorem 1.4 P { τ 0 < ∞} = 0 if θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 , and ( r 1 + 1 − θ 1 )( r 2 + 1 − θ 2 ) > κ 1 κ 2 . (1.4) Before considering critical cases, we ﬁrst state the following condition, which means that the drift term pla ys a dominant role when the pro cesses are near zero. Condition 1.5 (i) F or b 10 b 20  = 0 , r 10 − 1 < min  r 11 − ϱ 1 j , j ∈ { b 1 j  = 0 , j = 1 , 2 }  (1.5) and r 20 − 1 < min  r 21 − ϱ 2 j , j ∈ { b 2 j  = 0 , j = 1 , 2 }  , (1.6) where ϱ i 1 = 2 and ϱ i 2 = α i for i = 1 , 2 ; (ii) In addition to ( r 1 + 1 − θ 1 ) /κ 1 < r 1 /r 2 , either (1.5) holds and (1.6) is not satisﬁed for b 10 b 20  = 0 , or b 20 = 0 and (1.5) holds for b 10  = 0 ; (iii) In addition to ( r 2 + 1 − θ 2 ) /κ 2 < r 2 /r 1 , either (1.6) holds and (1.5) is not satisﬁed for b 10 b 20  = 0 , or b 10 = 0 and (1.6) holds for b 20  = 0 . Theorem 1.6 Supp ose that θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 , and ( r 1 + 1 − θ 1 )( r 2 + 1 − θ 2 ) = κ 1 κ 2 . (1.7) W e also assume that one of the follo wing holds: (i) Condition 1.5(i) is satisﬁed and ( a 1 /b 1 ) 1 / ( r 1 +1 − θ 1 ) ( a 2 /b 2 ) 1 /κ 2 > 1; (1.8) (ii) b 12 = b 22 = 0 , r 10 − 1 = r 11 − 2 = r 20 − 1 = r 21 − 2 =: r , θ 1 = θ 2 , κ 1 = κ 2 , and a 1 a 2 ≥ b 1 b 2 . Then P { τ 0 < ∞} = 0 . 5 Theorem 1.6(ii) implies that P { τ 0 < ∞} = 0 when the inequalit y in (1.8) is replaced by an equality . Under (1.7), w e hav e ( r 1 + 1 − θ 1 ) − 1 = κ 1 κ 2 r 2 +1 − θ 2 and then (1.8) is equiv alent to ( a 2 /b 2 ) 1 r 2 +1 − θ 2 ( a 1 /b 1 ) 1 κ 1 > 1, whic h is a symmetric form of (1.8). Next, we study the extinction b eha vior for ( X, Y ). F or the one-dimensional nonlinear CSBP (with a 1 X θ 1 s Y κ 1 s replaced by a 1 X θ 1 s in (1.3)), it reaches 0 with a p ositive probability when r 1 < ( θ 1 − 1) ∧ 0. The following theorem shows that a similar result still holds for the tw o-dimensional system (1.3) in this case. Theorem 1.7 Supp ose that one of the follo wing holds: (i) r 1 ≤ θ 1 − 1 and r 1 < 0 ; (ii) r 2 ≤ θ 2 − 1 and r 2 < 0 . Then P { τ 0 < ∞} > 0 . Remark 1.8 Observe that conditions (i) and (ii) of Theorem 1.7 only concern pro cesses X and Y , resp ectively . This suggests that P { τ 0 ( X ) < ∞} > 0 under condition (i) and P { τ 0 ( Y ) < ∞} > 0 under condition (ii). In the following, we consider the case θ 1 − 1 < r 1 < 0 and θ 2 − 1 < r 2 < 0. W e ﬁrst state the follo wing condition. Condition 1.9 One of the following holds: (i) Either r 1 + 1 r 2 + 1 ∨ r 1 r 2 < κ 2 r 2 + 1 − θ 2 (1.9) or r 2 + 1 r 1 + 1 ∨ r 2 r 1 < κ 1 r 1 + 1 − θ 1 ; (1.10) (ii) Either 1 − θ 1 κ 1 − r 2 < κ 2 r 2 + 1 − θ 2 ∧ κ 2 − r 1 1 − θ 2 ∧ (1 − θ 1 ) ∧ (2 − κ 2 ) (1.11) or 1 − θ 2 κ 2 − r 1 < κ 1 r 1 + 1 − θ 1 ∧ κ 1 − r 2 1 − θ 1 ∧ (1 − θ 2 ) ∧ (2 − κ 1 ) . (1.12) The assumptions in Condition 1.9 are related to the choices of test functions (see (3.82) and Lemmas 3.12 and 3.14) and ma y not allow intuitiv e in terpretations. Theorem 1.10 Supp ose that θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 and one of the follo wing holds: (i) Condition 1.5 holds and ( r 1 + 1 − θ 1 )( r 2 + 1 − θ 2 ) < κ 1 κ 2 ; (1.13) 6 (ii) F or i = 1 , 2 , r i 0 − 1 ≥ min  r i 1 − ϱ ij , j ∈ { b ij  = 0 , j = 1 , 2 }  , b i 0  = 0 (1.14) with ϱ i 1 = 2 and ϱ i 2 = α i , and (1.13) and Condition 1.9 holds; (iii) F or i = 1 , 2 , (1.14) holds, r 1 + 1 − θ 1 < κ 2 and r 2 + 1 − θ 2 < κ 1 . Then P { τ 0 < ∞} > 0 . Theorem 1.10(i) concerns the case in whic h drift terms dominate when the pro cess is near zero. Theorems 1.10(ii) and (iii) concern the case where diﬀusive terms dominate when the pro cess is close to zero. When r 1 = r 2 , the inequalit y (1.13) implies Condition 1.9(i). In the follo wing, we present a result when the inequality in (1.13) is replaced b y the equality . Theorem 1.11 Supp ose that θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 and (1.7) holds. W e also assume that Condition 1.5 holds and ( a 1 /b 1 ) 1 / ( r 1 +1 − θ 1 ) ( a 2 /b 2 ) 1 /κ 2 < 1 . (1.15) Then P { τ 0 < ∞} > 0 . Giv en that (1.7) holds, the inequality (1.15) is equiv alen t to ( a 2 /b 2 ) 1 r 2 +1 − θ 2 ( a 1 /b 1 ) 1 κ 1 < 1, whic h is a symmetric form of (1.15). It is more challenging and in teresting to establish the conditions for P { τ 0 < ∞} = 1, which will b e c onsidered in future work. Com bining Prop osition 1.2, Theorems 1.3–1.4, 1.6–1.7 and 1.10–1.11 w e hav e the follo wing remark. Remark 1.12 Supp ose that r 1 = r 2 := r . (1) P { τ 0 < ∞} = 0 if one of the follo wing conditions: (i) r ≥ 0 ; (ii) ( θ 1 − 1) ∨ ( θ 2 − 1) < r < 0 and ( r + 1 − θ 1 )( r + 1 − θ 2 ) > κ 1 κ 2 . (2) P { τ 0 < ∞} > 0 if one of the follo wing conditions: (i) r ≤ [( θ 1 − 1) ∨ ( θ 2 − 1)] and r < 0 ; (ii) ( θ 1 − 1) ∨ ( θ 2 − 1) < r < 0 , ( r + 1 − θ 1 )( r + 1 − θ 2 ) < κ 1 κ 2 , and in addition to either Condition 1.5(i) or (1.14) for i = 1 , 2 . (3) If Condition 1.5(i) holds, ( θ 1 − 1) ∨ ( θ 2 − 1) < r < 0 and ( r + 1 − θ 1 )( r + 1 − θ 2 ) = κ 1 κ 2 , then (i) P { τ 0 < ∞} = 0 when ( a 1 /b 1 ) 1 / ( r +1 − θ 1 ) ( a 2 /b 2 ) 1 /κ 2 > 1 ; (ii) P { τ 0 < ∞} > 0 when ( a 1 /b 1 ) 1 / ( r +1 − θ 1 ) ( a 2 /b 2 ) 1 /κ 2 < 1 . A t the end of this section, we give a brief description of the approaches to pro ofs of theo- rems. W e use the generalized Chen’s criteria to establish the assertions on non-extinction and extinction with p ositiv e probability (see Section 2). These Chen’s criteria are ﬁrst prop osed in [7, 8] and can also b e found in [9, Theorems 2.25 and 2.27]. F or the pro ofs of Prop osition 1.2, 7 Theorems 1.3–1.4 and 1.6, w e apply the non-extinction criteria giv en in [32] (see, e.g., Prop osi- tion 2.1) to test functions that are of t wo-v ariable p o wer or logarithmic t yp e (see the functions g and ˜ g deﬁned in Lemmas 3.5 and 3.6). F or the pro ofs of Theorems 1.7 and 1.10–1.11, we ﬁrst establish the extinction criteria, whic h is giv en in Prop osition 2.2 and a generalized Chen’s criterion. The key test function for the pro of of Theorem 1.7 is chosen as a tw o-v ariable p ow er function (see (3.18)). W e select a tw o-v ariable p o wer t yp e test function in the pro of of Theo- rems 1.10–1.11 (see (3.82) with the functions h deﬁned in (3.36) and (3.49) and Lemmas 3.12, 3.14 and 3.16 for diﬀeren t conditions). F or the case where the drift term dominates (Condition 1.5) as the system approaches zero, the test functions h and ˜ h are giv en by (3.36) and (3.49), resp ectiv ely , and condition (iii) in Proposition 2.2 is v eriﬁed b y Lemmas 3.10 and 3.11. F or the case where diﬀusive terms dominate (see (1.14)) and Condition 1.9, w e mo dify the function h deﬁned in (3.36). Under Condition 1.9(i), we ﬁrst replace y ρ 2 with y − y 1 − ε for a suﬃcien tly small y , ε > 0 in (3.36) and then c hange it into ˜ h ( y ) given by (3.51) (see (3.55)). F or Condition 1.9(ii), we also c hange ρ 1 ≥ 1 to 0 < ρ 1 < 1 and smo oth it by replacing x ρ 1 with ( x + y δ ) ρ 1 for some δ > 1 (see (3.67)). F or condition (iii) in Theorem 1.10, the function h is giv en in Lemma 3.16. The k ey condition (iii) in Prop osition 2.2 is v eriﬁed by Lemmas 3.12, 3.14, and 3.16. Similarly , the results can be pro v ed for the cases of b 10 , b 20 < 0 and b 10 b 20 < 0, and we omit them here. Let C 2 ((0 , ∞ )) and C 2 ((0 , ∞ ) × (0 , ∞ )) denote the second-order con tinuous diﬀeren tiable function spaces on (0 , ∞ ) and (0 , ∞ ) × (0 , ∞ ), resp ectiv ely . The remainder of the pap er is arranged as follows. W e state the extinction criteria for the t wo-dimensional pro cess in Section 2. The pro ofs for the assertions in this section are given in Section 3. 