Transition probabilities for degenerate diffusions arising in population genetics

Abstract

We provide a detailed description of the structure of the transition probabilities and of the hitting distributions on boundary components of a manifold with corners for a degenerate strong Markov process arising in population genetics. The Markov processes that we study are a generalization of the classical Wright–Fisher process. The main ingredients in our proofs are based on the analysis of the regularity properties of solutions to a forward Kolmogorov equation defined on a compact manifold with corners, which is degenerate in the sense that it is not strictly elliptic and the coefficients of the first order drift term have mild logarithmic singularities.

Appendix

We establish auxiliary results used in the proof of Lemma 5.3. The first result is a Landis-type growth lemma which was proved by Safonov [24], in the two-dimensional case (that is, we take $n=2$ and $m=0$ in the statement of Lemma A.1). Given $r,\mu >0$ and a bounded function u, we introduce the notation:

$$\begin{aligned} M(r;\mu )&:= \sup \{u(z):\, z\in \bar{Q}(r;\mu )\}\text { where,} \end{aligned}$$

(A.1)

$$\begin{aligned} Q(r;\mu )&:= (0,\mu r)^n \times (-\mu \sqrt{r},\mu \sqrt{r})^m. \end{aligned}$$

(A.2)

We consider a differential operator A defined on $B^{\infty }_R\subset S_{n,m}$ by

$$\begin{aligned} Au(z)= & {} \sum _{i,j=1}^n \sqrt{x_ix_j}a_{ij}(z) u_{x_ix_j} +\sum _{i=1}^n b_i(z)u_{x_i}+\sum _{i=1}^n \sum _{l=1}^m \sqrt{x_i}c_{il}(z)u_{x_iy_l}\nonumber \\&+ \sum _{l,k=1}^m d_{lk}(z) u_{y_ly_k} +\sum _{l=1}^m e_l(z) u_{y_l},\quad \forall \, u\in C^2(B^{\infty }_R), \end{aligned}$$

(A.3)

where we assume that the coefficients $\{b_i(z):1\le i\le n\}$ satisfy the cleanness condition:

$$\begin{aligned}&b_i = 0 \quad \hbox { on } \partial B^{\infty }_R\cap \{x_i=0\}, \end{aligned}$$

(A.4)

$$\begin{aligned}&\hbox {or } b_i > 0 \quad \hbox { on } \partial B^{\infty }_R\cap \{x_i=0\}. \end{aligned}$$

(A.5)

We denote by $I^T$ the set of indices $1\le i\le n$ such that property (A.4) holds and we denote by $I^{\pitchfork }$ the set of indices $1\le i\le n$ such that property (A.5). In this appendix it is assumed that neither set is empty. We also denote:

$$\begin{aligned} F_i&:=\partial B^{\infty }_R\cap \{x_i=0\},\quad \forall \, 1\le i\le n,\\ \partial ^T B^{\infty }_R&:= \bigcup _{i\in I^T} F_i. \end{aligned}$$

We can now state:

Lemma A.1

(Growth lemma) [24] Let n, m be non-negative integers such that $n\ge 2$ and let $0<R, \nu <1$. Suppose that the coefficients of the operator A in (A.3) are bounded continuous functions on $\bar{B}^{\infty }_R$ and that there are positive constants, $\delta$ and K, such that for all $1\le i, j \le n$ and $1\le l,k\le m$, we have that

$$\begin{aligned}&\Vert a_{ij}\Vert _{C({\bar{B}}^{\infty }_R)} +\Vert b_i\Vert _{C({\bar{B}}^{\infty }_R)}+ \Vert c_{il}\Vert _{C({\bar{B}}^{\infty }_R)} + \Vert d_{lk}\Vert _{C({\bar{B}}^{\infty }_R)} + \Vert e_l\Vert _{C({\bar{B}}^{\infty }_R)} \le K, \end{aligned}$$

(A.6)

$$\begin{aligned}&\sum _{i,j=1}^n a_{ij}(z)\xi _i \xi _j + \sum _{l,k=1}^m d_{lk}(z)\eta _l\eta _k \ge \delta (|\xi |^2 + |\eta |^2), \quad \forall \, z\in {\bar{B}}^{\infty }_R, \forall \,\xi \in \mathbb {R}^n, \forall \,\eta \in \mathbb {R}^m, \end{aligned}$$

