| Abstract: |
| We consider a backward parabolic SPDE on $\R^d$ whose solution is forced to stay between two barriers $\underline{v}$ and $\bar{v}$. More precisely, we consider the following SPDE with obstacles:
\begin{equation*}
\hspace{-0.5cm}\left\lbrace
\begin{split}
& du_t (x) + \big[
\; \frac{1}{2} \Delta u_t (x) + f_t(x,u_t (x),\nabla u_t (x))+ \mbox{div} g_t \left(x,u_t\left( x\right)
,\nabla u_t\left( x\right) \right) \, \big]\, dt\
& \hspace*{3cm} + h_t(x,u_t(x),\nabla u_t(x))\cdot \overleftarrow{dB}_t = - \nu^+ (dt,dx) + \nu^- (dt ,dx)\
& { u_T = \xi, }\
& {\underline{v}(t,x)\leq u_t (x)\leq\bar{v} (t,x)\ \ dt\times dx \ a.e.}\
\end{split}
\right.
\end{equation*}
with the Skhorod minimal conditions:
\begin{equation*}
\int_{0}^{T}\int_{\mathbb{R}^{d}}\left( \tilde{u}_{s}\left( x\right)
-\underline{v}_{s}\left( x\right) \right) \nu^+ \left( dsdx\right) =\int_{0}^{T}\int_{\mathbb{R}^{d}}\left(
\overline{v}_{s} \left( x\right)
-\tilde{u}_s \left( x\right) \right) \nu^- \left( dsdx\right) =0, \; a.s.
\end{equation*}
where $\tilde{u}$ is a {\it quasi-continuous} version of $u$.\
In this equation, the random measures $\nu^+$ and $\nu^-$ pushes the solution when it reaches the barriers.\
We prove existence and uniqueness of the solution under usual Lipschitz hypotheses.
The proof is based
on a probabilistic interpretation of the Laplacian and the divergence term thanks to a forward Brownian motion $W$ independent of $B$. Namely we prove that the solution $(u,\nu^+ ,\nu^-)$ of our SPDE with obstacles can be expressed in terms of the solution $\left(Y_t, Z_t, K_t^+,K_t^- \right)_{t \in [0,T]}$ of the following doubly reflected backward doubly stochastic differential equation :
\begin{equation}
\hspace{-0.5cm}\left\lbrace
\begin{split}
\label{RBDSDE}
Y_{t}& = \Phi\left( W_T\right) +\int\limits_{t}^{T}f_r
\left(W_{r}\right) dr +\frac{1}{2}\int\limits_{t}^{T}g_r\left(W_{r}\right) *dW + \int\limits_{t}^{T}h_r
\left(W_{r}\right)\cdot \overleftarrow{dB}_r \&\quad\quad -\sum\limits_{i}\int\limits_{t}^{T}Z_{i,r}dW_{r}^{i}+ K_T^+ - K_t ^+-K_T^-+K_t^- \,\
Y_T &=\xi (W_T)
\end{split}
\right.
\end{equation}
with $\underline{v}(t,W_t )=L_t\leq Y_t\leq S_t =\overline{v}(t,W_t),\; \forall \, t \in [0,T] $, $\left(K_t^+\right)_{t \in[0,T]}$ and $\left(K_t^-\right)_{t \in[0,T]}$ are increasing continuous processes s.t. \begin{equation}
\label{reflection:minimum}
\displaystyle \int_{0}^{T}(Y_s-L_s)dK^+_s = \int_0^T(S_t-Y_t) dK_t^-= 0.
\end{equation} |
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