2 Criteria for extinction In this section, w e establish some criteria that will be used to pro ve the theorems in Section 1. In the follo wing, let (( x t , y t )) t ≥ 0 b e a t w o-dimensional c` adl` ag process with deterministic x 0 , y 0 > 0, where ( x t ) t ≥ 0 and ( y t ) t ≥ 0 are tw o nonnegative pro cesses deﬁned b efore the minimum of their ﬁrst times of hitting zero or explosion. Let L denote the op erator such that for each g ∈ C 2 ((0 , ∞ ) × (0 , ∞ )), the pro cess t 7→ M g t ∧ γ m,n is a martingale , (2.1) where M g t := g ( x t , y t ) − g ( x 0 , y 0 ) − Z t 0 L g ( x s , y s )d s and γ m,n := τ − 1 /m ∧ τ + n with τ − 1 /m := τ − 1 /m ( x ) ∧ τ − 1 /m ( y ) and τ + n := τ + n ( x ) ∧ τ + n ( y ). In this section, let τ 0 := τ 0 ( x ) ∧ τ 0 ( y ) and τ ∞ := lim n →∞ τ + n . The following t wo criteria on non-extinction and extinction of pro cess (( x t , y t )) t ≥ 0 , which generalize Chen’s criteria for the uniqueness problem of Mark ov jump pro cesses. F or the SDE system, the op erator L can b e obtained b y Itˆ o’s formula. The following Prop osition 2.1 given in [32] is a criterion to establish the non-extinction asser- tion and test functions g often tak e either the form of t wo-dimensional p o wer type functions with negativ e p ow er or the form of logarithm type functions, which is a key point. This prop osition is used to pro ve Prop osition 1.2, Theorems 1.3–1.4 and 1.6. Prop osition 2.1 ([32, Prop osition 2.1]) F or any ﬁxed n ≥ 1 , if there is a nonnegativ e function g ∈ C 2 ((0 , ∞ ) × (0 , ∞ )) and a constan t d n > 0 suc h that the follo wing hold: (i) lim x ∧ y → 0 g ( x, y ) = ∞ ; (ii) L g ( x, y ) ≤ d n g ( x, y ) for all 0 < x, y ≤ n , then P { τ 0 < ∞} = 0 . 8 The prop osition 2.2 gives a new criterion for extinction and can b e shown by adapting the approac h to Chen’s criteria. The k ey to applying Prop osition 2.2 is to select a suitable test function and to verify the key condition (iii) in the follo wing. This prop osition is used to pro ve Theorems 1.7, 1.10 and 1.11. The test functions g are often of the form of tw o-dimensional pow er t yp e functions with p ositiv e p o w ers, which are mostly obtained b y ad hoc guessing without m uch in tuitive interpretation, and the test functions in the previous pap ers are no longer appropriate. Note that the criteria and test functions in Theorems 1.7, 1.10, and 1.11 are diﬀerent from those in [32, 34] and other references. The pro of of Prop osition 2.2 is based on (2.1) and a martingale argumen t. Prop osition 2.2 Let v > 0 and 0 < x 0 , y 0 < v b e ﬁxed. Supp ose that there are a function g ∈ C 2 ((0 , ∞ ) × (0 , ∞ )) and a constant d > 0 such that (i) g ( x 0 , y 0 ) > 0 and C 0 := sup x,y > 0 g ( x, y ) < ∞ ; (ii) g ( x, y ) ≤ 0 for all ( x, y ) with x ∨ y ≥ v ; (iii) L g ( x, y ) ≥ dg ( x, y ) for all 0 < x, y ≤ v . Then P { τ 0 < ∞} ≥ g ( x 0 , y 0 ) /C 0 > 0 . Pr o of. In view of (2.1), for all large enough m ≥ 1, E  g ( x t ∧ γ m,v , y t ∧ γ m,v )  = g ( x 0 , y 0 ) + Z t 0 E  L g ( x s , y s )1 { s ≤ γ m,v }  d s. It then follo ws from integration by parts that e − dt E  g ( x t ∧ γ m,v , y t ∧ γ m,v )  = g ( x 0 , y 0 ) + Z t 0 e − ds d  E  g ( x s ∧ γ m,v , y s ∧ γ m,v )   + Z t 0 E  g ( x s ∧ γ m,n , y s ∧ γ m,v )  de − ds = g ( x 0 , y 0 ) + Z t 0 e − ds E  L g ( x s , y s )1 { s ≤ γ m,v }  d s − d Z t 0 e − ds E  g ( x s ∧ γ m,v , y s ∧ γ m,v )  d s. Under conditions (i) and (iii), w e hav e C 0 e − dt − g ( x 0 , y 0 ) ≥ e − dt E  g ( x t ∧ γ m,v , y t ∧ γ m,v )  − g ( x 0 , y 0 ) ≥ d Z t 0 e − ds E  g ( x s , y s )1 { s ≤ γ m,v }  d s − d Z t 0 e − ds E  g ( x s ∧ γ m,v , y s ∧ γ m,v )  d s = − d Z t 0 e − ds E  g ( x s ∧ γ m,v , y s ∧ γ m,v )1 { s>γ m,v }  d s ≥ − d Z t 0 e − ds E  g + ( x s ∧ γ m,v , y s ∧ γ m,v )1 { s>γ m,v }  d s, where g + ( u ) := g ( u ) ∨ 0. Letting t → ∞ we obtain g ( x 0 , y 0 ) ≤ d Z ∞ 0 e − ds E  g + ( x s ∧ γ m,v , y s ∧ γ m,v )1 { s>γ m,v }  d s = E h g + ( x γ m,v , y γ m,v ) d Z ∞ γ m,v e − ds d s i = E  g + ( x γ m,v , y γ m,v )e − dγ m,v  . 9 Then, b y the dominated conv ergence, w e get g ( x 0 , y 0 ) ≤ lim m →∞ E  g + ( x γ m,v , y γ m,v )e − dγ m,v  = E  g + ( x τ 0 ∧ τ + v , y τ 0 ∧ τ + v )e − d ( τ 0 ∧ τ + v )  = E h g + ( x τ 0 ∧ τ + v , y τ 0 ∧ τ + v )e − d ( τ 0 ∧ τ + v )  1 { τ 0 <τ + v } + 1 { τ 0 ≥ τ + v ,τ + v < ∞} + 1 { τ 0 = τ + v = ∞}  i = E h g + ( x τ 0 , y τ 0 )e − dτ 0 1 { τ 0 <τ + v } + g + ( x τ + v , y τ + v )e − dτ + v 1 { τ 0 ≥ τ + v ,τ + v < ∞} i ≤ C 0 P { τ 0 < τ + v } , (2.2) where conditions (i) and (ii) are used in the last inequality . Since g ( x 0 , y 0 ) > 0 under condition (i), then b y (2.2), P { τ 0 < ∞} ≥ P { τ 0 < τ + v } ≥ g ( x 0 , y 0 ) /C 0 > 0 , whic h completes the pro of. 2 3 Pro ofs of the main results In preparation for the main pro ofs, we ﬁrst introduce some notation and inequalities. F or any g ∈ C 2 ((0 , ∞ ) × (0 , ∞ )) and x, y , z ≥ 0 deﬁne K 1 z g ( x, y ) := g ( x + z , y ) − g ( x, y ) − g ′ x ( x, y ) z (3.1) and K 2 z g ( x, y ) := g ( x, y + z ) − g ( x, y ) − g ′ y ( x, y ) z . (3.2) By T a ylor’s formula, for any b ounded function g with contin uous second deriv ativ e, g ( x + z ) − g ( x ) − g ′ ( x ) z = z 2 Z 1 0 g ′′ ( x + z u )(1 − u )d u. (3.3) F or the Borel function g in (0 , ∞ ), replacing the v ariable z with z x we get Z ∞ 0 g ( z ) µ i (d z ) = x − α i Z ∞ 0 g ( z x ) µ i (d z ) , i = 1 , 2 , (3.4) where w e use the fact µ i (d( z x )) = x − α i µ i (d z ). The op erator L is given by L g ( x, y ) := a 1 x θ 1 y κ 1 g ′ x ( x, y ) + a 2 y θ 2 x κ 2 g ′ y ( x, y ) − b 10 x r 10 g ′ x ( x, y ) + b 11 x r 11 g ′′ xx ( x, y ) + b 12 x r 12 Z ∞ 0 K 1 z g ( x, y ) µ 1 (d z ) − b 20 y r 20 g ′ y ( x, y ) + b 21 y r 21 g ′′ y y ( x, y ) + b 22 y r 22 Z ∞ 0 K 2 z g ( x, y ) µ 2 (d z ) . (3.5) F or i = 1 , 2 and ρ < α i let c i ( ρ ) := Γ( α i − ρ ) Γ( α i )Γ(2 − ρ ) . By [34, Lemma 4.2] and (3.3), for ρ ∈ ( −∞ , 0) ∪ (0 , 1) ∪ (1 , α i ), c i ( ρ ) = [ ρ ( ρ − 1)] − 1 Z ∞ 0 [(1 + z ) ρ − 1 − ρz ] µ i (d z ) = Z ∞ 0 z 2 µ i (d z ) Z 1 0 (1 + z u ) ρ − 2 (1 − u )d u. (3.6) W e no w present a lemma for the pro ofs b elo w. 10 Lemma 3.1 (i) F or any u, v ≥ 0 and p, q > 1 with 1 /p + 1 /q = 1 , we ha v e u + v ≥ p 1 /p q 1 /q u 1 /p v 1 /q and u/p + v /q ≥ u 1 /p v 1 /q . (ii) F or x, y ≥ 0 , w e ha ve x p + y p ≥ ( x + y ) p for any 0 < p ≤ 1 and x p + y p ≥ 2 1 − p ( x + y ) p for an y p > 1 . (iii) Supp ose that p 1 , p 2 , p 3 , p 4 > 0 and c 1 , c 2 , c 3 > 0 . If p 3 /p 1 + p 4 /p 2 > 1 , then there is a constan t 0 < c < 1 such that c 1 x p 1 + c 2 y p 2 ≥ c 3 x p 3 y p 4 for all 0 < x, y < c . Pr o of. Assertion (i) follows immediately from the Y oung inequality , and assertion (ii) is obvious. Let 0 < δ < 1 satisfy 1 − p 4 /p 2 < δ < p 3 /p 1 . Then δ p 1 − p 3 < 0 and (1 − δ ) p 2 − p 4 < 0. By assertion (i), there are constan ts c 4 , c > 0 suc h that c 1 x p 1 + c 2 y p 2 ≥ c 4 x δ p 1 y (1 − δ ) p 1 = c 3 x p 3 y p 4 ( c 4 /c 3 ) x δ p 1 − p 3 y (1 − δ ) p 1 − p 4 ≥ c 3 x p 3 y p 4 for all 0 < x, y < c . This prov es assertion (iii). 2 3.1 Proofs of Prop osition 1.2 and Theorems 1.3–1.4 and 1.6 In this subsection, w e use Prop osition 2.1 to complete the pro ofs of Prop osition 1.2 and Theorems 1.3–1.4 and 1.6. The test functions g are presen ted in Lemmas 3.5 and 3.6. The k ey to the proofs is to verify condition (ii) in Prop osition 2.1 for whic h we need some lemmas in the following subsubsection. 3.1.1 Preliminaries The following lemma is used to pro ve Lemma 3.5 in the next subsection where the key to its pro of is applying Lemma 3.1(i). Lemma 3.2 Supp ose that r 2 > θ 2 − 1 , ρ 1 , ρ 2 > θ 1 ∨ θ 2 and r 2 + 1 − θ 2 κ 2 > 1 + ρ 2 + κ 1 − θ 2 1 + ρ 1 + κ 2 − θ 1 . (3.7) Then there are constan ts δ, C > 0 such that y θ 2 − 1 − ρ 2 x κ 2 + x θ 1 − 1 − ρ 1 y κ 1 ≥ C m δ y r 2 − ρ 2 for all m ≥ 1 and 0 < x, y ≤ m − 1 . Pr o of. Let p := 1 + ρ 1 + κ 2 − θ 1 1 + ρ 1 − θ 1 , q := 1 + ρ 1 + κ 2 − θ 1 κ 2 . Then p, q > 1 and p − 1 + q − 1 = 1. Moreo ver, by (3.7), we obtain 1 q = κ 2 1 + ρ 1 + κ 2 − θ 1 < r 2 + 1 − θ 2 1 + ρ 2 + κ 1 − θ 2 , and then δ := ( r 2 + 1 − θ 2 ) − 1 + ρ 2 + κ 1 − θ 2 q > 0 , 1 + ρ 1 + κ 2 − θ 1 p − (1 + ρ 1 − θ 1 ) = 0 . 