(A.7)

and the coefficients $\{b_i(z):1\le i\le n\}$ satisfy one of conditions (A.4) or (A.5). Let $i \in I^T$ and $j \in I^{\pitchfork }$. Then there is a positive constant, $\theta \in (0,1)$, depending only on $b_0,\delta ,K,\nu , m$, and n, such that if

$$\begin{aligned} u\in C^2(\bar{B}^{\infty }_R\backslash \partial ^T B^{\infty }_R)\cap C(\bar{B}^{\infty }_R\backslash (F_i\cap F_j)) \end{aligned}$$

(A.8)

is a solution such that

$$\begin{aligned} A u = 0&\quad \hbox { on } \quad \bar{B}^{\infty }_R\backslash \partial ^T B^{\infty }_R, \end{aligned}$$

(A.9)

$$\begin{aligned} 0\le u \le \nu&\quad \hbox { on } \quad \partial ^T B^{\infty }_R\backslash (F_i\cap F_j), \end{aligned}$$

(A.10)

$$\begin{aligned} 0 \le u \le 1&\quad \hbox { on } \quad {\bar{B}}^{\infty }_R\backslash (F_i\cap F_j), \end{aligned}$$

(A.11)

then u satisfies the growth property

$$\begin{aligned} M(r;1/2) \le \theta M(r; 1),\quad \forall \, r\in (0,R). \end{aligned}$$

(A.12)

The proof of Lemma A.1 is based on an induction argument, which applies a comparison principle, and on the following scaling property of the operator A defined in (A.3):

Remark A.2

(Scaling property of the operator A) For $\lambda >0$, we consider the rescaling described in (1.7). Using (A.9) we notice that the rescaled function $v(z')$ satisfies the equation $A' v(z') = 0$, for all $z'=(x',y')\in B^{\infty }_{R/\lambda }$, where the operator $A'$ is given by

$$\begin{aligned} A'v(z') =&\sum _{i,j=1}^n \sqrt{x'_ix'_j}a_{ij}(\lambda x', \sqrt{\lambda } y') v_{x'_ix'_j} +\sum _{i=1}^n b_i(\lambda x', \sqrt{\lambda } y') v_{x'_i}\\&+\sum _{i=1}^n \sum _{l=1}^m \sqrt{\lambda x'_i} c_{il}(\lambda x', \sqrt{\lambda } y') v_{x'_iy'_l}\\&+ \sum _{l,k=1}^m d_{lk}(\lambda x', \sqrt{\lambda } y') v_{y'_ly'_k} +\sum _{l=1}^m \sqrt{\lambda } e_l(\lambda x', \sqrt{\lambda } y') v_{y'_l}. \end{aligned}$$

Thus, by rescaling the solutions as in (1.7), the rescaled functions are solutions to an equation defined by a new operator, $A'$, that satisfies the same properties as the original operator, A, when we choose the parameter $\lambda$ in a bounded set. In the proof of Lemma A.1, we apply this scaling property for $\lambda \in (0,1)$.

Note that in the $n=2, m=0$ case we would have a function u defined on $[0,1]\times [0,1],$ which is not known to be (and in general will not be) continuous at (0, 0). This precludes a simple application of the maximum principle. If we can show that $u(1/2,z)\le \theta ,$ and $u(z,1/2)\le \theta$ for $z\in (0,1/2),$ in a manner that only depends on bounds on the coefficients of the operator and the other hypotheses on u,  then the scaling argument shows that, for $r\in (0,1),$

$$\begin{aligned} u(rz,r/2)\le \theta \text { and }u(r/2,rz)\le \theta \text { for }z\in (0,1/2), \end{aligned}$$

(A.13)

as well. In other words $u(x,y)\le \theta ,$ for $(x,y)\in (0,1/2)\times (0,1/2),$ which provides an effective replacement for the maximum principle. This is, in essence, how the argument below proceeds. The proof that we provide is a small modification of an argument communicated to us by Safonov [24].

Remark A.2 implies that to establish the growth property (A.12), we can assume without loss of generality that $R=1$ and $M(1;1)=1$, and it is sufficient to prove that there is a constant, $\theta \in (0,1)$, such that

$$\begin{aligned} M(1;1/2) \le \theta . \end{aligned}$$

(A.14)

Because our proof of Lemma A.1 relies on an induction argument, we begin with the proof of the base case in the induction.

Lemma A.3

(Growth lemma with $n=2$ and $m\ge 0$) [24] In the statement of Lemma A.1 assume that $n=2$, $m\ge 0$, $i=1$, and $j=2$. Then the conclusion of Lemma A.1 holds.

Proof

We establish property (A.14) in several steps.

