Backward Stochastic Differential Equations Driven by G-Brownian Motion with Uniformly Continuous Coefficients in (y, z)

Abstract

In this paper, we investigate backward stochastic differential equations driven by G-Brownian motion with uniformly continuous coefficients in (y, z). The existence and uniqueness of solutions are obtained via a method of Picard iteration, a linearization method and a monotone convergence argument. Furthermore, we establish the corresponding comparison theorem and related nonlinear Feynman–Kac formula.

Appendix: Comparison Theorem for PDE (4.7)

Theorem A.1

Suppose assumptions (A1)–(A3) hold. Let $u^1$ be a viscosity subsolution and $u^2$ be a viscosity supersolution to PDE (4.7) satisfying the polynomial growth condition, respectively. Then, $u^1\le u^2$ on $[0, T]\times \mathbb {R}^{n}$ provided that $u^1|_{t=T}\le u^2|_{t=T}$.

By doing the transformation $v(t,x):=u(T-t,x)$, we know that if we want to get the comparison theorem A.1 for PDE (4.7), we only need to prove the following comparison theorem A.2 for PDE

$$\begin{aligned} \left\{ \begin{array}{ll} &{}\partial _{t}v-F(t,x,v(t,x),D_{x}v,D^{2}_{x}v)=0\\ &{} v(0,x)=\phi (x),\\ \end{array} \right. \end{aligned}$$

(A.1)

where F is defined by (4.8). For the definition of the viscosity solution to PDE (A.1), see Definition 1.1 in Appendix C of [13].

Theorem A.2

Suppose assumptions (A1)–(A3) hold. Let $v^1$ be a viscosity subsolution and $v^2$ be a viscosity supersolution to PDE (A.1) satisfying the polynomial growth condition, respectively. Then, $v^1\le v^2$ on $[0, T]\times \mathbb {R}^{n}$ provided that $v^1|_{t=0}\le v^2|_{t=0}$.

Proof

The main idea comes from Theorem 2.2 in Appendix C of [13]. For reader’s convenience, we give a brief proof.

step1.

For a large constant $\lambda >0$ to be chosen in step 2, we set $\eta (x):=(1+|x|^{2})^{l/2}$ and

$$\begin{aligned} v_1(t,x):=v^1(t,x)\eta ^{-1} (x)e^{-\lambda t}, \ v_2(t,x):=v^2(t,x)\eta ^{-1} (x)e^{-\lambda t}, \end{aligned}$$

where $l\ge 2$ is chosen to be large enough such that $|v_{1}|+|v_{2}|\rightarrow 0$ uniformly as $|x|\rightarrow \infty$. It is easy to verify that, for $i=1,2$, $v_{i}$ is a bounded viscosity subsolution of

$$\begin{aligned} \partial _{t}v_{i}+\lambda v_i-F^*_i(t,x,v _{i},Dv_i,D^{2}v_{i})=0, \end{aligned}$$

where the function $F^*_1(t,x,v,p,X)=F^*(t,x,v,p,X), F^*_2(t,x,v,p,X)=-F^*(t,x,-v,-p,-X)$ and

$$\begin{aligned}&F^*(t,x,v,p,X)\\&\quad :=e^{-\lambda t}\eta ^{-1}(x)F(t,x,e^{\lambda t}v\eta (x),e^{\lambda t}(p\eta (x)+vD\eta (x) ),e^{\lambda t}(X\eta (x)\\&\qquad +p\otimes D\eta (x) +D\eta (x) \otimes p+vD^{2}\eta (x) )) \end{aligned}$$

for any $(x,v,p,X)\in \mathbb {R}^n\times \mathbb {R}\times \mathbb {R}^n\times \mathbb {S}_{n}$. Here, $p\otimes D\eta (x)=(p^iD_{x_{j}}\eta )^{n}_{i,j=1}$ and

$$\begin{aligned} D\eta (x)&=l \eta (x)(1+|x|^{2})^{-1}x,\ \ \ \ \\ \ D^{2}\eta (x)&=\eta (x)[l(1+|x|^{2})^{-1}I_{n}+l(l-2)(1+|x|^{2})^{-2}x\otimes x]. \end{aligned}$$

Obviously, for $l\ge 2$, $\eta ^{-1}(x)D\eta (x)$ and $\eta ^{-1}(x)D^{2}\eta (x)$ are uniformly bounded functions.

