Deep PDE solution to BSDE

Abstract

We numerically solve a high-dimensional backward stochastic differential equation (BSDE) by solving the corresponding partial differential equation (PDE) instead. To have a good approximation of the gradient of the solution of the PDE, we numerically solve a coupled PDE, consisting of the original semilinear parabolic PDE and the PDEs for its derivatives. We then prove the existence and uniqueness of the classical solution of this coupled PDE, and then show how to truncate the unbounded domain to a bounded one, so that the error between the original solution and that of the same coupled PDE but on the bounded domain, is small. We then solve this coupled PDE using neural networks, and proceed to establish a convergence of the numerical solution to the true solution. Finally, we test this on 100-dimensional Allen–Cahn equation, a nonlinear Black–Scholes equation and other examples. We also compare our results to the result of solving the BSDE directly.

Appendix

In this appendix, we present the proof of all the theorems and lemmas from the paper.

Proof for Theorem 4.1

First, we prove $\lim \nolimits _{\left| \left| x \right| \right| \rightarrow \infty }u(t,x)= 0$. Recall u is the solution of PDE:

$$\begin{aligned} -u_t+{\mathcal {L}}u=-f(t,x,u,\sigma \nabla u),~ (t,x) \in (0,T]\times {\mathbb {R}}^d, \quad u(0,x)=g(x), ~x\in {\mathbb {R}}^d, \end{aligned}$$

(15)

where

$$\begin{aligned} {\mathcal {L}} =\frac{1}{2}\sum _{i,j=1}^d(\sigma \sigma ^T)_{i,j}(t,x)\frac{\partial ^2}{\partial x_i \partial x_j}+\sum _{i=1}^db_i(t,x)\frac{\partial }{\partial x_i}. \end{aligned}$$

By Delarue (2002) [Lemma 2.1], there exists a unique classical solution of PDE (15). By Assumption 4.1, there exist constant vectors $K_1,K_2\in {\mathbb {R}}^d$, such that

$$\begin{aligned} f(t,x,r,0)+K_1^Tp\le f(t,x,r, p)\le f(t,x,r,0)+K_2^Tp \end{aligned}$$

for any $(t,x,r,p)\in [0,T]\times {\mathbb {R}}^d\times {\mathbb {R}}\times {\mathbb {R}}^d$. Consider two auxiliary problems

$$\begin{aligned} -{w_1}_t+{\mathcal {L}}{w_1} & = {} -f(t,x,{w_1},0)-K_1^T\nabla {w_1},~ (t,x) \in (0,T]\times {\mathbb {R}}^d, \quad {w_1}(0,x)\nonumber \\ & = {} g(x), ~x\in {\mathbb {R}}^d, \end{aligned}$$

(16)

and

$$\begin{aligned} -{w_2}_t+{\mathcal {L}}{w_2} & = {} -f(t,x,{w_2},0)-K_2^T\nabla {w_2},~ (t,x) \in (0,T]\times {\mathbb {R}}^d, \quad {w_2}(0,x)\nonumber \\ & = {} g(x), ~x\in {\mathbb {R}}^d. \end{aligned}$$

(17)

Also by Delarue (2002) [Lemma 2.1], there exist classical solutions for these two PDEs. By PDEs (15), (16) and (17), we have that

$$\begin{aligned}{} & {} u_t+F(t,x,u,\nabla u,\nabla ^2 u)=0,~ (t,x) \in (0,T]\times {\mathbb {R}}^d,\nonumber \\{} & {} {w_1}_t+F(t,x,{w_1},\nabla {w_1},\nabla ^2 {w_1})\le {w_1}_t-{\mathcal {L}}{w_1}-f(t,x,{w_1},0)-K_1^T\nabla {w_1} =0,\nonumber \\{} & {} {w_2}_t+F(t,x, {w_2},\nabla {w_2},\nabla ^2 {w_2})\ge {w_2}_t-{\mathcal {L}}{w_2}-f(t,x,{w_2},0)-K_2^T\nabla {w_2}=0, \end{aligned}$$

(18)

where $F(t,x,p, X,Y)=-{\mathcal {L}}p-f(t,x,p,X)$. Here, u is a solution to the PDE (18), $w_1$ is a sub-solution to the PDE and $w_2$ is a super-solution to the PDE. By Giga et al. (1991) [Theorem 4.1], we have the following comparison result:

$$\begin{aligned} w_1(t,x)\le u(t,x)\le w_2(t,x),~ (t,x)\in [0,T]\times {\mathbb {R}}^d. \end{aligned}$$

(19)

Note that $w_1$ and $w_2$ are classical solutions of semilinear PDEs. Together with integrable condition and uniformly Lipschitz condition of f(t, x, r, 0) a a function of r from Assumption 4.1, by Amann (1985)[Theorem 2.1], $w_1$ and $w_2$ are also global weak solutions of the corresponding semilinear PDEs. By Amann (1985)[Corollary 2.2], for all $t\in [0,T]$, we have that $w_1(t,x)$ and $w_2(t,x)$ vanish as $\left| \left| x \right| \right|$ approaches infinity. Combining this with the comparison result (19), we get that u(t, x) vanishes there as well for all $t\in [0,T]$.