Step 1

In this step we prove that there are positive constants, $h,\theta _1\in (0,1)$, such that

$$\begin{aligned} u(x,y) \le \theta _1,\quad \forall \, (x,y) \in \left( \bigcup _{r\in (0,1)} [0,hr]\times \left\{ \frac{r}{2}\right\} \right) \times \left[ -\frac{1}{2},\frac{1}{2}\right] ^m. \end{aligned}$$

(A.15)

We let $H\in (0,1)$ and we consider the set $W_2 := (0,H)\times (1/4,3/4)\times (-1,1)^m,$ and the barrier function

$$\begin{aligned} w_2(z) = \nu +16\left( x_2-\frac{1}{2}\right) ^2 + \sqrt{\frac{x_1}{H}} + \sum _{l=1}^m y_l^2, \end{aligned}$$

where we recall that $\nu$ belongs to (0, 1);  it is the constant appearing in (A.10). Direct calculations give us that

$$\begin{aligned} A w_2 = 32\left( x_2a_{22} + b_2\left( x_2-\frac{1}{2}\right) \right) + \frac{2b_1-a_{11}}{4\sqrt{Hx_1}} + \sum _{l=1}^m (d_{ll}+e_ly_l), \end{aligned}$$

and, using the assumption that $b_1=0$ along $\partial B^{\infty }_1\cap \{x_1=0\}$ together with (A.6) and (A.7), we see that we can find positive constants, C and H, such that

$$\begin{aligned} A w_2 \le -\frac{C}{\sqrt{Hx_1}} + C < 0\quad \hbox { on } W_2. \end{aligned}$$

We also see that on the boundary of the set $W_2$ the following hold. By property (A.10), we have that $u\le \nu \le w_2$ on $\partial W_2\cap \{x_1=0\}$. By property (A.11), we have that $u\le 1\le w_2$ on $\partial W_2\cap S_{n,m}$. Applying the comparison principle, it follows that $u\le w_2$ on $W_2$. From the choice of the barrier function $w_2$, we see that for all $\theta _1\in (\nu ,1)$, there is a positive constant $h\in (0,1)$ such that $u(x,y) \le \theta _1$, for all points $(x,y)\in \bar{W}_2$ with the property that $x_2=1/2$, $x_1\in [0,h]$, and $y_l\in [-h,h]$, for all $1\le l\le m$.

Applying Remark (A.2) with $\lambda := r\in (0,1)$ arbitrarily chosen, we obtain that $u(x,y) \le \theta _1$, for all points (x, y) with the property that there is $r\in (0,1)$ such that $x_2=r/2$, $x_1\in [0,hr]$, and $y_l\in [-h\sqrt{r},h\sqrt{r}]$, for all $1\le l\le m$. We note that the scaling property described in Remark A.2 continues to hold when we apply translations in the y-coordinates of solutions. From here we deduce that $u(x,y) \le \theta _1$, for all points (x, y) with the property that $y_l\in [-1/2,1/2]$, for all $1\le l\le m$, and there is $r\in (0,1)$ such that $x_2=r/2$ and $x_1\in [0,hr]$. This proves that (A.15) holds.

Step 2

In this step, we extend property (A.15) in the sense that we prove that for all $k\in (0,1/2)$ there is a constant, $\theta _2=\theta _2(k,\theta _1)\in (0,1)$, such that

$$\begin{aligned} u(x,y)\le \theta _2,\quad \forall \, (x,y)\in \left[ 0,\frac{1}{2}\right] \times \left[ k,\frac{1}{2}\right] \times \left[ -\frac{1}{2},\frac{1}{2}\right] ^m. \end{aligned}$$

(A.16)

For $k\in (0,1/2)$, we consider the sets $D_1:=[hk,1/2]\times [k,1/2]$ and $D_2 := [hk/2,1]\times [k/2,1]$, which have the property that $D_1\subset D_2$ and the following hold:

$$\begin{aligned} 0\le u\le \theta _1<1&\quad \hbox { on } \partial D_2\cap \{x_1=k/2\},\quad (\hbox {by Step }1 \hbox { applied with } r=k)\\ 0\le u\le 1&\quad \hbox { on } D_2,\quad (\hbox {by }(A.11))\\ Au = 0&\quad \hbox { on } D_2,\quad (\hbox {by }(A.9)). \end{aligned}$$

Because the operator A is strictly elliptic on $D_2$, the preceding properties of u show that we can apply the strong maximum principle to conclude that there is a positive constant, $\theta _3=\theta _3(A,k)\in (0,1)$, such that $u\le \theta _3$ on $D_1$. Combining this property with inequality (A.15) and letting $\theta _2:=\theta _1\vee \theta _3$, we obtain (A.16).

