Doubly reflected generalized BSDEs with jumps and an obstacle problem of parabolic IPDEs with nonlinear Neumann boundary conditions

Mohammed Elhachemy; Mohamed El Otmani

doi:10.1515/rose-2024-2002

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Doubly reflected generalized BSDEs with jumps and an obstacle problem of parabolic IPDEs with nonlinear Neumann boundary conditions

Mohammed Elhachemy and Mohamed El Otmani

Published/Copyright: February 28, 2024

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From the journal Random Operators and Stochastic Equations Volume 32 Issue 2

Abstract

A one-dimensional generalized backward stochastic differential equation with jumps and two barriers is the main objective of this paper. When the generators are monotone and the barriers are right continuous with left limits and completely separated, we prove the existence and uniqueness of a solution. As in application, we provide a probabilistic interpretation of a solution of a double obstacle problem of second-order parabolic integral-partial differential equations with nonlinear Neumann boundary conditions.

Keywords: Generalized BSDE with jumps; reflected; RCLL; viscosity solution; IPDE; Neumann boundary conditions

MSC 2020: 60H05; 60H10; 60H30; 35D40; 35R09

A Appendix

We consider the following generalized BSDE with jumps:

(A.1) { (i) Y ∈ S μ 2 , Z ∈ M μ 2 , d and V ∈ L μ 2 , (ii) Y t = ξ + ∫ t T f ⁢ ( s , Y s , Z s , V s ) ⁢ d s + ∫ t T g ⁢ ( s , Y s ) ⁢ d A s − ∫ t T Z s ⁢ d W s − ∫ t T ∫ E V s ⁢ ( e ) ⁢ N ̃ ⁢ ( d ⁢ s , d ⁢ e ) .

It is important to highlight that the uniqueness and existence of the solution for this generalized BSDE were ensured by [29, Theorem 55.2, p. 214]. In this appendix, we are going to establish the link between the solution of (A.1) and an IPDE with nonlinear Neumann boundary conditions. Initially, a comparison result is required in this context. However, it is worth noting that the results provided in [13, Theorems 2 and 3] remain applicable even in this particular case.

Now, taking into account the reflected SDE (4.2), for each t ∈ [ 0 , T ] , we consider the following Markovian generalized BSDE:

{ (i) Y t , x ∈ S μ 2 , Z t , x ∈ M μ 2 , d , V t , x ∈ L μ 2 , (ii) Y s t , x = H ⁢ ( X T t , x ) + ∫ s T f ⁢ ( r , X r t , x , Y r t , x , Z r t , x , V r t , x ) ⁢ d r + ∫ s T g ⁢ ( r , X r t , x , Y r t , x ) ⁢ d A r t , x − ∫ s T Z r t , x ⁢ d W r − ∫ s T ∫ E V r t , x ⁢ ( e ) ⁢ N ̃ ⁢ ( d ⁢ r , d ⁢ e ) , t ≤ s ≤ T .

Next, let v = v ⁢ ( t , x ) denote the solution of the following associated IPDE with nonlinear Neumann boundary condition:

(A.2) { − ∂ v ∂ t − L ⁢ v − f ⁢ ( t , x , v , ( ∇ v ⁢ σ ) , B ⁢ v ) = 0 for all ⁢ ( t , x ) ∈ [ 0 , T ] × G , v ⁢ ( T , x ) = H ⁢ ( x ) for all ⁢ x ∈ G , ∂ v ∂ n + g ⁢ ( t , x , v ) = 0 for all ⁢ x ∈ ∂ G ,

where ℒ, ℬ, ∂ ∂ n , 𝑓, 𝑔 and 𝐻 are the same as defined in (4.6), (4.7) and Assumption 3.

Subsequently, as in Lemma 4, it is straightforward to demonstrate that the deterministic function defined by

(A.3) v ⁢ ( t , x ) = Y t t , x

belongs to Π g c . Presently, let us concentrate on proving that 𝑣 is a unique viscosity solution of (A.2). Then we have the following theorem.