Next, we prove $\lim \nolimits _{\left| \left| x \right| \right| \rightarrow \infty }{\textbf{v}}(t,x)=0$ for any $t\in [0,T]$. Recall that $v_i$ are solution of PDEs (5). We have that $f_w(t,x,u,\sigma \nabla u)$ and $f_p(t,x,u,\sigma \nabla u)$ are bounded, since f is Lipschitz as a function of both w and p. $\frac{\partial \sigma }{\partial x_i}$ is bounded by Assumption 3.1. $f_{x_i}(t,x,u,\sigma \nabla u)\in L^{d+1}({\mathbb {R}}^d)$ by Assumption 4.1, and ${\mathcal {L}}_i u\in L^{d+1}({\mathbb {R}}^d)$ by the proof above. Together with the assumption that $g\in W^{2,d+1}({\mathbb {R}}^d)$, by Amann (1985)[Corollary 2.2], we have that $\lim \nolimits _{\left| \left| x \right| \right| \rightarrow \infty }{\textbf{v}}(t,x)=0$ for any $t\in [0,T]$. $\square$

Proof for Theorem 4.2

By Assumption 4.1 and Amann (1985) [Theorem 2.1 (i)(ii)], we have that

$$\begin{aligned} u\in C^1((0,T],L^{d+1}({\mathbb {R}}^d))\cap C((0,T],W^{2,d+1}({\mathbb {R}}^d))\cap C([0,T],W^{1,d+1}({\mathbb {R}}^d)). \end{aligned}$$

Since $u\in C([0,T],W^{1,d+1}({\mathbb {R}}^d))$ and u is a classical solution, then $t\mapsto u(t,\cdot )$ is a continuous function on the compact set and thus uniformly continuous. For fixed $\epsilon >0$, there exists $\delta _i>0$, such that for $|t_1-t_2|<\delta _i$, we have $||u(t_1,\cdot )-u(t_2,\cdot )||_{W^{1,d+1}}<\epsilon$. By Brezis (2011)[Corollary 9.13], we have that

$$\begin{aligned} ||u(t_1,\cdot )-u(t_2,\cdot )||_{L^\infty ({\mathbb {R}}^d)} \le C||u(t_1,\cdot )-u(t_2,\cdot )||_{W^{1,d+1}({\mathbb {R}}^d)}<C\epsilon . \end{aligned}$$

for all $t\in [0,T]$ and $C>0$ is a constant depends only on d. Since u is a classical solution, i.e., continuous with respect to t and x, we have

$$\begin{aligned} |u(t_1,x)-u(t_2,x)|<C\epsilon \end{aligned}$$

(20)

for $|t_1-t_2|<\delta _i$ and $x\in {\mathbb {R}}^d$. We consider a partition of [0, T], i.e., $t_0=0<t_1<\cdots <t_{k}=T$, such that $\max _{0\le j\le k-1} \left| t_{j+1} -t_j \right| \le \delta _i$. By Theorem 4.1, we have that for all $t\in [0,T]$, $\lim \nolimits _{\left| \left| x \right| \right| \rightarrow \infty }u(t,x)=0$. Thus for $j=0,\ldots ,k-1$, there exists $N_j>0$ such that for $\left| \left| x \right| \right| >N_j$, we have

$$\begin{aligned} |u(t_j,x)|<\epsilon . \end{aligned}$$

(21)

Combining (20) and (21), by triangle inequality, we have that

$$\begin{aligned} |u(t,x)|<(C+1)\epsilon \end{aligned}$$

for $(t,x)\in [t_j,t_{j+1}]\times {\mathbb {R}}^d\backslash Ball(0,N_j)$. We define $N_{max}=\max _{j=0,\ldots ,k-1}(N_j)$. Then we can define $U_1:=Ball(0,N_{max})$ and conclude that $|u(t,x)|<(C+1)\epsilon$ for $(t,x)\in [0,T]\times \partial U_1$.