Step 3

We now prove that there are positive constants, $k\in (0,1/2)$ and $\theta _4\in (0,1)$, such that

$$\begin{aligned} u(x,y)\le \theta _4,\quad \forall (x,y)\in \left\{ \frac{1}{2}\right\} \times (0,k)\times (-k,k)^m. \end{aligned}$$

(A.17)

For $k\in (0,1/2)$, we let $\theta _2=\theta _2(A,k)\in (0,1)$ be the constant in inequality (A.16). We consider the set $W_1 := (1/4, 3/4) \times (0,k) \times (-1,1)^m$ and the barrier function

$$\begin{aligned} w_1(z) = \theta _2 + (1-\theta _2)\left[ 16\left( x_1-\frac{1}{2}\right) ^2 + \beta \left( k-x_2\right) + \frac{1}{2}+\frac{1}{2}\sum _{l=1}^m y_l^2\right] , \end{aligned}$$

where the positive constant $\beta$ is suitably chosen below. Because $b_1>0$ along $\partial W_1\cap \{x_2=0\}$, we can choose k small enough such that there is a positive constant, $b_0$, with the property that $b_1\ge b_0$ on $\bar{W}_1$. Direct calculations give us that

$$\begin{aligned} Aw_1&\le (1-\theta _2)\left[ 32\left( x_2a_{22}+b_2\left( x_2-\frac{1}{2}\right) \right) -\beta b_0 + \frac{1}{2} \sum _{l=1}^m(d_{ll}+e_ly_l)\right] , \end{aligned}$$

and so we can choose $\beta =\beta (b_0,K)$ large enough so that $Aw_1 <0$ on $W_1$. We next prove that $u\le w_1$ on $\partial W_1 \cap S_{n,m}$. We recall that by adapting the argument of the proof of [8, Proposition 3.1.1] to the case of elliptic problem for the operator A, we do not need to have that $u\le w_1$ on the portion of the boundary $\partial W_1\cap \{x_2=0\}$ due to the regularity assumption (A.8) and to the fact that the operator A is degenerate as we approach this boundary portion of $W_1$. We see that on $\partial W_1\cap \{x_2=k\}$, we have that $w_1\ge \theta _2$ and $u\le \theta _2$ by inequality (A.16), and so $u\le w_1$ on $\partial W_1\cap \{x_2=k\}$. On the portion of the boundary $\partial W_1\cap \{x_1=1/4\hbox { or } x_1=3/4\}$, it is clear that $u\le w_1$ because on this set we have that $w_1\ge 1$ by construction and $u\le 1$ by (A.11). Thus, the comparison principle implies that $u\le w_1$ on $W_1$. In particular, when $z=(x,y)$ has the property that $x_1=1/2$, $x_2\in (0,k)$, and $y_l\in (0,k)$, for all $1\le l\le m$, we have that

$$\begin{aligned} u(z) \le \theta _2+(1-\theta _2)\left( \beta k+ \frac{1}{2}+\frac{1}{2} mk^2\right) . \end{aligned}$$

We can choose the positive constant k small enough such that there is a positive constant $\theta _4\in (0,1)$ with the property that $u(x,y)\le \theta _4$, for all (x, y) such that $x_1=1/2$ and $x_2,y_l\in (0,k)$, for all $1\le l\le m$. We recall that the scaling property in Remark A.2 continues to hold when we apply translations in the y-coordinates of solutions. This concludes the proof of inequality (A.17).

Let $k\in (0,1/2)$ and $\theta _4$ be chosen as in Step 3 and let $\theta _2$ be chosen as in Step 2. Let $\theta :=\theta _2\vee \theta _4$. Inequalities (A.16) and (A.17) give us that $u(x,y)\le \theta$, for all (x, y) such that $y\in (0,k)^m$, $x_1=1/2$ and $x_2\in (0,1/2)$, or $x_2=1/2$ and $x_1\in (0,1/2)$. Remark A.2 implies that $u(x,y)\le \theta$, for all (x, y) such that there is $r\in (0,1)$ with the property that $y\in (0,\sqrt{r}k)^m$, $x_1=r/2$ and $x_2\in (0,r/2)$, or $x_2=r/2$ and $x_1\in (0,r/2)$. We recall that the scaling property in Remark A.2 continues to hold when we apply translations in the y-coordinates of solutions. From here we deduce that $u(x,y) \le \theta$, for all points $(x,y)\in Q(1;1/2)$, where we recall the definition of Q(1; 1 / 2) in (A.2). This concludes the proof of inequality (A.14), when $n=2$ and $m\ge 0$. $\square$