Theorem 4

Assume that Assumptions 1–5 hold. Then the deterministic function defined in (A.3) is a continuous viscosity solution of (A.2) (in the sense of Definition 2). Moreover, if, for each r > 0 , there exists a continuous function m r : R + → R + , m r ⁢ ( 0 ) = 0 , such that | f ⁢ ( t , x , y , z , q ) − f ⁢ ( t , x ′ , y , z , q ) | ≤ m r ⁢ ( | x − x ′ | ⁢ ( 1 + | z | ) ) , then the solution v ⁢ ( t , x ) is unique in the class of function of Π g c .

Proof

We will prove only the existence; the claim of uniqueness can be found in [22, Theorem 4.7, p. 48]. Initially, let us demonstrate that 𝑣 is a viscosity subsolution of (A.2). A similar argument will establish its role as a viscosity supersolution of (A.2). Let φ ∈ C 1 , 2 ⁢ ( [ 0 , T ] × G ) and ( t 0 , x 0 ) ∈ [ 0 , T ] × G such that φ ⁢ ( t 0 , x 0 ) = v ⁢ ( t 0 , x 0 ) and φ ⁢ ( t , x ) ≥ v ⁢ ( t , x ) for all ( t , x ) ∈ [ 0 , T ] × G ̄ .

Step 1. Suppose that x 0 ∈ G and that

− φ t ⁢ ( t 0 , x 0 ) − L ⁢ φ ⁢ ( t 0 , x 0 ) − f ⁢ ( t 0 , x 0 , v ⁢ ( t 0 , x 0 ) , ( ∇ φ ⁢ σ ) ⁢ ( t 0 , x 0 ) , B ⁢ φ ⁢ ( t 0 , x 0 ) ) > 0 ,

and we will find a contradiction. It follows from the continuity of 𝑓, 𝑔, 𝑏, 𝜎, 𝑐 and 𝜑 that there exist ε > 0 and η ε > 0 such that, for all ( t , x ) , t 0 ≤ t ≤ t 0 + η ε and { x : | x − x 0 | ≤ η ε } ⊂ G , we have v ⁢ ( t , x ) ≥ ℓ ⁢ ( t , x ) + ε and

(A.4) − φ t ⁢ ( t , x ) − L ⁢ φ ⁢ ( t , x ) − f ⁢ ( t , x , v ⁢ ( t , x ) , ( ∇ φ ⁢ σ ) ⁢ ( t , x ) , B ⁢ φ ⁢ ( t , x ) ) ≥ ε .

Define

(A.5) τ = inf { s ≥ t 0 : | X s t 0 , x 0 − x 0 | > η ε } ∧ ( t 0 + η ε ) .

Then, for all s ∈ [ t 0 , τ ] , we have

Y s t 0 , x 0 = Y τ t 0 , x 0 + ∫ s τ f ⁢ ( r , X r t 0 , x 0 , Y r t 0 , x 0 , Z r t 0 , x 0 , V r t 0 , x 0 ) ⁢ d r − ∫ s τ Z r t 0 , x 0 ⁢ d W r − ∫ s τ ∫ E V r t 0 , x 0 ⁢ ( e ) ⁢ N ̃ ⁢ ( d ⁢ r , d ⁢ e ) .

On the other hand, applying Itô’s formula to φ ⁢ ( s , X s t 0 , x 0 ) yields that