Let $i\in \{1,\ldots , d\}$. We now prove that for fixed $\epsilon >0$, there exists $U^i_2$ such that $|v_i(t,x)|<\epsilon$ for $(t,x)\in [0,T]\times \partial U^i_2$. As proved in Theorem 4.1, $v_i$ are global weak solution of PDEs (5), by Amann (1985) [Theorem 2.1 (i)(ii)]. So we have that

$$\begin{aligned} v_i\in C^1((0,T],L^{d+1}({\mathbb {R}}^d))\cap C((0,T],W^{2,{d+1}}({\mathbb {R}}^d))\cap C([0,T],W^{1,{d+1}}({\mathbb {R}}^d)). \end{aligned}$$

Since $v_i\in C([0,T],W^{1,d+1}({\mathbb {R}}^d))$ and $v_i$ is a classical solution, then $t\mapsto v_i(t,\cdot )$ is a continuous function on the compact set and thus uniformly continuous. For fixed $\epsilon >0$, there exists $\delta _i>0$, such that for $|t_1-t_2|<\delta _i$, we have $||v_i(t_1,\cdot )-v_i(t_2,\cdot )||_{W^{1,{d+1}}}<\epsilon$. By Brezis (2011)[Corollary 9.13], we have that

$$\begin{aligned} ||v_i(t_1,\cdot )-v_i(t_2,\cdot )||_{L^\infty ({\mathbb {R}}^d)} \le C||v_i(t_1,\cdot )-v_i(t_2,\cdot )||_{W^{1,{d+1}}({\mathbb {R}}^d)}<C\epsilon \end{aligned}$$

for all $t\in [0,T]$ and $C>0$ is a constant that depends only on d. Since $v_i$ is a classical solution, i.e., continuous with respect to t and x, we have

$$\begin{aligned} |v_i(t_1,x)-v_i(t_2,x)|<C\epsilon \end{aligned}$$

(22)

for $|t_1-t_2|<\delta _i$ and $x\in {\mathbb {R}}^d$. We consider a partition of [0, T], i.e., $t_0=0<t_1<\cdots <t_{k}=T$, such that $\max _{0\le j\le k-1} \left| t_{j+1} -t_j \right| \le \delta _i$. By Theorem 4.1, we have that for all $t\in [0,T]$, $\lim \nolimits _{\left| \left| x \right| \right| \rightarrow \infty }v_i(t,x)=0$. Thus for $j=0,\ldots ,k-1$, there exists $N^i_j>0$ such that for $\left| \left| x \right| \right| >N^i_j$, we have

$$\begin{aligned} |v_i(t_j,x)|<\epsilon . \end{aligned}$$

(23)

Combining (22) and (23), by triangle inequality, we have that

$$\begin{aligned} |v_i(t,x)|<(C+1)\epsilon \end{aligned}$$

for $(t,x)\in [t_j,t_{j+1}]\times {\mathbb {R}}^d\backslash {\text {Ball}}(0,N^i_j)$. We define $N^i_{\max }=\max _{j=0,\ldots ,k-1}(N^i_j)$. Then we can define $U^i_2:={\text {Ball}}(0,N^i_{\max })$ and we conclude that $|v_i(t,x)|<(C+1)\epsilon$ for $(t,x)\in [0,T]\times \partial U^i_2$. We define $U_2=\cup _{i=1}^d U^i_2$, then we have that $\left| \left| {\textbf{v}}(t,x) \right| \right| <(C+1)\epsilon$ for $(t,x)\in [0,T]\times \partial U_2$. Note that since $U_2^i$ are all ball center at 0, $U_2$ is also a ball center at 0. Combining result from (1) of this proof, we define $U= U_1\cup U_2$. We can conclude that $|u(t,x)|<2\epsilon$ and $\left| \left| {\textbf{v}}(t,x) \right| \right| <(C+1)\epsilon$ for $(t,x)\in [0,T]\times \partial U$. Furthermore, since $U_1$ and $U_2$ are balls centered at 0, U is still a ball center at 0. $\square$

Proof for Theorem 4.3

First, we prove this theorem for u. Let $\alpha (t,x)=\bar{u}(t,x)+\epsilon (e^{Lt}-\frac{1}{2})$, where L is the global Lipschitz constant of f(t, x, w, p) with respect to w. We then have that $\alpha (t,x)$ is the classical solution to the following PDE:

$$\begin{aligned} \begin{array}{lr} -\alpha _t+{\mathcal {L}}\alpha =-f(t,x,\bar{u},\sigma \nabla \alpha )-L\epsilon e^{Lt},~~~~ (t,x)\in (0,T]\times U,\\ \alpha (0,x)=g(x)+\epsilon \frac{1}{2},~~~~ x\in U,\\ \alpha (t,x)=\epsilon \left( e^{Lt}-\frac{1}{2}\right) ,~~~~(t,x)\in (0,T]\times \partial U. \end{array} \end{aligned}$$

(24)