We can now give

Proof of Lemma A.1

We assume without loss of generality that $i=1$, $j=2$, and $R=1$. We prove inequality (A.14) by an induction argument on n. The base case, $n=2$, was established in Lemma A.3. We next consider the induction step. We assume that (A.14) holds with n replaced by $n-1$ and we want to establish it for n. We prove this assertion in several steps, which are adaptations of the steps of the proof of Lemma A.3 to the multi-dimensional case ($n>2$). For the multi-dimensional case, we do not need an adaptation of Step 1 in Lemma A.3.

Step 1

Analogously to Step 2 in the proof of Lemma A.3, we prove that for all $k\in (0,1/2)$ there is a constant, $\theta _1=\theta _1(A,k)\in (0,1)$, such that

$$\begin{aligned} u(x,y)\le \theta _1,\quad \forall \, (x,y)\in \left[ 0,\frac{1}{2}\right] \times \left( \left[ 0,\frac{1}{2}\right] ^{n-1}\backslash \left[ 0,k\right] ^{n-1}\right) \times \left[ -\frac{1}{2},\frac{1}{2}\right] ^m.\quad \end{aligned}$$

(A.18)

For $k\in (0,1/2)$, we consider the sets

$$\begin{aligned} D_1&:= \left[ 0,\frac{1}{2}\right] \times \left( \left[ 0,\frac{1}{2}\right] ^{n-1}\backslash \left[ 0,k\right] ^{n-1}\right) \times \left[ -\frac{1}{2},\frac{1}{2}\right] ^m,\\ D_2&:= \left[ 0, 1\right] \times \left( [0,1]^n\backslash \left[ 0,k/2\right] ^{n-1}\right) \times \left[ -1,1\right] ^m, \end{aligned}$$

which have the property that $D_1\subset D_2$ and the following hold:

$$\begin{aligned} 0\le u\le \nu <1&\quad \hbox { on } \partial ^T D_2\backslash (F_1\cap F_2), \quad (\hbox {by } (A.10) \hbox { with } i=1 \hbox { and } j=2)\\ 0\le u\le 1&\quad \hbox { on } D_2\backslash (F_1\cap F_2),\quad (\hbox {by } (A.11))\\ Au = 0&\quad \hbox { on } D_2\backslash \partial ^T D_2,\quad (\hbox {by } (A.9)), \end{aligned}$$

where we let $\partial ^T D_2 := \partial ^T B^{\infty }_1\cap D_2$. Because on the set $D_1$ the operator A can be viewed as a degenerate operator of the form (A.3) defined on $S_{n-1,m+1}$, as opposed to $S_{n,m}$, we can apply the induction hypothesis and by covering $D_1$ by a finite number of balls, we obtain that there is a positive constant, $\theta _1=\theta _1(A,k)\in (0,1)$, such that $u\le \theta _1$ on $D_1$. This completes the proof of inequality (A.18).

Step 2

Analogously to Step 3 in Lemma A.3, we next prove that there are positive constants, $k\in (0,1/2)$ and $\theta _2\in (0,1)$, such that

$$\begin{aligned} u(x,y)\le \theta _2,\quad \forall (x,y)\in \left\{ \frac{1}{2}\right\} \times (0,k)^{n-1}\times (-k,k)^m. \end{aligned}$$

(A.19)

We fix $k\in (0,1/2)$ and let $\theta _1=\theta _1(A,k)\in (0,1)$ be chosen as in Step 1. We consider the set $W_1 := (1/4, 3/4) \times (0,k) \times (-1,1)^m$ and the barrier function

$$\begin{aligned} w_1(z) = \theta _2 + (1-\theta _2)\left[ 16\left( x_1-\frac{1}{2}\right) ^2 + \beta \left( k-x_2\right) + \frac{1}{2}+\frac{1}{2}\left( \sum _{i=3}^n x_i + \sum _{l=1}^m y_l^2\right) \right] , \end{aligned}$$

where $\beta$ is a positive constant. The argument of the proof of Step 3 Lemma A.3 immediately adapts to the present choice of the set $W_1$ and of the barrier function $w_1$, and we obtain that we can choose $\beta$ and k small enough so that there is a constant, $\theta _2\in (0,1)$, such that inequality (A.19) holds. This completes the argument of Step 2.