φ ⁢ ( τ , X τ t 0 , x 0 ) = φ ⁢ ( s , X s t 0 , x 0 ) + ∫ s τ ∂ φ ∂ r ⁢ ( r , X r t 0 , x 0 ) ⁢ d r + ∫ s τ ∇ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ d X r t 0 , x 0 + 1 2 ⁢ ∫ s τ D x 2 ⁢ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ ( σ ⁢ σ T ) ⁢ ( X r t 0 , x 0 ) ⁢ d r + ∫ s τ ∫ E [ φ ⁢ ( r , X r − t 0 , x 0 + c ⁢ ( X r − t 0 , x 0 , e ) ) − φ ⁢ ( r , X r − t 0 , x 0 ) − ∇ φ ⁢ ( r , X r − t 0 , x 0 ) ⁢ c ⁢ ( X r − t 0 , x 0 , e ) ] ⁢ N ⁢ ( d ⁢ s , d ⁢ e ) = φ ⁢ ( s , X s t 0 , x 0 ) + ∫ s τ [ ∂ φ ∂ r ⁢ ( r , X r t 0 , x 0 ) + ∇ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ b ⁢ ( X r t 0 , x 0 ) + 1 2 ⁢ D x 2 ⁢ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ ( σ ⁢ σ T ) ⁢ ( X r t 0 , x 0 ) ] ⁢ d r + ∫ s τ ∇ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ σ ⁢ ( X r t 0 , x 0 ) ⁢ d W r + ∫ s τ ∫ E [ φ ⁢ ( r , X r − t 0 , x 0 + c ⁢ ( X r − t 0 , x 0 , e ) ) − φ ⁢ ( r , X r − t 0 , x 0 ) ] ⁢ N ̃ ⁢ ( d ⁢ s , d ⁢ e ) + ∫ s τ ∫ E [ φ ⁢ ( r , X r − t 0 , x 0 + c ⁢ ( X r − t 0 , x 0 , e ) ) − φ ⁢ ( r , X r − t 0 , x 0 ) − ∇ φ ⁢ ( r , X r − t 0 , x 0 ) ⁢ c ⁢ ( X r − t 0 , x 0 , e ) ] ⁢ λ ⁢ ( d ⁢ e ) ⁢ d r .

Then

φ ⁢ ( s , X s t 0 , x 0 ) = φ ⁢ ( τ , X τ t 0 , x 0 ) − ∫ s τ [ ∂ φ ∂ r + L ⁢ φ ] ⁢ ( r , X r t 0 , x 0 ) ⁢ d r − ∫ s τ ∇ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ σ ⁢ ( X r t 0 , x 0 ) ⁢ d W r − ∫ s τ ∫ E [ φ ⁢ ( r , X r − t 0 , x 0 + c ⁢ ( X r − t 0 , x 0 , e ) ) − φ ⁢ ( r , X r − t 0 , x 0 ) ] ⁢ N ̃ ⁢ ( d ⁢ s , d ⁢ e ) , t 0 ≤ s ≤ τ .

Now, by assumption (A.4), we have

− [ ∂ φ ∂ s + L ⁢ φ ] ⁢ ( s , X s t 0 , x 0 ) − f ⁢ ( s , X s t 0 , x 0 , φ ⁢ ( s , X s t 0 , x 0 ) , ( ∇ φ ⁢ σ ) ⁢ ( s , X s t 0 , x 0 ) , B ⁢ φ ⁢ ( s , X s t 0 , x 0 ) ) ≥ ε .

Also, φ ⁢ ( τ , X τ t 0 , x 0 ) ≥ v ⁢ ( τ , X τ t 0 , x 0 ) = Y τ t 0 , x 0 . We deduce with the help of [13, Theorem 3] that

φ ⁢ ( t 0 , x 0 ) > φ ⁢ ( t 0 , X t 0 t 0 , x 0 ) − ε ⁢ ( τ − t 0 ) ≥ v ⁢ ( t 0 , x 0 ) ,

which contradicts our assumption.

Step 2. Now suppose that x 0 ∈ ∂ G and that

[ − φ t ⁢ ( t 0 , x 0 ) − L ⁢ φ ⁢ ( t 0 , x 0 ) − f ⁢ ( t 0 , x 0 , v ⁢ ( t 0 , x 0 ) , ( ∇ φ ⁢ σ ) ⁢ ( t 0 , x 0 ) , B ⁢ φ ⁢ ( t 0 , x 0 ) ) ] ∧ [ − ∂ φ ∂ n ⁢ ( t 0 , x 0 ) − g ⁢ ( t 0 , x 0 , φ ⁢ ( t 0 , x 0 ) ) ] > 0 ,

and we will find a contradiction. It follows from the continuity of 𝑓, 𝑔, 𝑏, 𝜎, 𝑐 and 𝜑 that there exist ε > 0 and η ε > 0 such that, for all ( t , x ) , t 0 ≤ t ≤ t 0 + η ε and { x : | x − x 0 | ≤ η ε } , we have