Now, prove that $v_{1}+v_{2}\le 0$. According to the proof of Theorem 2.2 in [13], it suffices to prove the conclusion under the additional assumptions: for each $\bar{\delta }>0$,

$$\begin{aligned} \partial _{t}v_{i}+\lambda v_{i} -F^*_{i}(t,x,v_{i},Dv_{i},D^{2}v_{i})\le -\bar{\delta }/T^{2},\ \text {and }\lim _{t\rightarrow T}v_{i}(t,x)=-\infty \ \text {uniformly on } \mathbb {R}^{n}. \end{aligned}$$

(A.2)

Assume the contrary that

$$\begin{aligned} \sup _{(t,x)\in [0,T)\times \mathbb {R}^{n}}(v_{1}(t,x)+v_{2}(t,x))>0. \end{aligned}$$

Notice that $\left( v_1(t,x)+v_2(t,x)\right) ^+\rightarrow 0$ uniformly as $|x|\rightarrow \infty$. Then, according to the proof of Theorem 2.2 in [13], for large enough $\alpha >0$, there exists some point $(t^{\alpha },x_{1}^{\alpha },x_{2}^{\alpha })$ inside a compact subset of $[0,T)\times \mathbb {R}^{2n}$ such that $v_{1}(t^{\alpha },x_{1}^{\alpha })+v_{2}(t^{\alpha },x_{2}^{\alpha })-\frac{\alpha }{2}|x_{1}^{\alpha }-x_{2}^{\alpha }|^2>0$ and

$$\begin{aligned} \ \lim _{\alpha \rightarrow \infty }\alpha |x_{1}^{\alpha }-x_{2}^{\alpha }|^2=0, \ \text {and}\ \lim _{\alpha \rightarrow \infty }(t^\alpha ,x_{1}^{\alpha },x_{2}^{\alpha })=(\hat{t},\hat{x},\hat{x})\ \hbox { for some}\ (\hat{t},\hat{x})\in (0,T)\times \mathbb {R}^n. \end{aligned}$$

Then, there exist $b_{i}^{\alpha }\in \mathbb {R}$, $X_{i}^{\alpha }\in \mathbb {S}_{n}$ such that $b_1^{\alpha }+b^{\alpha }_2=0$,

$$\begin{aligned} (b_{1}^{\alpha },p_{1}^{\alpha },X_{1}^{\alpha } )\in \mathcal {\bar{P}}^{2,+}v_{1}(t^{\alpha },x_{1}^{\alpha }), \ \ (b_{2}^{\alpha },p_{2}^{\alpha },X_{2}^{\alpha } )\in \mathcal {\bar{P}}^{2,+}v_{2}(t^{\alpha },x_{2}^{\alpha }), \end{aligned}$$

and

$$\begin{aligned} \left( \begin{array}[c]{cccc} X_1^{\alpha } &{} 0\\ 0 &{} X_2^{\alpha } \end{array} \right) \le A+\frac{1}{\alpha }A^{2}, \ A=\alpha \left( \begin{array}[c]{cccc} I_n &{} -I_n\\ -I_n &{} I_n \end{array}\right) , \end{aligned}$$

where $p_{1}^{\alpha }=\alpha (x_{1}^{\alpha }-x_{2}^{\alpha }), p_{2}^{\alpha }=\alpha (x_{2}^{\alpha }-x_{1}^{\alpha }).$ Obviously, $p_{1}^{\alpha }+p_{2}^{\alpha }=0$, $\lim _{\alpha \rightarrow \infty }|p_{i}^{\alpha }||x_{1}^{\alpha }-x_{2}^{\alpha }|=0$ and

$$\begin{aligned} \left( \begin{array}[c]{cccc} X_1^{\alpha } &{} 0\\ 0 &{} X_2^{\alpha } \end{array} \right) \le 3\alpha \left( \begin{array}[c]{cccc} I_n &{} -I_n\\ -I_n &{} I_n \end{array}\right) . \end{aligned}$$

(A.3)

Moreover, it follows from Eq. (A.2) that

$$\begin{aligned}&b_{1}^{\alpha }+\lambda v_{1}(t^\alpha ,x_1^{\alpha }) -F^*_{1}(t^{\alpha },x_1^{\alpha },v_{1}(t^\alpha ,x_1^{\alpha }),p_{1}^{\alpha },X_1^{\alpha })\le -\bar{\delta }/T^{2},\\&b_{2}^{\alpha }+\lambda v_{2}(t^\alpha ,x_2^{\alpha }) -F^*_{2}(t^{\alpha },x_2^{\alpha },v_{2}(t^\alpha ,x_2^{\alpha }),p_{2}^{\alpha },X_2^{\alpha })\le -\bar{\delta }/T^{2}. \end{aligned}$$