Since f(t, x, w, p) is Lipschitz with respect to w, we have that $|f(t,x,\bar{u},\sigma \nabla \alpha )-f(t,x,\alpha ,\sigma \nabla \alpha )|\le L|\bar{u}-\alpha |=L\epsilon \left( e^{Lt}-\frac{1}{2}\right)$, and thus

$$\begin{aligned} -f(t,x,\bar{u},\sigma \nabla \alpha )-L\epsilon e^{Lt}\le -f(t,x,\alpha ,\sigma \nabla \alpha )-L\epsilon \frac{1}{2}< -f(t,x,\alpha ,\sigma \nabla \alpha ). \end{aligned}$$

So for PDE of $\alpha$ (24), we have

$$\begin{aligned} -\alpha _t+{\mathcal {L}}\alpha =-f(t,x,\bar{u},\sigma \nabla \alpha )-L\epsilon e^{Lt}<-f(t,x,\alpha ,\sigma \nabla \alpha ). \end{aligned}$$

Let $H(t,x,\phi ,D\phi ,D^2\phi ):={\mathcal {L}}\phi +f(t,x,\phi ,\sigma \nabla \phi )$. As a classical solution to PDE (2), we have that $u_t(t,x)= {\mathcal {L}}u(t,x)+f(t,x,u(t,x),\sigma \nabla u(t,x))$ for $(t,x)\in (0,T] \times {\mathbb {R}}^d$. Then we have that

$$\begin{aligned} \alpha _t>H(t,x,\alpha ,D\alpha ,D^2\alpha ),~ u_t\le H(t,x,u,Du,D^2u),~(t,x)\in (0,T]\times U. \end{aligned}$$

Since we have that $||u||_{L^\infty ((0,T]\times \partial U)}< \frac{\epsilon }{2}<\epsilon \left( e^{Lt}-\frac{1}{2}\right) =\alpha (t,x)$ for $(t,x)\in (0,T]\times \partial U$ and $\alpha (0,x)=g(x)+\frac{1}{2}\epsilon >g(x)=u(0,x)$ for any $x\in U$. Thus, we have $\alpha >u$ on $(0,T]\times \partial U$ and $\{0\}\times U$. By comparison theorem (Friedman, 1964)[Theorem 2.16], we have that

$$\begin{aligned} u(t,x)<\alpha (t,x)=\bar{u}(t,x)+\epsilon \left( e^{Lt}-\frac{1}{2}\right) ~(t,x)\in (0,T]\times U. \end{aligned}$$

In the same way, we can prove that

$$\begin{aligned} u(t,x)>\bar{u}(t,x)-\epsilon \left( e^{Lt}-\frac{1}{2}\right) ,~(t,x)\in (0,T]\times U. \end{aligned}$$

Finally, we have $||\bar{u}-u||_{L^\infty ((0,T]\times U )}<\epsilon \left( e^{LT}-\frac{1}{2}\right) ,$ which is the desired result in (10) for u.

Next, we prove that $||\bar{v}_i-v_i||_{L^\infty ([0,T]\times U )}<\epsilon \left( e^{LT}-\frac{1}{2}\right)$ given $||v_i||_{L^\infty ([0,T]\times \partial U )}<\frac{\epsilon }{2}$. Let $i\in \{1,\ldots ,d\}$. Let $\alpha _i(t,x)=\bar{v}_i+\epsilon (e^{Lt}-\frac{1}{2})$, then $\alpha _i$ is the classical solution to the following PDE:

$$\begin{aligned}&-\partial _t \alpha _i+{\mathcal {L}}\alpha _i=-f_w(t,x,u,\sigma \nabla u)\left( \alpha _i-\epsilon \left( e^{Lt}-\frac{1}{2}\right) \right) -f_{x_i}(t,x,u,\sigma \nabla u)\\&\quad -f_p^T(t,x,u,\sigma \nabla u)\left( \frac{\partial \sigma }{\partial x_i}\nabla u+ \sigma \nabla \alpha _i\right) -{\mathcal {L}}_i u -L\epsilon e^{Lt},~~~~ (t,x) \in (0,T]\times U,\\&\alpha _i(0,x)=g_{x_i}(x)+\epsilon \frac{1}{2},~~~~x\in U,\\&\alpha _i(t,x)=\epsilon \left( e^{Lt}-\frac{1}{2}\right) ,~~~ (t,x)\in (0,T]\times \partial U. \end{aligned}$$

Since L is the Lipschitz constant of f with respect to w, then $|f_w(t,x,w,p)|\le L$ in $[0,T]\times {\mathbb {R}}^d\times {\mathbb {R}}$, and we have that

$$\begin{aligned} f_w\epsilon (e^{Lt}-\frac{1}{2})\le L\epsilon \left( e^{Lt}-\frac{1}{2}\right) < L\epsilon e^{Lt}. \end{aligned}$$