11 No w, by Lemma 3.1(i), x 1+ ρ 1 + κ 2 − θ 1 + y 1+ ρ 2 + κ 1 − θ 2 ≥ p 1 /p q 1 /q x (1+ ρ 1 + κ 1 − θ 1 ) /p y (1+ ρ 2 + κ 1 − θ 2 ) /q , and then ( x 1+ ρ 1 + κ 2 − θ 1 + y 1+ ρ 2 + κ 1 − θ 2 ) / ( y r 2 +1 − θ 2 x 1+ ρ 1 − θ 1 ) ≥ p 1 /p q 1 /q y − δ . It follo ws that for m ≥ 1 and 0 < x, y ≤ m − 1 , y θ 2 − 1 − ρ 2 x κ 2 + x θ 1 − 1 − ρ 1 y κ 1 = y r 2 − ρ 2 ( x 1+ ρ 1 + κ 2 − θ 1 + y 1+ ρ 2 + κ 1 − θ 2 ) / ( y r 2 +1 − θ 2 x 1+ ρ 1 − θ 1 ) ≥ p 1 /p q 1 /q y r 2 − ρ 2 y − δ ≥ p 1 /p q 1 /q m δ y r 2 − ρ 2 , whic h completes the pro of. 2 The following Lemmas 3.3 and 3.4 are used to establish Lemma 3.6 in the next subsubsection and the k ey to the pro ofs is to apply Lemma 3.1(i) again. F or θ 1 − 1 < r 1 and θ 2 − 1 < r 2 , under (1.7), there are constan ts ρ 1 , ρ 2 > θ 1 ∨ θ 2 suc h that r 1 + 1 − θ 1 κ 1 = κ 2 r 2 + 1 − θ 2 = 1 + ρ 1 + κ 2 − θ 1 1 + ρ 2 + κ 1 − θ 2 . (3.8) Lemma 3.3 Recalling a i and b i deﬁned in Section 1, supp ose that r 1 > θ 1 − 1 , r 2 > θ 2 − 1 and ( a 1 /b 1 ) 1 / ( r 1 +1 − θ 1 ) ( a 2 /b 2 ) 1 /κ 2 ≥ 1 , (3.9) and that (3.8) holds for ρ 1 , ρ 2 > θ 1 ∨ θ 2 . Then there is a constan t δ 0 > 0 suc h that a 1 ρ 1 δ 0 x θ 1 − 1 − ρ 1 y κ 1 + a 2 ρ 2 x κ 2 y θ 2 − 1 − ρ 2 ≥ b 1 ρ 1 δ 0 x r 1 − ρ 1 + b 2 ρ 2 y r 2 − ρ 2 , x, y > 0 . Pr o of. Let p 1 := 1 + ρ 2 + κ 1 − θ 2 1 + ρ 2 − θ 2 , q 1 := 1 + ρ 2 + κ 1 − θ 2 κ 1 , p 2 := 1 + ρ 1 + κ 2 − θ 1 κ 2 , q 2 := 1 + ρ 1 + κ 2 − θ 1 1 + ρ 1 − θ 1 . Then p 1 , q 1 , p 2 , q 2 > 1, p − 1 1 + q − 1 1 = 1 and p − 1 2 + q − 1 2 = 1. In view of (3.8), we obtain q 1 = (1 + ρ 1 + κ 2 − θ 1 ) / ( r 1 + 1 − θ 1 ). Combining this with (3.9) w e ha ve ( b 1 /a 1 ) q 1 ( b 2 /a 2 ) p 2 ≤ 1, which implies that ( b 1 ρ 1 ) q 1 ( b 2 ρ 2 ) p 2 ≤ ( a 1 ρ 1 ) q 1 ( a 2 ρ 2 ) p 2 = ( a 1 ρ 1 ) q 1 /p 1 ( a 2 ρ 2 ) p 2 /q 2 a 1 ρ 1 a 2 ρ 2 and then ( b 2 ρ 2 ) p 2 ( a 2 ρ 2 ) p 2 /q 2 a 1 ρ 1 p 2 /q 1 ≤ ( a 1 ρ 1 ) q 1 /p 1 a 2 ρ 2 q 1 /p 2 ( b 1 ρ 1 ) q 1 . Th us there is a constant δ 0 > 0 satisfying ( b 2 ρ 2 ) p 2 ( a 2 ρ 2 ) p 2 /q 2 a 1 ρ 1 p 2 /q 1 ≤ δ 0 ≤ ( a 1 ρ 1 ) q 1 /p 1 a 2 ρ 2 q 1 /p 2 ( b 1 ρ 1 ) q 1 . 12 It follo ws that ( a 1 ρ 1 δ 0 ) 1 /p 1 ( a 2 ρ 2 q 1 /p 2 ) 1 /q 1 ≥ b 1 ρ 1 δ 0 , ( a 1 ρ 1 δ 0 p 2 /q 1 ) 1 /p 2 ( a 2 ρ 2 ) 1 /q 2 ≥ b 2 ρ 2 . (3.10) By (3.8) again, 1 q 1 = κ 1 1 + ρ 2 + κ 1 − θ 2 = r 1 + 1 − θ 1 1 + ρ 1 + κ 2 − θ 1 , 1 p 2 = κ 2 1 + ρ 1 + κ 2 − θ 1 = r 2 + 1 − θ 2 1 + ρ 2 + κ 1 − θ 2 , and then ( r 1 + 1 − θ 1 ) − 1 + ρ 1 + κ 2 − θ 1 q 1 = 0 , 1 + ρ 2 + κ 1 − θ 2 p 1 − (1 + ρ 2 − θ 2 ) = 0 , ( r 2 + 1 − θ 2 ) − 1 + ρ 2 + κ 1 − θ 2 p 2 = 0 , 1 + ρ 1 + κ 2 − θ 1 q 2 − (1 + ρ 1 − θ 1 ) = 0 . No w by Lemma 3.1(i) and (3.10), p − 1 1 a 1 ρ 1 δ 0 y 1+ ρ 2 + κ 1 − θ 2 + p − 1 2 a 2 ρ 2 x 1+ ρ 1 + κ 2 − θ 1 ≥ p 1 /p 1 1 q 1 /q 1 1 ( p − 1 1 a 1 ρ 1 δ 0 y 1+ ρ 2 + κ 1 − θ 2 ) 1 /p 1 ( p − 1 2 a 2 ρ 2 x 1+ ρ 1 + κ 2 − θ 1 ) 1 /q 1 = ( a 1 ρ 1 δ 0 ) 1 /p 1 ( a 2 ρ 2 q 1 /p 2 ) 1 /q 1 x r 1 − 1 − θ 1 y 1+ ρ 2 − θ 2 ≥ b 1 ρ 1 δ 0 x r 1 − 1 − θ 1 y 1+ ρ 2 − θ 2 and q − 1 1 a 1 ρ 1 δ 0 y 1+ ρ 2 + κ 1 − θ 2 + q − 1 2 a 2 ρ 2 x 1+ ρ 1 + κ 2 − θ 1 ≥ p 1 /p 2 2 q 1 /q 2 2 ( q − 1 1 a 1 ρ 1 δ 0 y 1+ ρ 2 + κ 1 − θ 2 ) 1 /p 2 ( q − 1 2 a 2 ρ 2 x 1+ ρ 1 + κ 2 − θ 1 ) 1 /q 2 = ( a 1 ρ 1 δ 0 p 2 /q 1 ) 1 /p 2 ( a 2 ρ 2 ) 1 /q 2 y r 2 − 1 − θ 2 x 1+ ρ 1 − θ 1 ≥ b 2 ρ 2 y r 2 − 1 − θ 2 x 1+ ρ 1 − θ 1 . Then one can conclude the pro of. 2 Similar to Lemma 3.3 w e hav e the follo wing lemma. Lemma 3.4 If b oth the assumption of Theorem 1.6(ii) and (1.7) hold, and r > θ − 1 , then there is a constan t δ 0 > 0 suc h that δ 0 a 1 y 1+ κ − θ + a 2 x 1+ κ − θ ≥ δ 0 b 1 x κ y 1 − θ + b 2 y κ x 1 − θ , x, y > 0 . Pr o of. Let p := (1 + κ − θ ) / (1 − θ ) and q := p/ ( p − 1). As the essential same argumen t in the pro of of Lemma 3.3, we obtain p − 1 δ 0 a 1 x θ − 1 y κ + p − 1 a 2 x κ y θ − 1 ≥ δ 0 b 1 x κ + θ − 1 , q − 1 δ 0 a 1 y κ x θ − 1 + q − 1 a 2 y θ − 1 x κ ≥ b 2 y κ + θ − 1 for some δ 0 > 0, whic h implies the assertion. 2 3.1.2 Pro ofs of Prop osition 1.2 and Theorems 1.3–1.4 and 1.6 W e next apply Prop osition 2.1 to complete the pro ofs of Theorems 1.3–1.4 and 1.6, where the follo wing Lemmas 3.5 and 3.6 are applied to v alidate condition (ii) of Prop osition 2.1. 13 Lemma 3.5 Given g ( x, y ) := x − ρ 1 + y − ρ 2 , x, y > 0 , if the assumptions of either Theorem 1.3 or 1.4 hold, then there are constan ts ρ 1 , ρ 2 , C n > 0 suc h that L g ( x, y ) ≤ C n g ( x, y ) for all 0 < x, y ≤ n and n ≥ 1 . Pr o of. (i) W e ﬁrst prov e the result under the assumptions of Theorem 1.3. Since the pro ofs are similar, we only present that under condition (i) of Theorem 1.3. Observe that there are constan ts ρ 1 , ρ 2 > θ 1 ∨ θ 2 suc h that (3.7) holds. By (3.4) and (3.6), Z ∞ 0 [( x + z ) − ρ i − x − ρ i + ρ i z x − ρ i ] µ i (d z ) = c i ρ i ( ρ i + 1) x − α i − ρ i (3.11) with c i := c i ( − ρ i ). Then b y (3.5), for all 0 < x, y ≤ n , L g ( x, y ) = − a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + ρ 1 b 10 x r 10 − 1 − ρ 1 + b 11 ρ 1 ( ρ 1 + 1) x r 11 − 2 − ρ 1 + c 1 ρ 1 ( ρ 1 + 1) b 12 x r 12 − α 1 − ρ 1 + ρ 2 b 20 y r 20 − 1 − ρ 2 + b 21 ρ 2 ( ρ 2 + 1) y r 21 − 2 − ρ 2 + c 2 ρ 2 ( ρ 2 + 1) b 22 y r 22 − α 2 − ρ 2 ≤ − a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + ρ 1 ( ρ 1 + 1) x − ρ 1  b 10 x r 10 − 1 + b 11 x r 11 − 2 + b 12 c 1 x r 12 − α 1  + ρ 2 ( ρ 2 + 1) y − ρ 2  b 20 y r 20 − 1 + b 21 y r 21 − 2 + b 22 c 2 y r 22 − α 2  ≤ − a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + C 1 ,n x r 1 − ρ 1 + C 2 ,n y r 2 − ρ 2 (3.12) with C i,n := ρ i ( ρ i + 1)[ b i 0 n r i 0 − 1 − r i + b i 1 n r i 1 − 2 − r i + b i 2 n r i 2 − α i − r i ] for i = 1 , 2. By (3.7) and Lemma 3.2, − a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + C 2 ,n y r 2 − ρ 2 ≤ − [( a 1 ρ 1 ) ∧ ( a 2 ρ 2 )] · [ x θ 1 − 1 − ρ 1 y κ 1 + y θ 2 − 1 − ρ 2 x κ 2 ] + C 2 ,n y r 2 − ρ 2 ≤  C 2 ,n − C [( a 1 ρ 1 ) ∧ ( a 2 ρ 2 )] m δ  y r 2 − ρ 2 ≤ 0 , 0 < x, y ≤ m − 1 (3.13) for all large enough m , where δ , C > 0 are the constants determined in Lemma 3.2. F rom (3.12) and (3.13) it follo ws that L g ( x, y ) ≤ − a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + C 2 ,n y r 2 − ρ 2 + C 1 ,n n r 1 g ( x, y ) ≤ C 1 ,n n r 1 g ( x, y ) , 0 < x, y ≤ m − 1 (3.14) for all large enough m . Under the assumptions in condition (i), b y (3.12), we hav e L g ( x, y ) ≤ C 1 ,n x r 1 − ρ 1 + C 2 ,n y r 2 − ρ 2 ≤ [ C 1 ,n n r 1 + C 2 ,n m − r 2 ] g ( x, y ) (3.15) for all m − 1 ≤ y ≤ n and 0 < x ≤ n , and L g ( x, y ) ≤ − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + C 1 ,n x r 1 − ρ 1 + C 2 ,n y r 2 − ρ 2 ≤ y r 2 − ρ 2 [ C 2 ,n − a 2 ρ 2 y θ 2 − 1 − r 2 x κ 2 ] + C 1 ,n n r 1 g ( x, y ) ≤ y r 2 − ρ 2 [ C 2 ,n − a 2 ρ 2 m r 2 +1 − θ 2 0 m − κ 2 ] + C 1 ,n n r 1 g ( x, y ) ≤ C 1 ,n n r 1 g ( x, y ) , m − 1 ≤ x ≤ n, 0 < y ≤ m − 1 0 (3.16) for m ≥ 1 and large enough m 0 ≥ 1. Observe that L g ( x, y ) ≤ C n g ( x, y ) for some C n > 0 and all m − 1 ≤ x ≤ n and m − 1 0 ≤ y ≤ n . Combining (3.14)–(3.16) we get the assertion. (ii) No w we show the result under the assumptions of Theorem 1.4. Under (1.4), there are constan ts ρ 1 , ρ 2 > θ 1 ∨ θ 2 suc h that r 1 − 1 − θ 1 κ 1 > 1 + ρ 1 + κ 2 − θ 1 1 + ρ 2 + κ 1 − θ 2 > κ 2 r 2 − 1 − θ 2 . 14 Then b y Lemma 3.2 and the same arguments as in (3.13), we ha ve − a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 − a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 + C 1 ,n x r 1 − ρ 1 + C 2 ,n y r 2 − ρ 2 ≤ 0 , 0 < x, y ≤ m − 1 for large enough m ≥ 1. By arguments similar to those in (3.15) and (3.16), we ha ve L g ( x, y ) ≤ C n,m g ( x, y ) for some constan t C n,m > 0 for large enough m ≥ 1 when m − 1 ≤ x ≤ n and 0 < y ≤ n , or m − 1 ≤ y ≤ n and 0 < x ≤ n . Now L g ( x, y ) ≤ C n,m g ( x, y ) for 0 < x, y ≤ n by (3.12). 