Backward Stochastic Differential Equations With Continuous Coefficient

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.

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Acknowledgements

The author is grateful for the anonymous for their careful reading of the manuscript and their helpful suggestions.

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1. School of Mathematics and Statistics, Shandong Normal University, Jinan, 250014, Shandong, People's Republic of China

   Shengqiu Sun

2. School of Mathematics, Shandong University, Jinan, 250010, Shandong, People's Republic of China

   Shengqiu Sun

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Correspondence to Shengqiu Sun.

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Appendix: Comparison Theorem for PDE (4.7)

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. \(\square \)

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Sun, S. Backward Stochastic Differential Equations Driven by G-Brownian Motion with Uniformly Continuous Coefficients in (y,z). J Theor Probab 35, 370–409 (2022). https://doi.org/10.1007/s10959-020-01057-2

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• Received: 03 March 2020

• Revised: 26 October 2020

• Accepted: 02 November 2020

• Published: 19 November 2020

• Issue Date: March 2022

• DOI : https://doi.org/10.1007/s10959-020-01057-2

Keywords

• G-Brownian motion
• G-BSDEs
• Comparison theorem
• Feynman–Kac formula

Mathematics Subject Classification (2020)

• 60H10
• 60H30

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