Let $k\in (0,1/2)$ and $\theta _2$ be chosen as in Step 2 and let $\theta _1$ be chosen as in Step 1. Let $\theta :=\theta _1\vee \theta _2$. Inequalities (A.18) and (A.19) gives us that $u(x,y)\le \theta$, for all $(x,y)\in S_{n,m}\cap \partial (0,1/2)^n\times (-k,k)^m$. Remark A.2 implies that $u(x,y)\le \theta$, for all $(x,y)\in (0,1/2)^n\times (-k,k)^m$. We recall that the scaling property in Remark A.2 holds when we apply translations in the y-coordinates of solutions. From here we deduce that $u(x,y) \le \theta$, for all points $(x,y)\in Q(1;1/2)$, where we recall the definition of Q(1; 1 / 2) in (A.2). This completes the proof of inequality (A.14). $\square$

We next establish that the probability of hitting the intersection of a tangent and a transverse boundary component defines a function (by formula (A.20)) that satisfies the hypotheses of Lemma A.1.

Lemma A.4

(Regularity of the hitting probability) Suppose that the generalized Kimura operator satisfies the standard assumptions. Let $i\in I^T$ and $j\in I^{\pitchfork }$. Then the hitting probability,

$$\begin{aligned} u(p) := \mathbb {Q}^p(\omega (\tau _{\partial ^T P}) \in H_i\cap H_j),\quad \forall \, p \in P, \end{aligned}$$

(A.20)

belongs to the space of functions

$$\begin{aligned} C^{\infty }(P\backslash \partial ^T P) \cap C(P \backslash (H_i\cap H_j)) \end{aligned}$$

(A.21)

and satisfies properties

$$\begin{aligned} L u = 0&\quad \hbox { on } \quad P\backslash \partial ^T P, \end{aligned}$$

(A.22)

$$\begin{aligned} u=0&\quad \hbox { on } \quad \partial ^T P\backslash (H_i\cap H_j), \end{aligned}$$

(A.23)

$$\begin{aligned} 0 \le u \le 1&\quad \hbox { on } \quad P. \end{aligned}$$

(A.24)

Proof

From definition (A.20) of u, it is clear that property (A.24) holds. Applying Lemma 2.18 with $p\in \partial ^T P\backslash (H_i\cap H_j)$, we also have that u satisfies property (A.23). To prove the remaining assertions of Lemma A.4, we use an approximation argument. Let $k\in \mathbb {N}$, $\varphi _k:P\rightarrow [0,1]$, and $\psi _k:[0,\infty ]\rightarrow [0,1]$ be smooth functions such that

$$\begin{aligned} \begin{aligned}&\varphi _k = 1\quad \hbox { on }\quad P\cap \{p\in P: \hbox { dist}(p, H_i\cap H_j) \le 1/k\},\\&\varphi _k = 0\quad \hbox { on }\quad P \backslash \{p\in P: \hbox { dist}(p, H_i\cap H_j) \ge 2/k\},\\&\psi _k(t) = 1\quad \hbox { for all } t\ge 2/k, \quad \hbox { and }\quad \psi _k(t) = 0\quad \hbox { for all } t\le 1/k, \end{aligned} \end{aligned}$$

(A.25)

and let $\zeta _k(t,p) := \psi _k(t)\varphi _k(p)$, for all $(t,p)\in [0,\infty )\times P$. Let $v_k$ be the unique solution in the space of functions (4.3) to the non-homogeneous Dirichlet problem (4.1) with boundary condition $\zeta =\zeta _k$. The stochastic representation (4.5) gives us that

$$\begin{aligned} v_k(t,p) = \mathbb {E}_{\mathbb {Q}^p}\left[ \psi _k(t-t\wedge \tau _{\partial ^T P})\varphi _k(\omega (t\wedge \tau _{\partial ^T P}))\right] , \quad \forall \, (t,p)\in [0,\infty )\times P, \end{aligned}$$

(A.26)

where we recall that the stopping time $\tau _{\partial ^T P}$ is defined in (1.18) and the probability measure $\mathbb {Q}^p$ is the unique solution to the martingale problem in Definition 1.1.

We divide the proof into several steps.