According to the definition of $F^*_i$, we derive that

$$\begin{aligned} -2\bar{\delta }/T^{2}&\ge \lambda v_{2}(t^{\alpha },x_{2}^{\alpha })+F^*(t^{\alpha },x_{2}^{\alpha },-v_{2}(t^{\alpha },x_{2}^{\alpha }),p_{1}^{\alpha },-X_{2}^{\alpha })\\&\quad +\lambda v_{1}(t^{\alpha },x_{1}^{\alpha })-F^*(t^{\alpha },x_{1}^{\alpha },v_{1}(t^{\alpha },x_{1}^{\alpha }),p_{1}^{\alpha },X_{1}^{\alpha }),\\&:=I_{1}^{\alpha }+I_{2}^{\alpha }, \end{aligned}$$

where

$$\begin{aligned} I_{1}^{\alpha }&:=\lambda v_{2}(t^{\alpha },x_{2}^{\alpha })+F^*(t^{\alpha },x_{2}^{\alpha },-v_{2}(t^{\alpha },x_{2}^{\alpha }),p_{1}^{\alpha },-X_{2}^{\alpha }) +\lambda v_{1}(t^{\alpha },x_{1}^{\alpha })\\&\quad -F^*(t^{\alpha },x_{2}^{\alpha },v_{1}(t^{\alpha },x_{1}^{\alpha }),p_{1}^{\alpha },-X_{2}^{\alpha })\\ I_{2}^{\alpha }&:=F^*(t^{\alpha },x_{2}^{\alpha },v_{1}(t^{\alpha },x_{1}^{\alpha }),p_{1}^{\alpha },-X_{2}^{\alpha })- F^*(t^{\alpha },x_{1}^{\alpha },v_{1}(t^{\alpha },x_{1}^{\alpha }),p_{1}^{\alpha },X_{1}^{\alpha }). \end{aligned}$$

We claim that

$$\begin{aligned} I_{1}^{\alpha }\ge 0\ \text { as } \ \alpha \rightarrow \infty , \end{aligned}$$

(A.4)

and

$$\begin{aligned} I_{2}^{\alpha }\ge 0\ \text {as} \ \alpha \rightarrow \infty , \end{aligned}$$

(A.5)

whose proof will be given in step 2 and step 3, respectively. Then, we induce a contradiction. Consequently, we get that $v^1\le v^2.$ The proof is complete.

step2. The proof of (A.4).

Note that $|G(A)|\le \frac{1}{2}\bar{\sigma }^{2}\sqrt{d}|A|, \forall A\in \mathbb {S}_{d}.$ For large enough $\alpha ,$ any $v\ge \tilde{v}$, by the definition of $F^*$, assumptions (A2), (A3) and Remark 4.1, we can get

$$\begin{aligned}&F^*(t^{\alpha },x_{2}^{\alpha },v,p_{1}^{\alpha },-X_{2}^{\alpha })-F^*(t^{\alpha },x_{2}^{\alpha },\tilde{v},p_{1}^{\alpha },-X_{2}^{\alpha })\\&\quad \le C[(v-\tilde{v})+\rho (C(v-\tilde{v}))+\psi (C(v-\tilde{v}))]\\&\quad \le C+C(v-\tilde{v}), \end{aligned}$$

where the constant C is only dependent on $L,K,\bar{\sigma },d,T.$ Then, we have

$$\begin{aligned}&-\lambda v+F^*(t^{\alpha },x_{2}^{\alpha },v,p_{1}^{\alpha },-X_{2}^{\alpha })+\lambda \tilde{v}-F^*(t^{\alpha },x_{2}^{\alpha },\tilde{v},p_{1}^{\alpha },-X_{2}^{\alpha })\\&\quad \le -(\lambda -C)(v-\tilde{v})+C. \end{aligned}$$

By virtue of the boundness of the viscosity solution v and $\tilde{v}$, we can choose $\lambda$ large enough, so that the function

$$\begin{aligned} v\rightarrow -\lambda v+F^*(t^{\alpha },x_{2}^{\alpha },v,p_{1}^{\alpha },-X_{2}^{\alpha })\ \text {is non-increasing.} \end{aligned}$$

In view of $-v_{2}(t^{\alpha },x_{2}^{\alpha })<v_{1}(t^{\alpha },x_{1}^{\alpha })$, (A.4) is proved.

step3. The proof of (A.5).