Thus,

$$\begin{aligned}&-\partial _t \alpha _i+{\mathcal {L}}\alpha _i<-f_w(t,x,u,\sigma \nabla u)\alpha _i-f_{x_i}(t,x,u,\sigma \nabla u)\\&\quad -f_p^T(t,x,u,\sigma \nabla u)\left( \frac{\partial \sigma }{\partial x_i}\nabla u+ \sigma \nabla \alpha _i\right) -{\mathcal {L}}_i u. \end{aligned}$$

Define

$$\begin{aligned} H_i(t,x,\phi ,D\phi ,D^2\phi )&:={\mathcal {L}}\phi +f_w(t,x,u,\sigma \nabla u)\phi +f_{x_i}(t,x,u,\sigma \nabla u)\\&\quad +f_p^T(t,x,u,\sigma \nabla u)(\frac{\partial \sigma }{\partial x_i}\nabla u+ \sigma \nabla \phi )+{\mathcal {L}}_i u. \end{aligned}$$

It follows that

$$\begin{aligned} \partial _t \alpha _i>H_i(t,x,\alpha _i,D\alpha _i,D^2\alpha _i),~ v_{it}\le H_i(t,x,v_i,Dv_i,D^2v_i),~(t,x)\in (0,T]\times U. \end{aligned}$$

Since we have that $||v_i||_{L^\infty ((0,T]\times \partial U)}< \frac{\epsilon }{2}<\epsilon \left( e^{Lt}-\frac{1}{2}\right) =\alpha _i(t,x)$ on boundary $(0,T]\times \partial U$ and $\alpha _i(0,x)=g_{x_i}(x)+\frac{1}{2}\epsilon >g_{x_i}(x)=v_i(0,x)$ for any $x\in U$, we have $\alpha _i>v_i$ on $(0,T]\times \partial U$ and $\{0\}\times U$. By Comparison Theorem (Friedman, 1964)[Theorem 2.16], we have that

$$\begin{aligned} v_i(t,x)<\alpha _i(t,x)=\bar{v}_i(t,x)+\epsilon \left( e^{Lt}-\frac{1}{2}\right) ,~(t,x)\in (0,T]\times U. \end{aligned}$$

In the same way, we can prove that

$$\begin{aligned} v_i(t,x)>\bar{v}_i(t,x)-\epsilon \left( e^{Lt}-\frac{1}{2}\right) ,~(t,x)\in (0,T]\times U. \end{aligned}$$

The desired result in (10) for $v_i, \bar{v}_i$ now follows:

$$\begin{aligned} ||\bar{v}_i-v_i||_{L^\infty ((0,T]\times U )}<\epsilon \left( e^{LT}-\frac{1}{2}\right) . \end{aligned}$$

$\square$

Proof for Theorem 4.4

By Friedman (1964) [Theorems 7.4.6 and 7.4.10 ], there exists unique classical solution of PDE (11). Note that by Delarue (2002) equation 2.15, $u\ge {\bar{u}}$ on $[0,T]\times \partial \Omega$. By comparison theorem (Friedman, 1964)[Theorem 2.6.16], we have that $u\ge \bar{u}$ on $[0,T]\times \Omega$. So $u-\bar{u}=|u-\bar{u}|$, i.e., $|u(t,x)-\bar{u}(t,x)|$ is also twice continuously differentiable as a function of x. Now we want to show $|u-\bar{u}|$ is a super-solution of PDE $-u_t+{\mathcal {L}}u=f(t,x,u,\sigma \nabla u)$ with boundary condition $u-h$ and initial condition 0. By Assumption 4.2: for any $p_1,p_2\in {\mathbb {R}}^d$, $(t,x)\in [0,T]\times {\mathbb {R}}^d$ and $v_1,v_2\in {\mathbb {R}}$, $f(t,x,v_1,p_1)-f(t,x,v_2,p_2)\ge f(t,x,v_1-v_2,p_1-p_2)$. Therefore, the PDE that $u-\bar{u}$ satisfies is:

$$\begin{aligned} -(u-\bar{u})_t+{\mathcal {L}}(u-\bar{u})=f(t,x,u,\sigma \nabla u)-f(t,x,\bar{u},\sigma \nabla \bar{u})\ge f(t,x,u-\bar{u},\sigma \nabla (u-\bar{u})) \end{aligned}$$

on $(0,T]\times \Omega$, which means $|u-\bar{u}|$ is a super-solution.