2 Lemma 3.6 Supp ose that the assumptions of Theorem 1.6 hold. Under the conditions of Theorem 1.6(i) let ˜ g ( x, y ) := δ 0 x − ρ 1 + y − ρ 2 , x, y > 0 , and under the conditions of Theorem 1.6(ii) deﬁne a nonnegativ e function ˜ g ∈ C 2 ((0 , ∞ ) × (0 , ∞ )) suc h that ˜ g ( x, y ) := ( δ 0 + 1) ln n + δ 0 ln x − 1 + ln y − 1 , 0 < x, y < n. Then there are constants ρ 1 , ρ 2 , δ 0 , C n > 0 such that L ˜ g ( x, y ) ≤ C n ˜ g ( x, y ) for all 0 < x, y < n and n ≥ 1 . Pr o of. (i) W e ﬁrst pro v e it under the conditions in Theorem 1.6(i). Under (1.7), there are constan ts ρ 1 , ρ 2 > θ 1 ∨ θ 2 suc h that (3.8) holds. Let δ 0 > 0 b e the constant determined in the follo wing. Then b y (3.5) and (3.11), we obtain L ˜ g ( x, y ) = − J 1 ( x, y ) + J 2 ( x, y ) + J 3 ( x, y ) with J 1 ( x, y ) := δ 0 a 1 ρ 1 x θ 1 − 1 − ρ 1 y κ 1 + a 2 ρ 2 y θ 2 − 1 − ρ 2 x κ 2 , J 2 ( x, y ) := δ 0 b 10 ρ 1 x r 10 − 1 − ρ 1 + b 20 ρ 2 y r 20 − 1 − ρ 2 = δ 0 b 1 ρ 1 x r 1 − ρ 1 + b 2 ρ 2 y r 2 − ρ 2 under Condition 1.5(i) and J 3 ( x, y ) := δ 0 b 11 ρ 1 ( ρ 1 + 1) x r 11 − 2 − ρ 1 + b 12 δ 0 c 1 ( − ρ 1 ) ρ 1 ( ρ 1 + 1) x r 12 − α 1 − ρ 1 + b 21 ρ 2 ( ρ 2 + 1) y r 21 − 2 − ρ 2 + b 22 c 2 ( − ρ 2 ) ρ 2 ( ρ 2 + 1) y r 22 − α 2 − ρ 2 ≤ c 0 [ x r 11 − 2 + x r 12 − α 1 ] x − ρ 1 + c 0 [ y r 21 − 2 + y r 22 − α 2 ] y − ρ 2 for some constan t c 0 > 0. Under (1.8), there is a small enough constan t ε > 0 such that h a 1 (1 + ε ) b 1 i 1 / ( r 1 +1 − θ 1 ) h a 2 (1 + ε ) b 2 i 1 /κ 2 ≥ 1 . By Lemma 3.3, there is a constan t δ 0 > 0 such that (1 + ε ) J 2 ( x, y ) ≤ J 1 ( x, y ) for all x, y > 0. By Condition 1.5(i) again, there is a constan t c n > 0 suc h that − εδ 0 b 10 ρ 1 x r 10 − 1 + c 0 [ x r 11 − 2 + x r 12 − α 1 ] ≤ c n δ 0 , − εb 20 ρ 2 y r 20 − 1 + c 0 [ y r 21 − 2 + y r 22 − α 2 ] ≤ c n for all 0 < x, y ≤ n . Thus, for all 0 < x, y ≤ n , J 3 ( x, y ) − εJ 2 ( x, y ) ≤ c n ˜ g ( x, y ) and we hav e L ˜ g ( x, y ) ≤ c n ˜ g ( x, y ). (ii) W e then pro ve the result under the conditions in Theorem 1.6(ii). Let θ := θ 1 and κ := κ 1 . Then r = κ + θ − 1. By Lemma 3.4, there is a constant δ 0 > 0 suc h that I ( x, y ) := δ 0 a 1 y 1+ κ − θ + a 2 x 1+ κ − θ − δ 0 b 1 x κ y 1 − θ − b 2 y κ x 1 − θ ≥ 0 , x, y > 0 . 15 It th us follows from (3.5) that L ˜ g ( x, y ) = − δ 0 a 1 x θ − 1 y κ − a 2 y θ − 1 x κ + δ 0 ( b 10 x r 10 − 1 + b 11 x r 11 − 2 ) +( b 20 y r 20 − 1 + b 21 y r 21 − 2 ) = − x θ − 1 y θ − 1 I ( x, y ) ≤ 0 , whic h ends the pro of. 2 No w we are ready to complete the pro ofs of Prop osition 1.2 and Theorems 1.3, 1.4 and 1.6. Pr o of of Pr op osition 1.2. Let g b e the test function given in Lemma 3.5. By (3.12), L g ( x, y ) ≤ C 1 ,n x r 1 − ρ 1 + C 2 ,n y r 2 − ρ 2 ≤ [ C 1 ,n x r 1 + C 2 ,n y r 2 ] g ( x, y ) ≤ [ C 1 ,n n r 1 + C 2 ,n n r 2 ] g ( x, y ) for some constants C 1 ,n , C 2 ,n ≥ 0 and all 0 < x, y ≤ n . The conclusion follo ws from Prop osition 2.1. 2 Pr o of of The or ems 1.3 and 1.4. Let g b e the test function determined in Lemma 3.5. Then b y Prop osition 2.1 and Lemma 3.5 one completes the pro ofs. 2 Pr o of of The or em 1.6. Let ˜ g b e the test function deﬁned in Lemma 3.6. Then one completes the pro of combining Prop osition 2.1 and Lemma 3.6. 2 3.2 Proofs of Theorems 1.7, 1.10 and 1.11 In this subsection, w e use Prop osition 2.2 to establish the pro ofs of Theorems 1.7, 1.10 and 1.11. F or the pro of of Theorem 1.7, the test function g is tak en as a pow er function with p ositiv e pow er (see (3.18)). F or the pro ofs of Theorems 1.10 and 1.11, the test functions g ( x, y ) := c + h ( x, y ) for some constan t c > 0 (see (3.82)) and h are mo diﬁcations of − ( x ρ 1 + y ρ 2 ) ρ in diﬀeren t cases. The k ey to applying Prop osition 2.2 is v erifying assumption (iii) of Prop osition 2.2. F or the case of dominant drifts (Condition 1.5), the function h is c hosen as h ( x, y ) := − ( x ρ 1 + y ρ 2 ) ρ for ρ 1 , ρ 2 > 1 and 0 < ρ < 1 and the estimation of L h is the k ey to verifying assumption (iii) of Prop osition 2.2, which are giv en in Subsubsection 3.2.3. F or the case of dominant diﬀusion or jump terms (assumption (1.14)), the functions h are mo diﬁcations of − ( x ρ 1 + y ) ρ or − ( x + y ρ 2 ) ρ for ρ 1 , ρ 2 > 0 and 0 < ρ < 1 and the estimation of L h is also the key in c hecking assumption (iii) of Prop osition 2.2, which are given in Subsubsection 3.2.4. In Subsubsection 3.2.1, the pro of of Theorem 1.7 is presented. Three lemmas in Subsubsection 3.2.2 are used to estimate L h . The pro ofs of Theorems 1.10 and 1.11 are given at the end of Subsubsection 3.2.4. 3.2.1 Pro of of Theorem 1.7 In this subsubsection, w e pro ve Theorem 1.7 by applying Prop osition 2.2 to the test function g giv en by (3.18). Pr o of of The or em 1.7. Since the proofs are similar, w e only presen t that under condition (i). Let 0 < ρ < 1 ∧ ( − r 1 ) b e ﬁxed as follows. Since r 1 ≤ θ 1 − 1 and κ 1 > 0, there are constants 0 < v < 1 and d > 0 such that 2 − 1 b 1 (1 − ρ )[1 ∧ c 1 ( ρ )] x r 1 +1 − θ 1 − a 1 y κ 1 ≥ 0 , 2 − 1 b 1 ρ (1 − ρ )[1 ∧ c 1 ( ρ )] x ρ + r 1 − a 2 ≥ dv (3.17) for all 0 < x, y ≤ v , where c 1 ( ρ ) is giv en by (3.6). Let g ( x, y ) := v − x ρ − y , x, y > 0 . (3.18) 16 Then sup x,y > 0 g ( x, y ) ≤ v , and g ( x, y ) ≤ 0 for all x ≥ v or y ≥ v . By (3.6), x − ρ + α 1 Z ∞ 0 [( x + z ) ρ − x ρ − ρz x ρ − 1 ] µ 1 (d z ) = Z ∞ 0 [(1 + z ) ρ − 1 − ρz ] µ 1 (d z ) = ρ ( ρ − 1) c 1 ( ρ ) . Then it follo ws from (3.5) and (3.17) that L g ( x, y ) = − a 1 ρx ρ + θ 1 − 1 y κ 1 − a 2 y θ 2 x κ 2 + b 20 y r 20 + b 10 ρx ρ + r 10 − 1 + b 11 ρ (1 − ρ ) x ρ + r 11 − 2 + b 12 c 1 ( ρ ) ρ (1 − ρ ) x ρ + r 12 − α 1 ≥ − a 1 ρx ρ + θ 1 − 1 y κ 1 − a 2 y θ 2 x κ 2 + b 1 ρ (1 − ρ )[1 ∧ c 1 ( ρ )] x ρ + r 1 ≥ 2 − 1 b 1 ρ (1 − ρ )[1 ∧ c 1 ( ρ )] x ρ + r 1 − a 2 + ρx ρ + θ 1 − 1 h 2 − 1 b 1 (1 − ρ )[1 ∧ c 1 ( ρ )] x r 1 +1 − θ 1 − a 1 y κ 1 i ≥ dv ≥ dg ( x, y ) , 0 < x, y ≤ v . Using Prop osition 2.2 one gets P { τ 0 < ∞} ≥ g ( X 0 , Y 0 ) /v for 0 < X 0 , Y 0 < (2 − 1 v ) 1 /ρ . In general, X 0 , Y 0 > 0, w e conclude the pro of b y applying the strong Marko v property and the same argumen ts at the end of the pro of of [34, Theorem 1.3]. 2 3.2.2 Preliminaries The following Lemmas 3.7–3.9 are needed to estimate L h in Lemmas 3.10 and 3.11 with h deﬁned b y (3.36) and (3.49). Lemma 3.1(iii) and (i) are applied to show Lemmas 3.7 and 3.8, resp ectiv ely . Lemma 3.9 contains an estimate of the integral R ∞ 0 K i z h ( x, y ) µ i (d z ) for i = 1 , 2 in (3.5), whic h is applied to prov e Lemma 3.12. Lemma 3.7 Supp ose that r 1 + 1 − θ 1 κ 1 < r 1 + ρ 1 r 2 + ρ 2 < κ 2 r 2 + 1 − θ 2 (3.19) for θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 and ρ 1 , ρ 2 ≥ 1 . Then for any constants c 1 , c 2 , c 3 , c 4 > 0 there is a constan t 0 < c < 1 such that for all 0 < x, y ≤ c , c 1 x r 1 + ρ 1 + c 2 y r 2 + ρ 2 − c 3 x θ 1 − 1+ ρ 1 y κ 1 − c 4 y θ 2 − 1+ ρ 2 x κ 2 ≥ 0 . Pr o of. Under (3.19), we ha ve θ 1 − 1 + ρ 1 r 1 + ρ 1 + κ 1 r 2 + ρ 2 > 1 , κ 2 r 1 + ρ 1 + θ 2 − 1 + ρ 2 r 2 + ρ 2 > 1 . Th us, by Lemma 3.1(iii), there is a constant 0 < c < 1 such that 2 − 1 c 1 x r 1 + ρ 1 + 2 − 1 c 2 y r 2 + ρ 2 ≥ c 3 x θ 1 − 1+ ρ 1 y κ 1 , 2 − 1 c 1 x r 1 + ρ 1 + 2 − 1 c 2 y r 2 + ρ 2 ≥ c 4 y θ 2 − 1+ ρ 2 x κ 2 for all 0 < x, y ≤ c . This concludes the pro of. 2 Lemma 3.8 Supp ose that  a 1 (1 − ε 0 ) b 1  1 / ( r 1 +1 − θ 1 )  a 2 (1 − ε 0 ) b 2  1 /κ 2 < 1 (3.20) 17 for θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 and ε 0 ∈ (0 , 1) , and in addition,  a 1 / ˜ b 1  1 / ( r 1 +1 − θ 1 )  a 2 / ˜ b 2  1 /κ 2 < 1 (3.21) and r 1 + 1 − θ 1 κ 1 = r 1 + ρ 1 r 2 + ρ 2 = κ 2 r 2 + 1 − θ 2 , (3.22) for ρ 1 , ρ 2 > 1 , 0 < ρ < ρ − 1 1 ∧ ρ − 1 2 and ˜ b i := (1 − ε 0 )(1 − ρ i ρ ) b i , i = 1 , 2 . Then there is δ 0 > 0 suc h that δ 0 ˜ b 1 ρ 1 x r 1 + ρ 1 + ˜ b 2 ρ 2 y r 2 + ρ 2 > δ 0 a 1 ρ 1 x θ 1 + ρ 1 − 1 y κ 1 + a 2 ρ 2 y θ 2 + ρ 2 − 1 x κ 2 . Pr o of. Let p 1 := r 1 + ρ 1 θ 1 − 1 + ρ 1 , q 1 := r 2 + ρ 2 κ 1 , p 2 := r 1 + ρ 1 κ 2 , q 2 := r 2 + ρ 2 θ 2 − 1 + ρ 2 . By (3.22) w e hav e 1 q 1 = κ 1 r 2 + ρ 2 = r 1 + 1 − θ 1 r 1 + ρ 1 , 1 p 2 = κ 2 r 1 + ρ 1 = r 2 + 1 − θ 2 r 2 + ρ 2 . (3.23) It follo ws that p − 1 1 + q − 1 1 = 1 , p − 1 2 + q − 1 2 = 1 , (3.24) whic h implies that p 1 , q 1 , p 2 , q 2 > 1. Under (3.21) and (3.22), w e hav e 1 > [( a 1 / ˜ b 1 ) 1 / ( r 1 +1 − θ 1 ) ( a 2 / ˜ b 2 ) 1 /κ 2 ] r 1 + ρ 1 = ( a 1 / ˜ b 1 ) ( r 2 + ρ 2 ) /κ 1 ( a 2 / ˜ b 2 ) ( r 1 + ρ 1 ) /κ 2 = a q 1 1 a p 2 2 / ( ˜ b q 1 1 ˜ b p 2 2 ) , where (3.23) is used in the last equalit y . Then ( a 1 ρ 1 ) q 1 ( a 2 ρ 2 ) p 2 < ( ˜ b 1 ρ 1 ) q 1 ( ˜ b 2 ρ 2 ) p 2 = ( ˜ b 1 ρ 1 ) q 1 /p 1 ˜ b 2 ρ 2 · ( ˜ b 2 ρ 2 ) p 2 /q 2 ˜ b 1 ρ 1 , whic h implies that ( a 2 ρ 2 ) p 2 ( ˜ b 2 ρ 2 ) p 2 /q 2 ˜ b 1 ρ 1 p 2 /q 1 < ( ˜ b 1 ρ 1 ) q 1 /p 1 ˜ b 2 ρ 2 q 1 /p 2 ( a 1 ρ 1 ) q 1 . Then there is a constan t δ 0 > 0 suc h that that ( a 2 ρ 2 ) p 2 ( ˜ b 2 ρ 2 ) p 2 /q 2 ˜ b 1 ρ 1 p 2 /q 1 < δ 0 < ( ˜ b 1 ρ 1 ) q 1 /p 1 ˜ b 2 ρ 2 q 1 /p 2 ( a 1 ρ 1 ) q 1 . Therefore, ( q 1 /p 2 ) 1 /q 1 ( δ 0 ˜ b 1 ρ 1 ) 1 /p 1 ( ˜ b 2 ρ 2 ) 1 /q 1 > δ 0 a 1 ρ 1 , ( p 2 /q 1 ) 1 /p 2 ( δ 0 ˜ b 1 ρ 1 ) 1 /p 2 ( ˜ b 2 ρ 2 ) 1 /q 2 > a 2 ρ 2 . No w by Lemma 3.1(i) and (3.23) again p − 1 1 δ 0 ˜ b 1 ρ 1 x r 1 + ρ 1 + p − 1 2 ˜ b 2 ρ 2 y r 2 + ρ 2 18 ≥ p 1 /p 1 1 q 1 /q 1 1 ( p − 1 1 δ 0 ˜ b 1 ρ 1 x r 1 + ρ 1 ) 1 /p 1 ( p − 1 2 ˜ b 2 ρ 2 y r 2 + ρ 2 ) 1 /q 1 = ( q 1 /p 2 ) 1 /q 1 ( δ 0 ˜ b 1 ρ 1 ) 1 /p 1 ( ˜ b 2 ρ 2 ) 1 /q 1 x θ 1 + ρ 1 − 1 y κ 1 > δ 0 a 1 ρ 1 x θ 1 + ρ 1 − 1 y κ 1 and q − 1 1 δ 0 ˜ b 1 ρ 1 x r 1 + ρ 1 + q − 1 2 ˜ b 2 ρ 2 y r 2 + ρ 2 ≥ p 1 /p 2 2 q 1 /q 2 2 ( q − 1 1 δ 0 ˜ b 1 ρ 1 x r 1 + ρ 1 ) 1 /p 2 ( q − 1 2 ˜ b 2 ρ 2 y r 2 + ρ 2 ) 1 /q 2 = ( p 2 /q 1 ) 1 /p 2 ( δ 0 ˜ b 1 ρ 1 ) 1 /p 2 ( b 2 ρ 2 ) 1 /q 2 y θ 2 + ρ 2 − 1 x κ 2 > a 2 ρ 2 y θ 2 + ρ 2 − 1 x κ 2 . This completes the pro of. 2 T o estimate the integral R ∞ 0 K i z h ( x, y ) µ i (d z ) for i = 1 , 2 in (3.5) with the function h deﬁned in (3.36), (3.49) and in Lemmas 3.12 and 3.14 for diﬀerent cases, w e introduce functions K ( v , z ) and M ( x, y ) in the follo wing. Giv en 0 < ρ < 1 and ρ 1 ≥ 1, for 0 ≤ v ≤ 1 and z > 0 let K ( v , z ) := −  v [(1 + z ) ρ 1 − 1] + 1  ρ + 1 + z v ρρ 1 . (3.25) Then for M ( x, y ) := − ( x ρ 1 + y ) ρ , w e hav e K 1 z x M ( x, y ) = ( x ρ 1 + y ) ρ K ( x ρ 1 x ρ 1 + y , z ). It th us follows from (3.4) that Z ∞ 0 K 1 z M ( x, y ) µ 1 (d z ) = x − α 1 Z ∞ 0 K 1 z x M ( x, y ) µ 1 (d z ) = x − α 1 ( x ρ 1 + y ) ρ Z ∞ 0 K ( x ρ 1 x ρ 1 + y , z ) µ 1 (d z ) . (3.26) In the follo wing we give an estimate on R ∞ 0 K ( v , z ) µ 1 (d z ). The estimate mainly relies on T a ylor’s form ula in (3.3) and change of v ariable similar to (3.4). Lemma 3.9 (i) There is a constant ρ 0 ≥ 2 suc h that for all ρ 1 > ρ 0 , all ˜ ρ ∈ (0 , 1) and ρ := ˜ ρ/ρ 1 , Z ∞ 0 K ( v , z ) µ 1 (d z ) ≥ ρ (1 − ρ ) ρ 2 1 v 2 d 1 − ρρ 1 ( ρ 1 − 1) v ˜ d 1 (3.27) for some constan ts d 1 , ˜ d 1 > 0 . (ii) F or an y ρ 1 > 1 and 0 < ρρ 1 < 1 , and for all δ > 0 we hav e Z ∞ 0 K ( v , z ) µ 1 (d z ) ≥ ρρ 1 (1 − ρρ 1 ) c 1 ( ρρ 1 ) v 2 − ρρ 1 ( ρ 1 − 1)[ v (1 − v ) d 1 ,δ + d 2 ,δ v ρ ] , (3.28) where d 1 ,δ := R δ 0 z 2 (1 + z ) ρ 1 − 2 µ 1 (d z ) , d 2 ,δ := R ∞ δ z 2 µ 1 (d z ) R 1 0 (1 + uz ) ρρ 1 − 2 d u and function c 1 is deﬁned in (3.6). Pr o of. (i) Observe that K ′′ z z ( v , z ) = ρ (1 − ρ ) ρ 2 1 v 2 (1 + z ) 2 ρ 1 − 2  v [(1 + z ) ρ 1 − 1] + 1  ρ − 2 − ρρ 1 ( ρ 1 − 1) v (1 + z ) ρ 1 − 2  v [(1 + z ) ρ 1 − 1] + 1  ρ − 1 =: ρ (1 − ρ ) ρ 2 1 H 1 ( v , z ) − ρρ 1 ( ρ 1 − 1) H 2 ( v , z ) . (3.29) Then H 1 ( v , z ) ≥ v 2 (1 + z ) ρρ 1 − 2 and H 2 ( v , z ) ≤ v (1 + z ) ρ 1 − 2 for 0 ≤ v ≤ 1. Now by (3.3), Z 1 0 K ( v , z ) µ 1 (d z ) = Z 1 0 z 2 µ 1 (d z ) Z 1 0 K ′′ z z ( v , z u )(1 − u )d u 19 ≥ ρ (1 − ρ ) ρ 2 1 v 2 Z 1 0 z 2 µ 1 (d z ) Z 1 0 (1 + z u ) ρρ 1 − 2 (1 − u )d u − ρρ 1 ( ρ 1 − 1) v Z 1 0 z 2 µ 1 (d z ) Z 1 0 (1 + z u ) ρ 1 − 2 d u =: ρ (1 − ρ ) ρ 2 1 v 2 d 1 − ρρ 1 ( ρ 1 − 1) v ¯ d 1 , 0 ≤ v ≤ 1 . (3.30) Substituting z by z v − 1 /ρ 1 w e obtain Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 H 1 ( v , z u )(1 − u )d u = v ( α 1 − 2) /ρ 1 Z ∞ v 1 /ρ 1 z 2 µ 1 (d z ) Z 1 0 H 1 ( v , z v − 1 /ρ 1 u )(1 − u )d u =: v α 1 /ρ 1 ˜ H 1 ( v ) and Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 H 2 ( v , z u )(1 − u )d u = v ( α 1 − 2) /ρ 1 Z ∞ v 1 /ρ 1 z 2 µ 1 (d z ) Z 1 0 H 2 ( v , z uv − 1 /ρ 1 )(1 − u )d u =: v α 1 /ρ 1 ˜ H 2 ( v ) . It is elemen tary to see that lim v → 0 ˜ H 1 ( v ) = Z ∞ 0 z 2 µ 1 (d z ) Z 1 0 ¯ H 1 ,ρ 1 ( z u )(1 − u )d u and lim v → 0 ˜ H 2 ( v ) = Z ∞ 0 z 2 µ 1 (d z ) Z 1 0 ¯ H 2 ,ρ 1 ( z u )(1 − u )d u, where ¯ H 1 ,ρ 1 ( z ) := z 2 ρ 1 − 2 ( z ρ 1 + 1) ρ − 2 and ¯ H 2 ,ρ 1 ( z ) := z ρ 1 − 2 ( z ρ 1 + 1) ρ − 1 . Then Z 1 0 z 2 µ 1 (d z ) Z 1 0 ¯ H 2 ,ρ 1 ( z u )(1 − u )d u ≤ Z 1 0 z ρ 1 µ 1 (d z ) = α 1 ( α 1 − 1) ( ρ 1 − α 1 )Γ( α 1 )Γ(2 − α 1 ) and Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 ¯ H 2 ,ρ 1 ( z u )(1 − u )1 { z u ≤ 1 } d u ≤ Z ∞ 1 z ρ 1 µ 1 (d z ) Z z − 1 0 u ρ 1 − 2 d u = ( ρ 1 − 1) − 1 Z ∞ 1 z µ 1 (d z ) . By the dominated con vergence, for ρρ 1 = ˜ ρ ∈ (0 , 1), lim ρ 1 →∞ Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 ¯ H 1 ,ρ 1 ( z u )(1 − u )1 { z u> 1 } d u = Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 ( z u ) ˜ ρ − 2 (1 − u )1 { z u> 1 } d u = Z 1 0 u ˜ ρ − 2 (1 − u )d u Z ∞ u − 1 z ˜ ρ µ 1 (d z ) = α 1 ( α 1 − 1) ( α 1 − ˜ ρ )Γ( α 1 )Γ(2 − α 1 ) Z 1 0 u α 1 − 2 (1 − u )d u and lim ρ 1 →∞ Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 ¯ H 2 ,ρ 1 ( z u )(1 − u )1 { z u> 1 } d u 20 = α 1 ( α 1 − 1) ( α 1 − ˜ ρ )Γ( α 1 )Γ(2 − α 1 ) Z 1 0 u α 1 − 2 (1 − u )d u. Since ρ (1 − ρ ) ρ 2 1 − ρρ 1 ( ρ 1 − 1) = ρρ 1 (1 − ρρ 1 ) > 0, then there is a constant ρ 0 > 0 such that for all ρ 1 > ρ 0 , lim v → 0 ρ (1 − ρ ) ρ 2 1 ˜ H 1 ( v ) > lim v → 0 ρρ 1 ( ρ 1 − 1) ˜ H 2 ( v ). Now there is a constant v 0 := v 0 ( ρ 1 ) ∈ (0 , 1) suc h that we hav e ρ (1 − ρ ) ρ 2 1 ˜ H 1 ( v ) ≥ ρρ 1 ( ρ 1 − 1) ˜ H 2 ( v ) , 0 < v ≤ v 0 . (3.31) F or v 0 < v ≤ 1, we hav e H 2 ( v , z ) ≤ v v ρ − 1 0 (1 + z ) ρρ 1 − 2 and then Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 H 2 ( v , z u )(1 − u )d u ≤ v v ρ − 1 0 Z 1 0 d u Z ∞ 1 z 2 (1 + z u ) ρρ 1 − 2 µ 1 (d z ) . Replacing the v ariable z with y /u and using the fact µ 1 (d( y /u )) = u α 1 µ 1 (d y ) w e hav e Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 H 2 ( v , z u )(1 − u )d u ≤ v v ρ − 1 0 Z 1 0 u α 1 − 2 d u Z ∞ u y 2 (1 + y ) ρρ 1 − 2 µ 1 (d y ) ≤ v v ρ − 1 0 Z 1 0 u α 1 − 2 d u Z ∞ 0 z 2 (1 + z ) ρρ 1 − 2 µ 1 (d z ) =: v v ρ − 1 0 c 0 , v 0 < v ≤ 1 . Com bining this with (3.29) and (3.31) we obtain Z ∞ 1 K ( v , z ) µ 1 (d z ) = Z ∞ 1 z 2 µ 1 (d z ) Z 1 0 K ′′ z z ( v , z u )(1 − u )d u ≥ − ρρ 1 ( ρ 1 − 1) v v ρ − 1 0 c 0 for all 0 < v ≤ 1. T ogether this with (3.30) one gets (3.27). (ii) By (3.29), K ′′ z z ( v , z ) = ρρ 1 (1 − ρρ 1 ) v 2 (1 + z ) 2 ρ 1 − 2  v [(1 + z ) ρ 1 − 1] + 1  ρ − 2 − ρρ 1 ( ρ 1 − 1) v (1 − v )(1 + z ) ρ 1 − 2  v [(1 + z ) ρ 1 − 1] + 1  ρ − 2 =: ρρ 1 (1 − ρρ 1 ) K 1 ( v , z ) − ρρ 1 ( ρ 1 − 1) K 2 ( v , z ) . (3.32) F or 0 ≤ v ≤ 1, K 1 ( v , z ) ≥ v 2  (1 + z ) ρ 1  ρ − 2 (1 + z ) 2 ρ 1 − 2 = v 2 (1 + z ) ρρ 1 − 2 . By (3.6), Z ∞ 0 z 2 µ 1 (d z ) Z 1 0 K 1 ( v , uz )(1 − u )d u ≥ c 1 ( ρρ 1 ) v 2 (3.33) and for δ > 0 Z δ 0 z 2 µ 1 (d z ) Z 1 0 K 2 ( v , uz )(1 − u )d u ≤ v (1 − v ) Z δ 0 z 2 (1 + z ) ρ 1 − 1 µ 1 (d z ) =: v (1 − v ) d 1 ,δ . (3.34) Moreo ver, for 0 ≤ v ≤ 1, K 2 ( v , z ) ≤ v (1 − v )  v (1 + z ) ρ 1  ρ − 1  (1 − v )  − 1 (1 + z ) ρ 1 − 1 = v ρ (1 + z ) ρρ 1 − 2 and then for all δ > 0, Z ∞ δ z 2 µ 1 (d z ) Z 1 0 K 2 ( v , uz )(1 − u )d u ≤ v ρ Z ∞ δ z 2 µ 1 (d z ) Z 1 0 (1 + uz ) ρρ 1 − 2 d u =: d 2 ,δ v ρ . (3.35) 21 By (3.3) and (3.32), K ( v , z ) = z 2 Z 1 0 K ′′ z z ( v , z u )(1 − u )d u = ρρ 1 (1 − ρρ 1 ) z 2 Z 1 0 K 1 ( v , uz )(1 − u )d u − ρρ 1 ( ρ 1 − 1) z 2 Z 1 0 K 2 ( v , uz )(1 − u )d u. Com bining this with (3.33)–(3.35) we obtain (3.28). 