(A.6) [ − φ t ⁢ ( t , x ) − L ⁢ φ ⁢ ( t , x ) − f ⁢ ( t , x , v ⁢ ( t , x ) , ( ∇ φ ⁢ σ ) ⁢ ( t , x ) , B ⁢ φ ⁢ ( t , x ) ) ] ∧ [ − ∂ φ ∂ n ⁢ ( t , x ) − g ⁢ ( t , x , φ ⁢ ( t , x ) ) ] ≥ ε .

Let 𝜏 the stopping time defined as above in (A.5) and note that, for all s ∈ [ t 0 , τ ] , we have

Y s t 0 , x 0 = Y τ t 0 , x 0 + ∫ s τ f ⁢ ( r , X r t 0 , x 0 , Y r t 0 , x 0 , Z r t 0 , x 0 , V r t 0 , x 0 ) ⁢ d r + ∫ s τ g ⁢ ( r , X r t 0 , x 0 , Y r t 0 , x 0 ) ⁢ d A r t 0 , x 0 − ∫ s τ Z r t 0 , x 0 ⁢ d W r − ∫ s τ ∫ E V r t 0 , x 0 ⁢ ( e ) ⁢ N ̃ ⁢ ( d ⁢ r , d ⁢ e ) .

On the other hand, applying Itô’s formula to φ ⁢ ( s , X s t 0 , x 0 ) , we end up with

φ ⁢ ( s , X s t 0 , x 0 ) = φ ⁢ ( τ , X τ t 0 , x 0 ) − ∫ s τ [ ∂ φ ∂ r + L ⁢ φ ] ⁢ ( r , X r t 0 , x 0 ) ⁢ d r + ∫ s τ ∂ φ ∂ n ⁢ ( r , X r t 0 , x 0 ) ⁢ d A r t 0 , x 0 − ∫ s τ ∇ φ ⁢ ( r , X r t 0 , x 0 ) ⁢ σ ⁢ ( X r t 0 , x 0 ) ⁢ d W r − ∫ s τ ∫ E [ φ ⁢ ( r , X r − t 0 , x 0 + c ⁢ ( X r − t 0 , x 0 , e ) ) − φ ⁢ ( r , X r − t 0 , x 0 ) ] ⁢ N ̃ ⁢ ( d ⁢ s , d ⁢ e ) , t 0 ≤ s ≤ τ .

Now, by assumption (A.6), we have

( − [ ∂ φ ∂ s + L ⁢ φ ] ⁢ ( s , X s t 0 , x 0 ) − f ⁢ ( s , X s t 0 , x 0 , φ ⁢ ( s , X s t 0 , x 0 ) , ( ∇ φ ⁢ σ ) ⁢ ( s , X s t 0 , x 0 ) , B ⁢ φ ⁢ ( s , X s t 0 , x 0 ) ) ) ∧ [ − ∂ φ ∂ n ⁢ ( s , X s t 0 , x 0 ) − g ⁢ ( s , X s t 0 , x 0 , φ ⁢ ( s , X s t 0 , x 0 ) ) ] ≥ ε .

Also, φ ⁢ ( τ , X τ t 0 , x 0 ) ≥ v ⁢ ( τ , X τ t 0 , x 0 ) = Y τ t 0 , x 0 . We deduce with the help of [13, Theorem 3] that

φ ⁢ ( t 0 , x 0 ) > φ ⁢ ( t 0 , X t 0 t 0 , x 0 ) − ε ⁢ ( τ − t 0 ) ≥ Y t 0 t 0 , x 0 = v ⁢ ( t 0 , x 0 ) ,

which leads to a contradiction. ∎

Communicated by: Vyacheslav L. Girko

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Received: 2023-02-02

Accepted: 2024-01-14

Published Online: 2024-02-28

Published in Print: 2024-06-01

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Keywords for this article

Generalized BSDE with jumps; reflected; RCLL; viscosity solution; IPDE; Neumann boundary conditions