Next, we find a sub-solution of the same PDE. We define a function $W(t,x):=\sum _{i=1}^d W_i$, where $W_i,~i=1,\ldots , d$, is defined as

$$\begin{aligned} W_i(t,x_i)=\frac{1}{\sqrt{t+\epsilon _i}}e^{(-\gamma _i\frac{(-k_i-\sqrt{M^2+1}+\sqrt{x_i^2+1})^2}{t+\epsilon _i})},~(t,x_i)\in [0,T]\times {\mathbb {R}},~i=1, \ldots , d, \end{aligned}$$

(25)

in which the parameters $\epsilon _i\ge 0$,$\gamma _i\ge 0$ and $k_i\ge 0$ needs to be such that $W_i$ satisfy the inequality

$$\begin{aligned} -W_{it}+{\mathcal {L}}W_i\le q_i\frac{\partial W_i}{\partial x_i}-L W_i+\frac{B}{d}. \end{aligned}$$

Here, $B\ge 0$ is the lower bound of f(t, x, 0, 0), L is the Lipschitz constant of f with respect to u, and $q:=(q_1,\ldots ,q_n)^T$ is a column vector in ${\mathbb {R}}^d$, which satisfies $f(t,x,v,0)+q^T p\le f(t,x,v, p)$ for all $(t,x,v,p)\in [0,T]\times {\mathbb {R}}^d\times {\mathbb {R}}\times {\mathbb {R}}^d$.

Given a structure of $W_i$ in (25) form, we can find $-W_{it}+{\mathcal {L}}W_i-q_i\sigma _i\frac{\partial W_i}{\partial x_i}+L W_i$ explicitly on $[0,T]\times [-M,M]$:

$$\begin{aligned}&-W_{it}+{\mathcal {L}}W_i-q_i\sigma _i\frac{\partial W_i}{\partial x_i}+L W_i\nonumber \\&\quad =\frac{W_i}{(t+\epsilon _i)^{5/2}} \Bigg [-\gamma _i Y_i^2+\frac{1}{2}(t+\epsilon _i)+\frac{2(b_i-q_i\sigma _i)x_iY_i\gamma _i(t+\epsilon _i)}{\sqrt{1+x_i^2}}+L(t+\epsilon _i)^2\nonumber \\&\qquad +\frac{4\sigma ^T\sigma _{ii}x_i^2Y_i^2\gamma _i^2}{1+x_i^2}-\frac{2\sigma ^T\sigma _{ii}x_i^2\gamma _i(t+\epsilon _i)}{1+x_i^2}+\frac{2\sigma ^T\sigma _{ii}x_i^2 Y_i \gamma _i(t+\epsilon _i)}{(1+x_i^2)^{3/2}} -\frac{2\sigma ^T\sigma _{ii}Y_i \gamma _i(t+\epsilon _i)}{\sqrt{1+x_i^2}}\Bigg ]\nonumber \\&\quad =\frac{W_i}{(t+\epsilon _i)^{5/2}}\Bigg [ Y_i^2\left( -\gamma _i+\frac{4\sigma ^T\sigma _{ii}x_i^2\gamma _i^2}{1+x_i^2}\right) +Y_i\gamma _i\left( \frac{2(b_i-q_i\sigma _i)x_i}{\sqrt{1+x_i^2}}+\frac{2\sigma ^T\sigma _{ii}x_i^2}{(1+x_i^2)^{3/2}}-\frac{2\sigma ^T\sigma _{ii}}{\sqrt{1+x_i^2}}\right) \nonumber \\&\qquad \times (t+\epsilon _i)+\left( \frac{1}{2}+L(t+\epsilon _i)-\frac{2\sigma ^T\sigma _{ii}x_i^2\gamma _i}{1+x_i^2}\right) (t+\epsilon _i) \Bigg ], \end{aligned}$$

(26)

where $Y_i:=-k_i-\sqrt{M^2+1}+\sqrt{x_i^2+1}\le -k_i$. Take $\gamma _i=\frac{1}{4||\sigma ^T\sigma _{ii}||_{L^\infty ([0,T]\times {\mathbb {R}}^d)}}$. Recall that by Assumption 3.1.1 we have that $\sigma$ is bounded; therefore, $0<\gamma _i<\infty .$ Then, $-\gamma _i+\frac{4\sigma ^T\sigma _{ii}x_i^2\gamma _i^2}{1+x_i^2}< 0$. Since $B\ge 0$, we only need to find parameters that make (26) less equal than 0. Because $\frac{W_i}{(t+\epsilon _i)^{5/2}}>0$, it is equivalent to finding parameters such that

$$\begin{aligned} \begin{aligned}&Y_i^2\left( -\gamma _i+\frac{4\sigma ^T\sigma _{ii}x_i^2\gamma _i^2}{1+x_i^2}\right) +Y_i\gamma _i\left( \frac{2(b_i-q_i\sigma _i)x_i}{\sqrt{1+x_i^2}}+\frac{2\sigma ^T\sigma _{ii}x_i^2}{(1+x_i^2)^{3/2}}-\frac{2\sigma ^T\sigma _{ii}}{\sqrt{1+x_i^2}}\right) (t+\epsilon _i)\\&\quad +\left( \frac{1}{2}+L(t+\epsilon _i)-\frac{2\sigma ^T\sigma _{ii}x_i^2\gamma _i}{1+x_i^2}\right) (t+\epsilon _i)\le 0, \end{aligned} \end{aligned}$$