2 3.2.3 Equations (1.3) with dominate drift terms In this subsubsection w e consider equations (1.3) under Condition 1.5 where the drifts dominate. F or 0 < ρ < 1 and ρ 1 , ρ 2 ≥ 1 deﬁne h ( x, y ) = − ( x ρ 1 + y ρ 2 ) ρ , x, y > 0 , (3.36) whic h is the k ey test function to pro ve Theorems 1.10(i) and 1.11 under Condition 1.5. In the follo wing we estimate L h in Lemmas 3.10 and 3.11 for verifying assumption (iii) of Prop osition 2.2. W e ﬁrst present a preliminary estimate on L h by (3.5) and Lemma 3.9. Recall (3.25). By (3.26) and Lemma 3.9(i), for all large enough ρ 1 ≥ 2 and ρρ 1 ∈ (0 , 1), there is a constan t ˜ d 1 > 0 suc h that Z ∞ 0 K 1 z h ( x, y ) µ 1 (d z ) = x − α 1 ( x ρ 1 + y ρ 2 ) ρ Z ∞ 0 K ( x ρ 1 x ρ 1 + y ρ 2 , z ) µ 1 (d z ) ≥ − ρρ 1 ( ρ 1 − 1) ˜ d 1 x ρ 1 − α 1 ( x ρ 1 + y ρ 2 ) ρ − 1 . Similarly , for all large enough ρ 2 ≥ 2 and ρρ 2 ∈ (0 , 1), there is a constan t ˜ d 2 > 0 suc h that Z ∞ 0 K 2 z h ( x, y ) µ 2 (d z ) ≥ − ρρ 2 ( ρ 2 − 1) ˜ d 2 y ρ 2 − α 2 ( x ρ 1 + y ρ 2 ) ρ − 1 . It th us follows from (3.5) that L h ( x, y ) ≥ ρ ( x ρ 1 + y ρ 2 ) ρ − 1 h b 10 ρ 1 x r 10 − 1+ ρ 1 + b 20 ρ 2 y r 20 − 1+ ρ 2 − a 1 ρ 1 x θ 1 − 1+ ρ 1 y κ 1 − a 2 ρ 2 x κ 2 y θ 2 − 1+ ρ 2 − ρ 1 ( ρ 1 − 1)[ b 11 x r 11 − 2+ ρ 1 + b 12 ˜ d 1 x r 12 − α 1 + ρ 1 ] − ρ 2 ( ρ 2 − 1)[ b 21 y r 21 − 2+ ρ 2 + b 22 ˜ d 2 y r 22 − α 2 + ρ 2 ] i . (3.37) By Lemma 3.9(ii), for 0 < ρρ 1 < 1 and ρ 1 > 1, Z ∞ 0 K 1 z h ( x, y ) µ 1 (d z ) = x − α 1 ( x ρ 1 + y ρ 2 ) ρ Z ∞ 0 K ( x ρ 1 x ρ 1 + y ρ 2 , z ) µ 1 (d z ) ≥ ρρ 1 x − α 1 h (1 − ρρ 1 ) c 1 ( ρρ 1 ) x 2 ρ 1 ( x ρ 1 + y ρ 2 ) ρ − 2 − ( ρ 1 − 1) d 1 ,δ x ρ 1 y ρ 2 ( x ρ 1 + y ρ 2 ) ρ − 2 − ( ρ 1 − 1) d 2 ,δ x ρρ 1 i . Similar results hold for R ∞ 0 K 2 z h ( x, y ) µ 2 (d z ) b y symmetry . Thus, we also hav e L h ( x, y ) ≥ ρ ( x ρ 1 + y ρ 2 ) ρ − 1 h b 10 ρ 1 x r 10 − 1+ ρ 1 − a 1 ρ 1 x θ 1 − 1+ ρ 1 y κ 1 − a 2 ρ 2 x κ 2 y θ 2 − 1+ ρ 2 22 − ρ 1 ( ρ 1 − 1)[ b 11 x r 11 − 2+ ρ 1 + b 12 ˜ d 1 x r 12 − α 1 + ρ 1 ] i + ρρ 2 ( x ρ 1 + y ρ 2 ) ρ − 2 h (1 − ρρ 2 )[ b 21 y r 21 − 2+2 ρ 2 + b 22 c 2 ( ρρ 2 ) y r 22 − α 2 +2 ρ 2 ] − ( ρ 2 − 1) x ρ 1 [ b 21 y r 21 − 2+ ρ 2 + b 22 ˜ d 1 ,δ y r 22 − α 2 + ρ 2 ] i − ρρ 2 ( ρ 2 − 1) b 22 ˜ d 2 ,δ y r 22 − α 2 + ρρ 2 , (3.38) where ˜ d 1 ,δ , ˜ d 2 ,δ > 0 satisfy lim δ →∞ ˜ d 2 ,δ = 0. Based on the ab ov e e stimation we state the follo wing lemma, whose pro of utilizes Lemmas 3.1 and 3.7. Lemma 3.10 Supp ose that θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 and the assumptions of Theorem 1.10(i) hold. Then there are constan ts ρ 1 , ρ 2 , ρ > 0 and C, c > 0 such that L h ( x, y ) ≥ C , 0 < x, y ≤ c. (3.39) Pr o of. W e ﬁrst establish the pro of under Condition 1.5(i). Let ρ 1 , ρ 2 ≥ 2 b e large enough constan ts satisfying (3.19). Let 0 < ρ < 1 b e small enough suc h that r 10 − 1 + ρ 1 ρ ≤ 0 and r 20 − 1 + ρ 2 ρ ≤ 0. By (3.37), L h ( x, y ) ≥ ρ ( x ρ 1 + y ρ 2 ) ρ − 1 [ J 1 ( x, y ) − J 2 ( x, y ) − J 3 ( x ) − J 4 ( y )] , 0 < x, y < 1 , (3.40) where J 1 ( x, y ) := b 10 ρ 1 x r 10 − 1+ ρ 1 + b 20 ρ 2 y r 20 − 1+ ρ 2 , J 2 ( x, y ) := a 1 ρ 1 x θ 1 − 1+ ρ 1 y κ 1 + a 2 ρ 2 x κ 2 y θ 2 − 1+ ρ 2 (3.41) and J 3 ( x ) := ρ 1 ( ρ 1 − 1)[ b 11 x r 11 − 2+ ρ 1 + b 12 ˜ d 1 x r 12 − α 1 + ρ 1 ] , J 4 ( y ) := ρ 2 ( ρ 2 − 1)[ b 21 y r 21 − 2+ ρ 2 + b 22 ˜ d 2 y r 22 − α 2 + ρ 2 ] (3.42) for some constan ts ˜ d 1 , ˜ d 2 > 0. By Lemma 3.7, there is a constan t 0 < c 1 < 1 suc h that 4 − 1 J 1 ( x, y ) ≥ J 2 ( x, y ) , 0 < x, y ≤ c 1 . (3.43) By Condition 1.5(i), there is a constan t 0 < c 2 ≤ c 1 suc h that 8 − 1 J 1 ( x, y ) ≥ J 3 ( x ) and 8 − 1 J 1 ( x, y ) ≥ J 4 ( y ) for all 0 < x, y ≤ c 2 . Com bining these with (3.40) and (3.43) we obtain L h ( x, y ) ≥ 2 − 1 ρ ( x ρ 1 + y ρ 2 ) ρ − 1 J 1 ( x, y ) for all 0 < x, y ≤ c 2 . By Lemma 3.1(ii), w e hav e J 1 ( x, y ) = b 10 ρ 1 x r 10 − 1+ ρ 1 ρ x ρ 1 (1 − ρ ) + b 20 ρ 2 y r 20 − 1+ ρ 2 ρ y ρ 2 (1 − ρ ) ≥ C [ x ρ 1 (1 − ρ ) + y ρ 2 (1 − ρ ) ] ≥ C ( x ρ 1 + y ρ 2 ) 1 − ρ (3.44) with C := ( b 10 ρ 1 ) ∧ ( b 20 ρ 2 ). Th us, L h ( x, y ) ≥ 2 − 1 ρC for all 0 < x, y ≤ c 2 , whic h gives (3.39). Since the pro ofs under Condition 1.5(ii)-(iii) are similar, we only presen t that under Condition 1.5(ii). If Condition 1.5(ii) and inequality (1.13) hold, there are large enough constants ρ 1 , ρ 2 ≥ 2 suc h that (3.19) holds and ρ 1 /ρ 2 < r 1 /r 2 . Moreo ver, for all 0 < ρ < 1, ρ 1 r 1 + 2 ρ 1 + r 2 + ρ 2 r 2 + 2 ρ 2 > 1 , ρ 1 (2 − ρ ) r 1 + 2 ρ 1 + r 2 + ρρ 2 r 2 + 2 ρ 2 > 1 . (3.45) 23 In the follo wing, let 0 < ρ < 1 b e small enough such that r 10 − 1 + ρ 1 ρ = r 1 + ρ 1 ρ ≤ 0 and r 2 + ρ 2 ρ ≤ 0. Recall (3.41) and (3.42). By (3.38), L h ( x, y ) ≥ ρ ( x ρ 1 + y ρ 2 ) ρ − 1  b 10 ρ 1 x r 1 + ρ 1 − J 2 ( x, y ) − J 3 ( x )  + ρ ( x ρ 1 + y ρ 2 ) ρ − 2  ˜ c 1 y r 2 +2 ρ 2 − ˜ c 2 x ρ 1 y r 2 + ρ 2  − ˜ c 0 ˜ d 2 ,δ y r 2 + ρρ 2 , where ˜ c 0 := ρρ 2 ( ρ 2 − 1) b 22 , ˜ c 1 := ρ 2 (1 − ρρ 2 )[ b 21 ∧ ( b 22 c 2 ( ρρ 2 ))] and ˜ c 2 := ρ 2 ( ρ 2 − 1)[ b 21 + b 22 ˜ d 1 ,δ ]. Then for 0 < σ 1 < b 10 ρ 1 and 0 < σ 2 < ˜ c 1 , L h ( x, y ) ≥ ρ ( x ρ 1 + y ρ 2 ) ρ − 1  I 1 ( x, y ) − J 2 ( x, y ) − J 3 ( x )  + ρ ( x ρ 1 + y ρ 2 ) ρ − 2  I 2 ( x, y ) − ( ˜ c 2 + σ 2 ) x ρ 1 y r 2 + ρ 2  − ˜ c 0 ˜ d 2 ,δ y r 2 + ρρ 2 , (3.46) where I 1 ( x, y ) := ( b 10 ρ 1 − σ 1 ) x r 1 + ρ 1 + σ 2 y r 2 + ρ 2 and I 2 ( x, y ) := σ 1 x r 1 +2 ρ 1 + ( ˜ c 1 − σ 2 ) y r 2 +2 ρ 2 . Since lim δ →∞ ˜ d 2 ,δ = 0, there is a constan t δ > 0 such that I 2 ( x, y ) ≥ 8 ˜ c 0 ˜ d 2 ,δ y r 2 +2 ρ 2 . By Lemma 3.7, (3.19) and (3.45), there is a constan t 0 < c 3 < 1 suc h that for all 0 < x, y ≤ c 3 w e hav e I 1 ( x, y ) ≥ 4 J 2 ( x, y ) , I 2 ( x, y ) ≥ 2( ˜ c 2 + σ 2 ) x ρ 1 y r 2 + ρ 2 , I 2 ( x, y ) ≥ 8 ˜ c 0 ˜ d 2 ,δ x ρ 1 (2 − ρ ) y r 2 + ρρ 2 . (3.47) It follo ws from Lemma 3.1(ii) that I 2 ( x, y ) − 2 ˜ c 0 ˜ d 2 ,δ y r 2 + ρρ 2 ( x ρ 1 + y ρ 2 ) 2 − ρ ≥ I 2 ( x, y ) − 4 ˜ c 0 ˜ d 2 ,δ [ y r 2 + ρρ 2 x ρ 1 (2 − ρ ) + y r 2 +2 ρ 2 ] ≥ 0 (3.48) for all 0 < x, y ≤ c 3 . By (1.5), there is a constan t 0 < c 4 < c 3 suc h that 4 − 1 I 1 ( x, y ) ≥ J 3 ( x ) for all 0 < x, y ≤ c 4 . Com bining this with (3.46)–(3.48), one obtains L h ( x, y ) ≥ 2 − 1 ρ ( x ρ 1 + y ρ 2 ) ρ − 1 I 1 ( x, y ) , 0 < x, y ≤ c 4 . Similarly to (3.44), w e hav e I 1 ( x, y ) ≥ [( b 10 ρ 1 − σ 1 ) ∧ σ 2 ]( x ρ 1 + y ρ 2 ) 1 − ρ , whic h gives (3.39). 2 Let δ 0 > 0 b e the constant determined in Lemma 3.8. F or ρ 1 , ρ 2 , ρ > 0 deﬁne ˜ h ( x, y ) := − ( δ 0 x ρ 1 + y ρ 2 ) ρ , x, y > 0 . (3.49) Similarly to Lemma 3.10, w e hav e the follo wing estimate of L ˜ h . Lemma 3.11 According to the assumptions of Theorem 1.11, there are constants ρ 1 , ρ 2 , δ 0 , ρ > 0 and C, c > 0 such that L ˜ h ( x, y ) ≥ C , 0 < x, y ≤ c. (3.50) Pr o of. Using Lemma 3.8, the pro of is a mo diﬁcation of that of Lemma 3.10, and we omit it here. 2 3.2.4 Equations (1.3) with dominan t diﬀusion or jump terms In this subsubsection w e study the SDE system (1.3) with dominant diﬀusion or jump terms, that is, under the assumption (1.14). The key function h is deﬁned in Lemma 3.12 under Condition 1.9(i), in Lemma 3.14 under Condition 1.9(ii), and in Lemma 3.16 under assumption (iii) of Theorem 1.10. The estimates of L h , k ey to verifying assumption (iii) of Prop osition 2.2, are giv en in Lemmas 3.12, 3.14 and 3.16 for diﬀerent cases. Applying these estimates and 24 Prop osition 2.2, the pro ofs of Theorems 1.10 and 1.11 are giv en at the end of this subsubsection. F or ε ∈ (0 , 1) deﬁne a nonnegativ e b ounded function ˜ h ( y ) := y − (1 + ε ) − 1 y 1+ ε (1 + y ) − ε , y > 0 . (3.51) Then ˜ h ′ ( y ) = 1 − y ε (1 + y ) − ε + ε (1 + ε ) − 1 y 1+ ε (1 + y ) − 1 − ε ≥ 0 , ˜ h ′′ ( y ) = − εy ε − 1 (1 + y ) − 2 − ε . (3.52) Th us, there is a constant c 0 = c 0 ( ε ) ∈ (0 , 1) suc h that ˜ h ( y ) ≥ 2 − 1 y , 2 − 1 ≤ ˜ h ′ ( y ) ≤ 1 , − ˜ h ′′ ( y ) ≥ 2 − 1 εy ε − 1 , 0 < y ≤ 2 c 0 . (3.53) It follo ws that for all 0 < y ≤ c 0 w e hav e ˜ h ( y + y z u ) ≤ y + y z u ≤ 2 y ≤ 4 ˜ h ( y ) , 0 < y ≤ c 0 , 0 < z , u ≤ 1 . (3.54) Lemma 3.12 Supp ose that (1.14) holds for i = 1 , 2 , θ 1 − 1 < r 1 < 0 , θ 2 − 1 < r 2 < 0 and (1.13) and that condition 1.9(i) holds. In addition, under (1.9) let h ( x, y ) := − [ x ρ 1 + ˜ h ( y )] ρ , x, y > 0 , (3.55) and under (1.10) let h ( x, y ) := − [ y ρ 2 + ˜ h ( x )] ρ , x, y > 0 . Then there are constants ρ 1 , ρ 2 , ρ, ε > 0 and C , c > 0 such that L h ( x, y ) ≥ C for all 0 < x, y ≤ c . Before pro ving Lemma 3.12, we ﬁrst sho w a preliminary estimate of L h b y (3.5) in the follo wing lemma. Lemma 3.13 Supp ose that (1.14) holds for i = 1 , 2 . Let