Note that $G(A)\le \frac{1}{2}\overline{\sigma }^2\mathrm {tr}[A]$ for any $A\ge 0$. By virtue of (A.3) and assumption (A2), we can get

$$\begin{aligned}&G(\sigma ^{T}(t^{\alpha },x_{1}^{\alpha })X_{1}^{\alpha }\sigma (t^{\alpha },x_{1}^{\alpha })-\sigma ^{T}(t^{\alpha },x_{2}^{\alpha })(-X_{2}^{\alpha })\sigma (t^{\alpha },x_{2}^{\alpha }))\\&\quad =G\left( \left( \begin{array}[c]{cccc} \sigma ^{T}(t^{\alpha },x_{1}^{\alpha }) &{} \sigma ^{T}(t^{\alpha },x_{2}^{\alpha })\\ \end{array} \right) \left( \begin{array}[c]{cccc} X_1^{\alpha } &{} 0\\ 0 &{} X_2^{\alpha } \end{array} \right) \left( \begin{array}[c]{cccc} \sigma (t^{\alpha },x_{1}^{\alpha }) \\ \sigma (t^{\alpha },x_{2}^{\alpha }) \end{array} \right) \right) \\&\quad \le 3\alpha G\left( \left( \begin{array}[c]{cccc} \sigma ^{T}(t^{\alpha },x_{1}^{\alpha }) &{} \sigma ^{T}(t^{\alpha },x_{2}^{\alpha })\\ \end{array} \right) \left( \begin{array}[c]{cccc} I_{n} &{} -I_{n}\\ -I_{n} &{} I_{n} \end{array} \right) \left( \begin{array}[c]{cccc} \sigma (t^{\alpha },x_{1}^{\alpha }) \\ \sigma (t^{\alpha },x_{2}^{\alpha }) \end{array} \right) \right) \\&\quad =3\alpha G\left( (\sigma (t^{\alpha },x_{1}^{\alpha })-\sigma (t^{\alpha },x_{2}^{\alpha }))^{T}(\sigma (t^{\alpha },x_{1}^{\alpha })-\sigma (t^{\alpha },x_{2}^{\alpha })\right) \\&\quad \le \frac{3}{2}\bar{\sigma }^{2}\alpha |\sigma (t^{\alpha },x_{1}^{\alpha })-\sigma (t^{\alpha },x_{2}^{\alpha })|^{2}\\&\quad \le \frac{3}{2}\bar{\sigma }^{2}L^{2}\alpha |x_{1}^{\alpha }-x_{2}^{\alpha }|^{2}. \end{aligned}$$

Note that $|G(A)|\le \frac{1}{2}\bar{\sigma }^{2}\sqrt{d}|A|, \forall A\in \mathbb {S}_{d}.$ According to the definition of $F^*$, assumptions (A2), (A3) and Remark 4.1, we can obtain

$$\begin{aligned}&I_{2}^{\alpha }\ge -C[|x_{1}^{\alpha }-x_{2}^{\alpha }|+|p_{1}^{\alpha }||x_{1}^{\alpha }-x_{2}^{\alpha }|+|v_{1}(t^{\alpha },x_{1}^{\alpha })||x_{1}^{\alpha }-x_{2}^{\alpha }|\\&\quad +\rho (C|v_{1}(t^{\alpha },x_{1}^{\alpha })||x_{1}^{\alpha }-x_{2}^{\alpha }|) +\psi (C(|p_{1}^{\alpha }||x_{1}^{\alpha }-x_{2}^{\alpha }|+|v_{1}(t^{\alpha },x_{1}^{\alpha })||x_{1}^{\alpha }-x_{2}^{\alpha }|))\\&\quad +\alpha |x_{1}^{\alpha }-x_{2}^{\alpha }|^{2}], \end{aligned}$$

where the constant C is only dependent on $L,K,\bar{\sigma },d,T.$ Note that $\lim _{\alpha \rightarrow \infty }|x_{1}^{\alpha }-x_{2}^{\alpha }|=0,$ $\lim _{\alpha \rightarrow \infty }$ $|v_{1}(t^\alpha ,x^\alpha )||x_{1}^{\alpha }-x_{2}^{\alpha }|=0,$ $\lim _{\alpha \rightarrow \infty }|p_{1}^{\alpha }||x_{1}^{\alpha }-x_{2}^{\alpha }|=0, \lim _{\alpha \rightarrow \infty }\alpha |x_{1}^{\alpha }-x_{2}^{\alpha }|^{2}=0$ and $\rho (\cdot )$ and $\psi (\cdot )$ are continuous functions. Letting $\alpha \rightarrow \infty$, we can get $I_{2}^{\alpha }\ge 0.$ (A.5) is proved. $\square$