Step 1

(Convergence as $t\rightarrow \infty$ and $k\rightarrow \infty$) From the definition of the function $\psi _k$ we have that

$$\begin{aligned} \begin{aligned} \psi _k(t-t\wedge \tau _{\partial ^TP})&=1\quad \hbox { if }\tau _{\partial ^TP}< t-1/k,\\ \psi _k(t-t\wedge \tau _{\partial ^TP})&=0\quad \hbox { if }t-2/k <\tau _{\partial ^TP}, \end{aligned} \end{aligned}$$

(A.27)

from which it follows that

$$\begin{aligned} \left| v_k(t,p) - \mathbb {E}_{\mathbb {Q}^p}\left[ \varphi _k(\omega (t\wedge \tau _{\partial ^T P}))\mathbf {1}_{\{\tau _{\partial ^TP}< t-1/k\}}\right] \right| \le \mathbb {Q}^p(t-1/k \le \tau _{\partial ^TP} < t-2/k). \end{aligned}$$

Letting t tend to $\infty$, we see that the right-hand side of the preceding inequality tends to 0 because

$$\begin{aligned} \sum _{a=0}^{\infty } \mathbb {Q}^p(a/k \le \tau _{\partial ^TP}< (a+1)/k) = \mathbb {Q}^p(\tau _{\partial ^TP}< \infty ) <\infty . \end{aligned}$$

Thus, we have that the sequence of functions $v_k(t,\cdot )$ converges pointwise on P, as $t\rightarrow \infty$, for each fixed $k\in \mathbb {N}$, and we denote

$$\begin{aligned} u_k(p):=\lim _{t\rightarrow \infty } v_k(t,p) = \mathbb {E}_{\mathbb {Q}^p}\left[ \varphi _k(\omega (\tau _{\partial ^T P}))\right] , \quad \forall \, k\in \mathbb {N},\quad \forall \, p \in P. \end{aligned}$$

(A.28)

From the definition of the cutoff functions $\varphi _k$ in (A.25) and of the hitting probability u in (A.20), we have that

$$\begin{aligned} \lim _{k\rightarrow \infty } u_k(p) = u(p),\quad \forall \, p \in P. \end{aligned}$$

We use this construction of the function u to prove that it belongs to the space of functions (A.21) and satisfies properties (A.22) and (A.23) in the following steps.

Step 2

(Proof of $u\in C^2(P\backslash \partial ^T P)$ and u satisfies (A.22)) Note that $|v_k(t,p)|\le 1$, for all $(t,p)\in [0,\infty )\times P$ and for all $k\in \mathbb {N}$. Let U be a relatively open set in P such that $\hbox {dist}(\bar{U}, P\backslash \partial ^TP)>0$. We can apply [23, Theorem 1.2] to conclude that, for all $l\in \mathbb {N}$, there is a positive constant, $C=C(L,l)$, such that

$$\begin{aligned} \Vert v_k(t,\cdot )\Vert _{C^l(\bar{U})} \le C,\quad \forall \, t\in [0,\infty ),\quad \forall \, k\in \mathbb {N}. \end{aligned}$$

The construction of the function u in Step 1, the preceding estimate which is uniform in $t\in [0,\infty )$ and $k\in \mathbb {N}$, and an application of the Arzelà-Ascoli Theorem imply that u belongs to $C^{\infty }(P\backslash \partial ^T P)$, since the relatively open set $U\subset P\backslash \partial ^T P$ such that $\hbox {dist}(\bar{U},\partial ^TP)>0$ was arbitrarily chosen. Moreover, because the functions $v_k$ are solutions to the parabolic equation (4.1), the preceding observation implies that the hitting probability u satisfies equation (A.22).

Step 3

(Proof of $u\in C(P\backslash (H_i\cap H_j))$ and of (A.23)) To establish that $u\in C(P\backslash (H_i\cap H_j))$ and satisfies (A.23), it is sufficient to prove that

$$\begin{aligned} \lim _{q\rightarrow p} u(q) =0,\quad \forall \, p\in \partial ^T P\backslash (H_i\cap H_j). \end{aligned}$$

Let $p\in \partial ^T P \backslash (H_i\cap H_j)$. Then there is a positive constant r such that $B_r(p)\subset P\backslash (H_i\cap H_j)$. Let $\varsigma$ be the first hitting time of the Kimura diffusion on the set $(\partial B_r(p)\cap \partial ^T P)\cup (\partial B_r(p)\cap {{\mathrm{int}}}(P))$, where we recall that $B_r(p)$ is the relatively open ball centered at p of radius r with respect to the Riemannian metric induced by the generalized Kimura operator on P. The strong Markov property of Kimura diffusions established in Corollary 1.4 and the stochastic representation (A.20) of the function u give us that

$$\begin{aligned} u(q) = \mathbb {E}_{\mathbb {Q}^q}\left[ \xi (\omega (\varsigma ))\right] ,\quad \forall \, q\in \bar{B}_r(p). \end{aligned}$$