(27)

for $[0,T]\times [-M,M]$. This is a quadratic function of $Y_i$, which has negative leading coefficient and $Y_i\le -k_i$. With fixed $\epsilon _i=1$, we can always find $k_i$ large enough such that (27) hold. Therefore,

$$\begin{aligned} -W_{it}+{\mathcal {L}}W_i-q_i\sigma _i\frac{\partial W_i}{\partial x_i}+L W_i\le 0 \le \frac{B}{d}. \end{aligned}$$

(28)

Next, we prove that W satisfies: $-u_t+{\mathcal {L}}u\le f(t,x,u,\nabla u)$. From (28), we have that W satisfies the following inequality:

$$\begin{aligned} -W_t+{\mathcal {L}}W=\sum _{i=1}^d -W_{it}+{\mathcal {L}}W_i\le \sum _{i=1}^d q_i\sigma _i\frac{\partial W_i}{\partial x_i} -LW+B. \end{aligned}$$

(29)

By definition (25), we have that $W_i\ge 0$, and therefore $W=|W|$. So (29) becomes:

$$\begin{aligned} \begin{aligned} -W_t+{\mathcal {L}}W&< \sum _{i=1}^d q_i\sigma _i\frac{\partial W_i}{\partial x_i} -L|W|+f(t,x,0,0)\le \sum _{i=1}^d q_i\sigma _i\frac{\partial W_i}{\partial x_i}+ f(t,x,W,0)\\&\le q^T\sigma \nabla W+f(t,x,W,0)\le f(t,x,W,\sigma \nabla W). \end{aligned} \end{aligned}$$

We define a function

$$\begin{aligned} \alpha (t,x):=\sup _{(t^\prime ,x^\prime )\in ((0,T]\times \partial \Omega ) \cup (\{0\}\times \Omega )}\left\{ \frac{|u(t^\prime ,x^\prime )-\bar{u}(t^\prime ,x^\prime )|}{W(t^\prime ,x^\prime )}\right\} W(t,x). \end{aligned}$$

Therefore, since $\alpha$ is proportional to W, it is also a sub-solution of $-\alpha _t+{\mathcal {L}}\alpha \le f(t,x,\alpha ,\sigma \nabla \alpha )$ and is greater than $u-\bar{u}$ on boundary and is strictly positioned at the initial time $t=0$. From Comparison Theorem (Friedman, 1964)[Theorem 2.6.16], we have that

$$\begin{aligned} |u(t,x)-\bar{u}(t,x)|&\le \sup _{(t^\prime ,x^\prime )\in ((0,T]\times \partial \Omega ) \cup (\{0\}\times \Omega )}\left\{ \frac{|u(t^\prime ,x^\prime )-\bar{u}(t^\prime ,x^\prime )|}{W(t^\prime ,x^\prime )}\right\} W(t,x)\\&\le ||u-\bar{u}||_{L^\infty ([0,T]\times \partial \Omega )} \cdot \frac{W(t,x)}{\inf _{(t^\prime ,x^\prime )\in [0,T]\times \partial \Omega } W(t^\prime ,x^\prime )}\\&\le ||u-\bar{u}||_{L^\infty ([0,T]\times \partial \Omega )} \cdot \frac{W(t,x)}{\sum _{i=1}^d\inf _{(t^\prime ,x^\prime )\in [0,T]\times \partial \Omega } W_i(t^\prime ,x^\prime )}\\&\le ||u-\bar{u}||_{L^\infty ([0,T]\times \partial \Omega )} \cdot \frac{W(t,x)}{\sum _{i=1}^d\inf _{{t^\prime \in [0,T]}} W_i(t^\prime ,M)}\\&\le C||u-\bar{u}||_{L^\infty ([0,T]\times \partial \Omega )}W(t,x)\\&\le C ||u-\bar{u}||_{L^\infty ([0,T]\times \partial \Omega )}\sum _{i=1}^d \frac{1}{\sqrt{t+1}}e^{\left( -\gamma _i\frac{(-k_i-\sqrt{M^2+1}+\sqrt{x_i^2+1})^2}{t+1}\right) }. \end{aligned}$$

This shows (13).