h b e the function deﬁned in (3.55) for 0 < ρ, ε < 1 and ρ 1 > 1 . Then there are constan ts ˜ c 1 , ˜ c 2 , ˜ c 3 , ˜ c 4 > 0 suc h that L h ( x, y ) ≥ − 8 − 1 ˜ c 2 ρx r 1 + ρρ 1 + ρ [ x ρ 1 + ˜ h ( y )] ρ − 1  ˜ c 1 y r 2 +1+ ε − a 1 ρ 1 x ρ 1 − 1+ θ 1 y κ 1 − 2 a 2 x κ 2 y θ 2  + ρ [ x ρ 1 + ˜ h ( y )] ρ − 2  ˜ c 2 x r 1 +2 ρ 1 + ˜ c 3 y r 2 +2 − ˜ c 4 x r 1 + ρ 1 ˜ h ( y )  , 0 < x, y ≤ c 0 , (3.56) where c 0 ∈ (0 , 1) is the constan t app earing in (3.53) . Pr o of. Observe that h ′ x ( x, y ) = − ρρ 1 [ x ρ 1 + ˜ h ( y )] ρ − 1 x ρ 1 − 1 , h ′ y ( x, y ) = − ρ [ x ρ 1 + ˜ h ( y )] ρ − 1 ˜ h ′ ( y ) (3.57) and h ′′ xx ( x, y ) = ρρ 1 [ x ρ 1 + ˜ h ( y )] ρ − 2 [(1 − ρρ 1 ) x 2 ρ 1 − 2 − ( ρ 1 − 1) x ρ 1 − 2 ˜ h ( y )] . (3.58) Moreo ver, by (3.53), h ′′ y y ( x, y ) = ρ (1 − ρ )[ x ρ 1 + ˜ h ( y )] ρ − 2 | ˜ h ′ ( y ) | 2 − ρ [ x ρ 1 + ˜ h ( y )] ρ − 1 ˜ h ′′ ( y ) ≥ 2 − 2 ρ (1 − ρ )[ x ρ 1 + ˜ h ( y )] ρ − 2 + 2 − 1 ερy ε − 1 [ x ρ 1 + ˜ h ( y )] ρ − 1 , 0 < y ≤ 2 c 0 . (3.59) 25 Let K ( v , z ) be the function deﬁned in (3.25). By (3.26) and (3.28), we hav e Z ∞ 0 K 1 z h ( x, y ) µ 1 (d z ) = x − α 1 [ x ρ 1 + ˜ h ( y )] ρ Z ∞ 0 K ( x ρ 1 x ρ 1 + ˜ h ( y ) , z ) µ 1 (d z ) ≥ ρρ 1 x − α 1 h (1 − ρρ 1 ) c 1 ( ρρ 1 ) x 2 ρ 1 [ x ρ 1 + ˜ h ( y )] ρ − 2 − ( ρ 1 − 1) d 1 ,δ x ρ 1 ˜ h ( y )[ x ρ 1 + ˜ h ( y )] ρ − 2 − ( ρ 1 − 1) d 2 ,δ x ρρ 1 i , (3.60) where the constan ts d 1 ,δ , d 2 ,δ > 0 satisfy lim δ →∞ d 2 ,δ = 0. Com bining with (3.53), (3.54) and (3.59), for 0 < z , u ≤ 1 and 0 < y ≤ c 0 w e hav e h ′′ y y ( x, y + y z u ) ≥ 2 − 2 ρ (1 − ρ )[ x ρ 1 + 4 ˜ h ( y )] ρ − 2 + 2 − 2 ρε [ x ρ 1 + 4 ˜ h ( y )] ρ − 1 y ε − 1 ≥ 2 2 ρ − 6 ρ (1 − ρ )[ x ρ 1 + ˜ h ( y )] ρ − 2 + 2 2 ρ − 4 ρε [ x ρ 1 + ˜ h ( y )] ρ − 1 y ε − 1 . (3.61) Since h ′′ y y ( x, y ) ≥ 0 for all y > 0, then b y (3.3), Z ∞ 0 K 2 y z h ( x, y ) µ 2 (d z ) = y 2 Z ∞ 0 z 2 µ 2 (d z ) Z 1 0 h ′′ y y ( x, y + y z u )(1 − u )d u ≥ y 2 Z 1 0 z 2 µ 2 (d z ) Z 1 0 h ′′ y y ( x, y + y z u )(1 − u )d u ≥ ˆ c 1 ρεy 2  (1 − ρ )[ x ρ 1 + ˜ h ( y )] ρ − 2 + [ x ρ 1 + ˜ h ( y )] ρ − 1 y ε − 1  , 0 < y ≤ c 0 , for ˆ c 1 := 2 2 ρ − 6 R 1 0 z 2 µ 2 (d z ), where (3.61) is used in the last inequality . Then b y (3.4), Z ∞ 0 K 2 z h ( x, y ) µ 2 (d z ) = y − α 2 Z ∞ 0 K 2 y z h ( x, y ) µ 2 (d z ) ≥ ˆ c 1 ρεy 2 − α 2  (1 − ρ )[ x ρ 1 + ˜ h ( y )] ρ − 2 + [ x ρ 1 + ˜ h ( y )] ρ − 1 y ε − 1  , 0 < y ≤ c 0 . (3.62) F urther com bining (3.5) with (3.57)–(3.60) and (3.62) one obtains L h ( x, y ) ≥ ρ [ x ρ 1 + ˜ h ( y )] ρ − 1  b 21 2 − 1 εy r 21 − 1+ ε + b 22 ˆ c 1 εy r 22 +1 − α 2 + ε − a 1 ρ 1 x ρ 1 − 1+ θ 1 y κ 1 − a 2 ˜ h ′ ( y ) x κ 2 y θ 2  + ρ [ x ρ 1 + ˜ h ( y )] ρ − 2 h b 11 ρ 1 (1 − ρρ 1 ) x r 11 +2 ρ 1 − 2 + b 12 ρ 1 (1 − ρρ 1 ) c 1 ( ρρ 1 ) x r 12 +2 ρ 1 − α 1 + b 21 (1 − ρ )2 − 2 y r 21 + b 22 (1 − ρ ) ˆ c 1 εy r 22 +2 − α 2 − b 11 ρ 1 ( ρ 1 − 1) x r 11 + ρ 1 − 2 ˜ h ( y ) − b 12 ρ 1 ( ρ 1 − 1) d 1 ,δ x r 12 + ρ 1 − α 1 ˜ h ( y ) i − b 12 ρρ 1 ( ρ 1 − 1) d 2 ,δ x r 12 + ρρ 1 − α 1 , 0 < x, y ≤ c 0 . Since lim δ →∞ d 2 ,δ = 0, there is a large enough constan t δ > 0 such that b 12 ρ 1 ( ρ 1 − 1) d 2 ,δ ≤ 8 − 1 ˜ c 2 with ˜ c 2 := ρ 1 (1 − ρρ 1 )[ b 11 ∧ ( b 12 c 1 ( ρρ 1 ))]. By (3.52), ˜ h ′ ( y ) ≤ 2 for all y > 0 w e get (3.56) with ˜ c 1 := ε [( b 21 2 − 1 ) ∧ ( b 22 ˆ c 1 )] , ˜ c 3 := [ b 21 (1 − ρ )2 − 2 ] ∧ [ b 22 ˆ c 1 ε ] and ˜ c 4 := ρ 1 ( ρ 1 − 1)[ b 11 + b 12 d 1 ,δ ]. This ends the pro of. 2 No w we are ready to complete the pro of of Lemma 3.12 applying Lemma 3.1(ii)–(iii). Pr o of of the L emma 3.12. Since the pro ofs are similar, we only present that under (1.9). By (1.9) and (1.13), w e obtain 1 < κ 2 ( r 2 + 1) r 2 + 1 − θ 2 − r 1 , ( r 1 + 1 − θ 1 ) κ 1 < κ 2 r 2 + 1 − θ 2 , r 1 r 2 < κ 2 ( r 2 + 1) r 2 + 1 − θ 2 − r 1 . 26 Then there is a constan t ρ 1 suc h that 1 < ρ 1 , r 1 r 2 < ρ 1 , ( r 1 + 1 − θ 1 )( r 2 + 1) κ 1 − r 1 < ρ 1 < κ 2 ( r 2 + 1) r 2 + 1 − θ 2 − r 1 , whic h gives ρ 1 − 1 + θ 1 r 1 + ρ 1 + κ 1 r 2 + 1 > 1 , κ 2 r 1 + ρ 1 + θ 2 r 2 + 1 > 1 and r 1 + ρ 1 r 1 + 2 ρ 1 + 1 r 2 + 2 > 1 , r 1 + ρρ 1 r 1 + 2 ρ 1 + 2 − ρ r 2 + 2 > 1 (3.63) for small enough 0 < ρ < 1 with r 1 + ρρ 1 < 0 , r 2 + ρ < 0 . (3.64) Moreo ver, there is a suﬃciently small constant 0 < ε < 1 such that ρ 1 − 1 + θ 1 r 1 + ρ 1 + κ 1 r 2 + 1 + ε > 1 , κ 2 r 1 + ρ 1 + θ 2 r 2 + 1 + ε > 1 . (3.65) By Lemma 3.1(ii), x r 1 + ρρ 1 = [ x ρ 1 + ˜ h ( y )] ρ − 2 x r 1 + ρρ 1 [ x ρ 1 + ˜ h ( y )] 2 − ρ ≤ 2[ x ρ 1 + ˜ h ( y )] ρ − 2 [ x r 1 +2 ρ 1 + x r 1 + ρρ 1 ˜ h ( y ) 2 − ρ ] and x r 1 + ρ 1 = [ x ρ 1 + ˜ h ( y )] − 1 [ x r 1 +2 ρ 1 + x r 1 + ρ 1 ˜ h ( y )] . Th us, it follows from Lemma 3.13 that L h ( x, y ) ≥ ρ [ x ρ 1 + ˜ h ( y )] ρ − 1  4 − 1 ˜ c 2 x r 1 + ρ 1 + ˜ c 1 y r 2 +1+ ε − a 1 ρ 1 x ρ 1 − 1+ θ 1 y κ 1 − 2 a 2 x κ 2 y θ 2  + ρ [ x ρ 1 + ˜ h ( y )] ρ − 2  2 − 1 ˜ c 2 x r 1 +2 ρ 1 + ˜ c 3 y r 2 +2 − (˜ c 4 + 4 − 1 ˜ c 2 ) x r 1 + ρ 1 ˜ h ( y ) − 4 − 1 ˜ c 2 x r 1 + ρρ 1 ˜ h ( y ) 2 − ρ  ≥ ρ [ x ρ 1 + ˜ h ( y )] ρ − 1  4 − 1 ˜ c 2 x r 1 + ρ 1 + ˜ c 1 y r 2 +1+ ε − a 1 ρ 1 x ρ 1 − 1+ θ 1 y κ 1 − 2 a 2 x κ 2 y θ 2  + ρ [ x ρ 1 + ˜ h ( y )] ρ − 2  2 − 1 ˜ c 2 x r 1 +2 ρ 1 + ˜ c 3 y r 2 +2 − (˜ c 4 + 4 − 1 ˜ c 2 ) x r 1 + ρ 1 y − 4 − 1 ˜ c 2 x r 1 + ρρ 1 y 2 − ρ  =: ρ [ x ρ 1 + ˜ h ( y )] ρ − 1 I 1 ( x, y ) + ρ [ x ρ 1 + ˜ h ( y )] ρ − 2 I 2 ( x, y ) , 0 < x, y ≤ c 0 , (3.66) where we use the fact ˜ h ( y ) ≤ y for all y > 0 in the second inequalit y . By Lemma 3.1(iii), (3.63) and (3.65), there is a constan t 0 < c 1 ≤ c 0 suc h that I 1 ( x, y ) ≥ 0 , I 2 ( x, y ) ≥ 4 − 1 [˜ c 2 x r 1 +2 ρ 1 + ˜ c 3 y r 2 +2 ] , 0 < x, y < c 1 . Th us, it follows from (3.64), (3.66), the fact ˜ h ( y ) ≤ y , and Lemma 3.1(ii) that L h ( x, y ) ≥ 4 − 1 ρ [ x ρ 1 + y ] ρ − 2 [˜ c 2 x ρ 1 (2 − ρ ) x r 1 + ρρ 1 + ˜ c 3 y 2 − ρ y r 2 + ρ ] ≥ 4 − 1 ρ (˜ c 2 ∧ ˜ c 3 )[ x ρ 1 + y ] ρ − 2 [ x ρ 1 (2 − ρ ) + y 2 − ρ ] ≥ 8 − 1 ρ (˜ c 2 ∧ ˜ c 3 ) , 0 < x, y < c 1 , whic h completes the pro of. 2 27 Lemma 3.14 Given θ 1 − 1 < r 1 < 0 and θ 2 − 1 < r 2 < 0 , under Condition 1.9(ii), supp ose that (1.13) holds and (1.14) holds for i = 1 , 2 . Under (1.11) let ˆ h ( x, y ) := ( x + y δ ) ρ 1 + ˜ h ( y ) , h ( x, y ) := − ˆ h ( x, y ) ρ , x, y > 0 , (3.67) and under (1.12) let ˆ h ( x, y ) := ( y + x δ ) ρ 2 + ˜ h ( x ) , h ( x, y ) := − ˆ h ( x, y ) ρ , x, y > 0 . Then there are constants ρ 1 , ρ 2 , ρ, ε > 0 and C , c > 0 such that L h ( x, y ) ≥ C for all 0 < x, y ≤ c . Before pro ving Lemma 3.14, we ﬁrst give an estimate of L h by (3.5) in the following lemma. Lemma 3.15 Let 0 < ρ, ρ 1 , ε < 1 and δ > 1 satisfy ρ 1 δ > 1 , 0 < ε < δ ρ 1 − 1 and δ ρρ 1 < 1 . Let h b e the function deﬁned by (3.67) . Then there are constants c 1 , ˜ c 1 , ˜ c 2 , ˜ c 3 > 0 suc h that L h ( x, y ) ≥ ρ ˆ h ( x, y ) ρ − 1 [ I 1 ( x, y ) − I 2 ( x, y )] − b 22 ˜ c 1 y r 22 , 0 < x, y ≤ c 1 , where I 1 ( x, y ) := ˜ c 2  ( x + y δ ) ρ 1 − 2 x r 11 + ( x + y δ ) ρ 1 − α 1 x r 12  + ˜ c 3 y ε +1 [ y r 21 − 2 + y r 22 − α 2 ] and I 2 ( x, y ) := a 1 ρ 1 x θ 1 y κ 1 ( x + y δ ) ρ 1 − 1 + 2 a 2 y θ 2 x κ 2 . Pr o of. Observe that h ′ x ( x, y ) = − ρρ 1 ˆ h ( x, y ) ρ − 1 ( x + y δ ) ρ 1 − 1 , h ′′ xx ( x, y ) ≥ ρρ 1 (1 − ρ 1 ) ˆ h ( x, y ) ρ − 1 ( x + y δ ) ρ 1 − 2 (3.68) and h ′ y ( x, y ) = − ρ ˆ h ( x, y ) ρ − 1  ρ 1 δ ( x + y δ ) ρ 1 − 1 y δ − 1 + ˜ h ′ ( y )  , h ′′ y y ( x, y ) ≥ ρ ˆ h ( x, y ) ρ − 1  − ρ 1 ( δ − 1) δ ( x + y δ ) ρ 1 − 1 y δ − 2 − ˜ h ′′ ( y )  . Since ρ 1 δ > 1 and 0 < ε < δ ρ 1 − 1, there is a suﬃcien tly small constant 0 < c 1 < 1 / 2 suc h that for all 0 < y ≤ 2 c 1 y δ − 1 ( x + y δ ) ρ 1 − 1 ≤ y δ − 1+ δ ( ρ 1 − 1) = y δ ρ 1 − 1 ≤ ( ρ 1 δ ) − 1 and ( x + y δ ) ρ 1 − 1 y δ − 2 ≤ y δ ρ 1 − 2 = y ε − 1 y δ ρ 1 − ε − 1 ≤ 4 − 1 ε [ ρ 1 ( δ − 1) δ ] − 1 y ε − 1 . Th us, by (3.53), h ′ y ( x, y ) ≥ − 2 ρ ˆ h ( x, y ) ρ − 1 , h ′′ y y ( x, y ) ≥ 2 − 2 ρε ˆ h ( x, y ) ρ − 1 y ε − 1 , 0 < y ≤ 2( c 0 ∧ c 1 ) , (3.69) where the constan t c 0 is determined in (3.53). Recall the function K ( v , z ) deﬁned in (3.25). By (3.29), for 0 ≤ v ≤ 1 w e hav e K ′′ z z ( v , z ) ≥ ρρ 1 (1 − ρ 1 ) v (1 + z ) ρ 1 − 2  v [(1 + z ) ρ 1 − 1] + 1  ρ − 1 ≥ ρρ 1 (1 − ρ 1 ) v (1 + z ) ρρ 1 − 2 28 and then b y (3.3) and (3.6), Z ∞ 0 K ( v , z ) µ 1 (d z ) = Z ∞ 0 z 2 µ 1 (d z ) Z 1 0 K ′′ z z ( v , z u )(1 − u )d u ≥ ρρ 1 (1 − ρ 1 ) v Z ∞ 0 z 2 µ 1 (d z ) Z 1 0 (1 + z u ) ρρ 1 − 2 (1 − u )d u = ρρ 1 (1 − ρ 1 ) c 1 ( ρρ 1 ) v . (3.70) No w, by the arguments for the ﬁrst equation of (3.60), Z ∞ 0 K 1 z h ( x, y ) µ 1 (d z ) = − ˆ h ( x, y ) ρ ( x + y δ ) − α 1 Z ∞ 0 K ( ( x + y δ ) ρ 1 ˆ h ( x, y ) , z ) µ 1 (d z ) ≥ ρρ 1 (1 − ρ 1 ) c 1 ( ρρ 1 ) ˆ h ( x, y ) ρ − 1 ( x + y δ ) ρ 1 − α 1 . (3.71) By (3.3) and (3.69), for 0 < y ≤ c 0 ∧ c 1 Z c 0 0 K 2 z h ( x, y ) µ 2 (d z ) = Z c 0 0 z 2 µ 2 (d z ) Z 1 0 h ′′ y y ( x, y + uz )(1 − u )d u ≥ 2 − 2 ρε Z c 0 0 z 2 µ 2 (d z ) Z 1 0 ˆ h ( x, y + uz ) ρ − 1 ( y + uz ) ε − 1 (1 − u )d u. (3.72) By (3.54), for all 0 < y ≤ c 0 and 0 < z , u ≤ 1, ˆ h ( x, y + y uz ) = [ x + ( y + y uz ) δ ] ρ 1 + ˜ h ( y + y uz ) ≤ ( x + 2 δ y δ ) ρ 1 + 4 ˜ h ( y ) ≤ 2 δ ρ 1 +1 ˆ h ( x, y ) . No w replacing the v ariable z with y v in (3.72) and using the fact µ 2 (d( y v )) = y − α 2 µ 2 (d v ), for 0 < y ≤ c 0 ∧ c 1 w e hav e Z c 0 0 K 2 z h ( x, y ) µ 2 (d z ) ≥ 2 − 2 ερy 2 − α 2 Z c 0 /y 0 v 2 µ 2 (d v ) Z 1 0 ˆ h ( x, y + y uv ) ρ − 1 ( y + y uv ) ε − 1 (1 − u )d u ≥ 2 ε − 3 ερy ε +1 − α 2 Z 1 0 z 2 µ 2 (d z ) Z 1 0 ˆ h ( x, y + y uz ) ρ − 1 (1 − u )d u ≥ 2 ε − 3+( δ ρ 1 +1)( ρ − 1) ερy ε +1 − α 2 ˆ h ( x, y ) ρ − 1 Z 1 0 z 2 µ 2 (d z ) =: 2 − 2 ρε ˜ c ˆ h ( x, y ) ρ − 1 y ε +1 − α 2 . On the other hand, for all 0 < x, y ≤ c 0 , Z ∞ c 0 K 2 z h ( x, y ) µ 2 (d z ) ≥ − Z ∞ c 0 ˆ h ( x, y + z ) ρ µ 2 (d z ) ≥ − Z ∞ c 0 ˆ h ( c 0 , c 0 + z ) ρ µ 2 (d z ) =: − ˜ c 1 with ˜ c 1 < ∞ b y the assumption of δ ρρ 1 < 1. Then Z ∞ 0 K 2 z h ( x, y ) µ 2 (d z ) ≥ 2 − 2 ρε ˜ c ˆ h ( x, y ) ρ − 1 y ε +1 − α 2 − ˜ c 1 , 0 < x, y ≤ c 0 ∧ c 1 . (3.73) It th us follows from (3.5), (3.68)–(3.69), (3.71) and (3.73) that L h ( x, y ) ≥ ρ ˆ h ( x, y ) ρ − 1 h ρ 1 (1 − ρ 1 )  b 11 ( x + y δ ) ρ 1 − 2 x r 11 + b 21 c 1 ( ρρ 1 )( x + y δ ) ρ 1 − α 1 x r 12  +2 − 2 ε [ b 21 y r 21 + ε − 1 + b 22 ˜ cy r 22 + ε +1 − α 2 ] 29 − a 1 ρ 1 x θ 1 y κ 1 ( x + y δ ) ρ 1 − 1 − 2 a 2 y θ 2 x κ 2 i − b 22 ˜ c 1 y r 22 ≥ ρ ˆ h ( x, y ) ρ − 1 [ I 1 ( x, y ) − I 2 ( x, y )] − b 22 ˜ c 1 y r 22 , 0 < x, y ≤ c 0 ∧ c 1 with ˜ c 2 := ρ 1 (1 − ρ 1 )[ b 11 ∧ ( b 21 c 1 ( ρρ 1 ))] and ˜ c 3 := 2 − 2 ε [ b 21 ∧ ( b 22 ˜ c )]. No w we are ready to conclude the pro of of Lemma 3.14 by Lemma 3.1(ii) again. Pr o of of L emma 3.14. Since the pro ofs are similar, w e only state that under (1.11). Under (1.11), there is a constan t 0 < ρ 1 < 1 satisfying ρ 1 < (1 − θ 1 ) ∧ (2 − κ 2 ) (3.74) and ρ 1 > 1 − θ 1 κ 1 − r 2 , ρ 1 < κ 2 r 2 + 1 − θ 2 , ρ 1 < κ 2 − r 1 1 − θ 2 , whic h implies that there is a constant δ > ρ − 1 1 > 1 suc h that δ (1 − ρ 1 − θ 1 ) < κ 1 − 1 − r 2 , r 2 + 1 − θ 2 − δ κ 2 < 0 and r 1 < κ 2 − ρ 1 + θ 2 /δ, (3.75) resp ectiv ely . Moreo ver, there is a small enough constant 0 < ε < δ ρ 1 − 1 suc h that r 2 + ε < 0, δ (1 − ρ 1 − θ 1 ) < κ 1 − 1 − r 2 − ε (3.76) and r 2 + ε + 1 − θ 2 − δ κ 2 < 0 . (3.77) Let 0 < ρ < 1 b e small enough satisfying ρρ 1 + r 1 < 0 , r 2 + ρ + ε < 0 (3.78) and δ ρρ 1 < 1. Let I 1 and I 2 b e the functions deﬁned in Lemma 3.15. As x ρ 1 ≤ y ≤ 1, by Lemma 3.1(ii) and the fact ˜ h ( y ) ≤ y for y > 0 we get ˆ h ( x, y ) ≤ x ρ 1 + y δ ρ 1 + y ≤ 3 y and then ˆ h ( x, y ) ρ − 1 I 1 ( x, y ) ≥ ˜ c 3 ˆ h ( x, y ) ρ − 1 y ε +1 [ y r 21 − 2 + y r 22 − α 2 ] ≥ 3 ρ − 1 ˜ c 3 y ε + ρ [ y r 21 − 2 + y r 22 − α 2 ] = 3 ρ − 1 ˜ c 3 y ε + ρ [ y r 21 − 2 + y r 22 − α 2 ] . Since y ≤ x ρ 1 ≤ 1, w e get ˆ h ( x, y ) ≤ x ρ 1 + y δ ρ 1 + y ≤ x ρ 1 + 2 y ≤ 3 x ρ 1 , x + y δ ≤ x + x ρ 1 δ ≤ 2 x and then ˆ h ( x, y ) ρ − 1 I 1 ( x, y ) ≥ ˜ c 2 ˆ h ( x, y ) ρ − 1  ( x + y δ ) ρ 1 − 2 x r 11 + ( x + y δ ) ρ 1 − α 1 x r 12  30 ≥ 3 ρ − 1 ˜ c 2 x ρρ 1 − ρ 1  2 ρ 1 − 2 x ρ 1 − 2+ r 11 + 2 ρ 1 − α 1 x ρ 1 − α 1 + r 12  . It follo ws from (1.14) that ˆ h ( x, y ) ρ − 1 I 1 ( x, y ) ≥ 3 ρ − 1 2 ρ 1 − 2 (˜ c 2 ∧ ˜ c 3 )[ x ρρ 1 + r 1 ∧ y r 2 + ρ + ε ] , 0 < x, y ≤ 1 . (3.79) By (3.74) and (3.76), there is a constan t 0 < c 2 ≤ c 0 ∧ c 1 suc h that ˜ c 3 y r 2 + ε +1 = ˜ c 3 y r 2 + ε +1 − δ ( ρ 1 − 1+ θ 1 ) y δ ( ρ 1 − 1+ θ 1 ) ≥ ˜ c 3 y r 2 + ε +1 − δ ( ρ 1 − 1+ θ 1 ) − κ 1 ( x + y δ ) ρ 1 − 1+ θ 1 y κ 1 ≥ 2 a 1 ρ 1 x θ 1 y κ 1 ( x + y δ ) ρ 1 − 1 , 0 < y ≤ c 2 . (3.80) Similarly , by (3.74), (3.75) and (3.77), there is a constant 0 < c 3 ≤ c 2 suc h that for all 0 < x, y ≤ c 3 ˜ c 3 y r 2 + ε +1 ( x + y δ ) 2 − ρ 1 ≥ ˜ c 3 y r 2 + ε +1 − θ 2 y θ 2 x κ 2 ( x + y δ ) 2 − ρ 1 − κ 2 ≥ ˜ c 3 y r 2 + ε +1 − θ 2 y θ 2 x κ 2 y δ (2 − ρ 1 − κ 2 ) = ˜ c 3 y r 2 + ε +1 − θ 2 − δ κ 2 y θ 2 x κ 2 y δ (2 − ρ 1 ) ≥ 4 a 2 2 1 − ρ 1 + θ 2 /δ x κ 2 y δ (2 − ρ 1 + θ 2 /δ ) and ˜ c 2 [ x r 11 + ( x + y δ ) 2 − α 1 x r 12 ] ≥ ˜ c 2 x r 1 +2 = ˜ c 2 x r 1 +2 − ( κ 2 +2 − ρ 1 + θ 2 /δ ) x κ 2 x 2 − ρ 1 + θ 2 /δ ≥ 4 a 2 2 1 − ρ 1 + θ 2 /δ x κ 2 x 2 − ρ 1 + θ 2 /δ . Th us, by Lemma 3.1(ii) again, I 1 ( x, y )( x + y δ ) 2 − ρ 1 ≥ 4 a 2 2 1 − ρ 1 + θ 2 /δ x κ 2 x 2 − ρ 1 + θ 2 /δ + 4 a 2 2 1 − ρ 1 + θ 2 /δ x κ 2 y δ (2 − ρ 1 + θ 2 /δ ) ≥ 4 a 2 x κ 2 ( x + y δ ) 2 − ρ 1 + θ 2 /δ ≥ 4 a 2 y θ 2 x κ 2 ( x + y δ ) 2 − ρ 1 , 0 < x, y ≤ c 3 . (3.81) Com bining (3.80) and (3.81) we get 2 − 1 I 1 ( x, y ) ≥ I 2 ( x, y ) for all 0 < x, y ≤ c 3 , whic h together with Lemma 3.15 and (3.79) implies L h ( x, y ) ≥ 3 ρ − 1 2 ρ 1 − 3 (˜ c 2 ∧ ˜ c 3 )[ x ρρ 1 + r 1 ∧ y r 2 + ρ + ε ] − b 22 ˜ c 1 y r 22 , 0 < x, y ≤ c 3 . By (3.78), there are constan ts C > 0 and 0 < c ≤ c 3 suc h that L h ( x, y ) ≥ C for all 0 < x, y ≤ c . 2 In the follo wing w e introduce the key test function h and the estimate on L h under the assumptions of Theorem 1.10(iii). Lemma 3.16 Recall the function ˜ h deﬁned b y (3.51) for 0 < ε < 2 − 1 ∧ ( r 2 − 1 − κ 1 ) ∧ ( r 1 − 1 − κ 2 ) . F or θ 1 , θ 2 > 0 let ˆ h θ 1 ,θ 2 ( x, y ) := x 1 − θ 1 + y 1 − θ 2 ; for θ 1 > 0 and θ 2 = 0 let ˆ h θ 1 , 0 ( x, y ) := x 1 − θ 1 + ˜ h ( y ) ; for θ 1 = 0 and θ 2 > 0 let ˆ h 0 ,θ 2 ( x, y ) := y 1 − θ 2 + ˜ h ( x ) ; for θ 1 = θ 2 = 0 let ˆ h 0 , 0 ( x, y ) := ˜ h ( x ) + ˜ h ( y ) . F or 0 < ρ < 1 let h ( x, y ) := − ˆ h θ 1 ,θ 2 ( x, y ) ρ . Then, under the assumptions of Theorem 1.10(iii), there are constan ts C , c, ε > 0 such that L h ( x, y ) ≥ C for all 0 < x, y ≤ c . Pr o of. The proof is similar to those of Lemmas 3.10 and 3.12. W e leav e the details to interested readers. 2 No w we are ready to complete the pro ofs of Theorems 1.10 and 1.11. Pr o of of The or em 1.10. Under the assumptions in Theorem 1.10(i), there are constants ρ, ρ 1 , ρ 2 determined in Lemma 3.10 and let the function h b e deﬁned by (3.36). Under the assumptions 31 of Theorem 1.10(ii), let h b e the function deﬁned in Lemmas 3.12 and 3.14. F or the assumptions of Theorem 1.10(iii), let the function h b e determined in Lemma 3.16. Let the constants C , c > 0 b e the constants determined in Lemmas 3.10, 3.12, 3.14 and 3.16. Set g ( x, y ) := | h ( c, 0) | ∧ | h (0 , c ) | + h ( x, y ) , x, y > 0 . (3.82) Then the condition (i) of Prop osition 2.2 is ob vious and g ( x, y ) ≤ 0 for all x ≥ c or y ≥ c . By Lemma 3.10, 3.12, 3.14 and 3.16, for all 0 < x, y ≤ c , L g ( x, y ) ≥ C ≥ C [ | h ( c, 0) | ∧ | h (0 , c ) | ] − 1 g ( x, y ) . Then b y Proposition 2.2, one gets P { τ 0 < ∞} ≥ g ( X 0 , Y 0 )[ | h ( c, 0) | ∧ | h (0 , c ) | ] − 1 for all small enough X 0 , Y 0 > 0. F or general, X 0 , Y 0 > 0, applying the strong Mark ov prop ert y and the same argumen ts at the end of the pro of of [34, Theorem 1.3] w e conclude the pro of. 2 Pr o of of The or em 1.11. Under the assumptions, let ρ, ρ 1 , ρ 2 , δ 0 b e the constan ts given in Lemma 3.11 and h := ˜ h b e the function giv en by (3.49). 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Extinction behaviour for mutually enhancing continuous-state population dynamics

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