(A.29)

where $\xi =0$ on $\partial ^T P\cap \partial B_r(p)$ and $\xi =u$ on $\partial B_r(p)\cap {{\mathrm{int}}}(P)$. Without loss of generality we can choose an adapted system of coordinates in a neighborhood of p such that p is equivalent to the origin in $\bar{S}_{n,m}$ and the generalized Kimura operator takes the form (1.1) on $B^{\infty }_R$, for some $R>0$ small enough, such that $b_1=0$ on $\partial B^{\infty }_R \cap \{x_1=0\}$. Then, in the adapted system of coordinates, the local stochastic representation (A.29) becomes

$$\begin{aligned} u(z) = \mathbb {E}_{\mathbb {Q}^z} \left[ \xi (\omega (\nu ))\right] ,\quad \forall \, z\in \bar{B}^{\infty }_{\rho }, \end{aligned}$$

(A.30)

where $0<\rho <R$, $\nu$ is the first hitting time of the boundary of the set $(\partial ^TB^{\infty }_{\rho }\cap \partial B^{\infty }_{\rho }) \cup (\partial B^{\infty }_{\rho }\cap S_{n,m})$, and $\xi =0$ on $\partial ^T B^{\infty }_{\rho }\cap \partial B^{\infty }_{\rho }$ and $\xi =u$ on $\partial B^{\infty }_{\rho }\cap S_{n,m}$. The constant $\rho$ is suitably chosen below. Our goal is to prove that

$$\begin{aligned} \lim _{z\rightarrow 0} u(z) = 0. \end{aligned}$$

(A.31)

Similarly to Step 1 in the proof of Lemma A.3, we choose the barrier function:

$$\begin{aligned} w(z) = \sqrt{\frac{x_1}{\rho }} +\sum _{i=2}^n x_i + \sum _{l=1}^m y_l,\quad \forall \, z=(x,y)\in {\bar{B}}^{\infty }_{\rho }. \end{aligned}$$

We clearly have that $w(0)=0$, $w(z)>0$, for all $z\in {\bar{B}}^{\infty }_{\rho }\backslash \{0\}$, and direct calculations give us that

$$\begin{aligned} Lw(z) = \frac{-a_{11}(z)+2b_1(z)}{4\sqrt{\rho x_1}} + \sum _{i=2}^n b_i(z) +\sum _{l=1}^m 2(d_{ll}(z)+e_l(z)y_l),\quad \forall \, z\in {\bar{B}}^{\infty }_{\rho }. \end{aligned}$$

Using the fact that $b_1=0$ along $\partial B^{\infty }_{\rho }\cap \{x_1=0\}$ together with the continuity and boundedness of the coefficients, and the fact that $a_{11}\ge \delta >0$ on ${\bar{B}}^{\infty }_{\rho }$ (by Assumption 1.2), it follows that we can choose $\rho >0$ small enough so that $Lw < 0$ on ${\bar{B}}^{\infty }_{\rho }$. Thus, applying Itô’s rule, we have that

$$\begin{aligned} \mathbb {E}_{\mathbb {Q}^z}\left[ w(\omega (\nu ))\right] \le w(z),\quad \forall \, z\in {\bar{B}}^{\infty }_{\rho }. \end{aligned}$$

(A.32)

Using the definition of the function $\xi$ and of the stopping time $\nu$, together with fact that $w(0)=0$ and $w(z)>0$, for all $z\in {\bar{B}}^{\infty }_{\rho }\backslash \{0\}$, we find that there is a positive constant, K, such that $\xi (\omega (\nu )) \le Kw(\omega (\nu ))$. Identity (A.30) together with inequality (A.32) yield

$$\begin{aligned} u(z) \le K w(z),\quad \forall \, z\in {\bar{B}}^{\infty }_{\rho }, \end{aligned}$$

which implies that

$$\begin{aligned} \limsup _{z\rightarrow 0} u(z) \le K\limsup _{z\rightarrow 0} w(z) = 0. \end{aligned}$$

Because the function u is non-negative, we see that the preceding inequality implies (A.31). This completes the proof of the fact that u is continuous at p, for all $p\in \partial ^T P\backslash (H_i\cap H_j)$.

Combining Steps 1, 2, and 3 completes the proof. $\square$