Next, we prove (14). Fix $i\in \{1,\ldots ,d\}$. Existence and uniqueness of the classical solution of PDE(12) follows, for example, from Friedman (1964)[Theorems 7.4.6 and 7.4.10]. Note that by Delarue (2002)[equation 2.15], $v_i\ge \bar{ v}_i$ on $[0,T]\times \partial \Omega$. By Comparison Theorem (Friedman, 1964)[Theorem 2.6.16], we have that $v_i\ge \bar{v}_i$ on $[0,T]\times \Omega$. So $v_i-\bar{v}_i=|v-\bar{v}_i|$, i.e., $|v_i(t,x)-\bar{v}_i(t,x)|$ is also twice continuously differentiable w.r.t. x. Note that $|v_i-\bar{v}_i|$ is a super-solution of PDE

$$\begin{aligned} -(v_{it}-{\bar{v}}_{it})+{\mathcal {L}}(v_i-\bar{v}_i)+f_w(t,x,{\bar{u}},\sigma \nabla {\bar{u}})(v_i-\bar{v}_i)+f_p^T(t,x,{\bar{u}},\sigma \nabla {\bar{u}})\sigma \nabla (v_i-\bar{v}_i)=0, \end{aligned}$$

(30)

with boundary condition $v_i-{\bar{v}}_i$ and initial condition 0. Next, we want to find a sub-solution of PDE (30). Similar to the construction in (25), we define $W^i$ as:

$$\begin{aligned} W^i=\sum _{j=1}^d \frac{1}{\sqrt{t+1}}e^{\left( -\gamma ^i\frac{(-k^i_j-\sqrt{M^2+1}+\sqrt{x_j^2+1})^2}{t+1}\right) }, \end{aligned}$$

which satisfy $-W^i_t+{\mathcal {L}}W^i+f_w(t,x,{\bar{u}},\sigma \nabla {\bar{u}})W^i+f_p^T(t,x,{\bar{u}},\sigma \nabla {\bar{u}})\sigma \nabla W^i\le 0$. Again, we define

$$\begin{aligned} \alpha ^i(t,x):=\sup _{(t^\prime ,x^\prime )\in ((0,T]\times \partial \Omega ) \cup (\{0\}\times \Omega )}\left\{ \frac{|v_i(t^\prime ,x^\prime )-\bar{v}_i(t^\prime ,x^\prime )|}{W^i(t^\prime ,x^\prime )}\right\} W^i(t,x). \end{aligned}$$

It follows that $\alpha$ is a sub-solution of $-\alpha ^i_t+{\mathcal {L}}\alpha ^i+ f_w(t,x,{\bar{u}},\sigma \nabla \bar{u})\alpha ^i+f_p^T(t,x,{\bar{u}},\sigma \nabla {\bar{u}})\sigma \nabla \alpha ^i\le 0$, and $\alpha \ge v_i-\bar{v}_i$ on the boundary $\partial \Omega$ and strictly positive at the initial time $t=0$. By the Comparison Theorem (Friedman, 1964)[Theorem 2.6.16], we have that

$$\begin{aligned} |v_i(t,x)-{\bar{v}}_i(t,x)|&\le \sup _{(t^\prime ,x^\prime )\in ((0,T]\times \partial \Omega ) \cup ({0}\times \Omega )}{\frac{|v_i(t^\prime ,x^\prime )-{\bar{v}}_i(t^\prime ,x^\prime )|}{W^i(t^\prime ,x^\prime )}}W_i(t,x)\\&\le ||v_i-{\bar{v}}_i||_{L^\infty ([0,T]\times \partial \Omega )} \cdot \frac{W_i(t,x)}{\inf {(t^\prime ,x^\prime )\in [0,T]\times \partial \Omega } W^i(t^\prime ,x^\prime )}\\&\le ||v_i-{\bar{v}}_i||_{L^\infty ([0,T]\times \partial \Omega )} \cdot \frac{W_i(t,x)}{\sum _{j=1}^d\inf _{(t^\prime ,x^\prime )\in [0,T]\times \partial \Omega } W^i_j(t^\prime ,x^\prime )}\\&\le ||v_i-{\bar{v}}_i||_{L^\infty ([0,T]\times \partial \Omega )} \cdot \frac{W_i(t,x)}{\sum _{j=1}^d\inf _{{t^\prime \in [0,T]}} W^i_j(t^\prime ,M)}\\&\le C||v_i-{\bar{v}}_i||_{L^\infty ([0,T]\times \partial \Omega )}W_i(t,x)\\&\le C ||v_i-{\bar{v}}_i||_{L^\infty ([0,T]\times \partial \Omega )}\sum _{j=1}^d \frac{1}{\sqrt{t+1}}e^{\left( -\gamma ^i\frac{(-k_j-\sqrt{M^2+1}+\sqrt{x_j^2+1})^2}{t+1}\right) }. \end{aligned}$$

This shows (14). $\square$