Please consider the following backward stochastic differential equation: $$ Y(s)=\xi+\int_{s}^{T}a(u)Y(u)+b(u)Y(u)Z(u)du-\int_{s}^{T}Z(u)dW(u)$$ Here $a(s)$, $b(s)$ are square-integrable stochastic processes adapted to the standard filtration generated by the Brownian motion $W(s)$. $\xi$ is adapted to the whole Brownian path.

My question: Is there any reference that discusses existence of a pair of solution $(Y,Z)$ to this BSDE?

Since Pardoux-Peng first paper on BSDEs which required Lipchitz's condition on the generator, many other mathematicians including Briand and Hu, Lepeltier and San Martin, Kobylanski, Bahlali, etc have attempted to weaken this hypothesis. Their hypothesis on the generator cover cases like linear growth in $y$,$z$; superlinear in $y$ and linear in $z$; superlinear in y and quadratic in $z$; logarithmic expression like $y \log\lvert y\rvert$, etc. None however has a term $yz$ in the generator. Closest I can find are the papers here which treat generator $a+b\vert y\rvert+c \lvert z\rvert+f(y)\lvert z \rvert^2$ and here, in which Bahlali consider the generator $\lvert z \rvert^2/y$.

Appreciate your insight to this problem.

  • $\begingroup$ @StephanSturm you are right. Terminal condition is usually random. I'll fix it. $\endgroup$ – ycloud77 Jan 2 at 19:54
  • $\begingroup$ Just a quick note: Looking at the "simple" case where $\xi = W_T$ this can be seen as FBSDE which (assuming Feynman-Kac) should correspond to the semilinear PDE $-u_t + \frac{1}{2}u_{xx} = uu_x$. A quick search in Polyanin-Zaitsev didn't produce any results. $\endgroup$ – Stephan Sturm Jan 2 at 22:26
  • $\begingroup$ (just to be clear, assuming $a=0, \, b=1$). $\endgroup$ – Stephan Sturm Jan 2 at 22:34
  • $\begingroup$ @StephanSturm Your case is true. But would you agree that for the general case of a(u), b(u) being random processes Feymann-Kac type arguments are unlikely to make the problem easier since they would lead to stochastic PDEs? Apologize if sth I said is wrong. I need existence of solution here while doing research but ultimately I'm not too familiar with PDEs or SPDEs. $\endgroup$ – ycloud77 Jan 3 at 7:54
  • $\begingroup$ If $a(u)$, $b(u)$ would be random processes, the according Feynman-Kac formula would be in a three dimensional state space $(x,a,b)$. But I think one has first to understand what is happening in easy (but not trivial) model cases. If the solution doesn't exist there, we cannot expect much in general. $\endgroup$ – Stephan Sturm Jan 4 at 6:37

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