Feynman-Kac theorem: probabilistic proof of existence of solution to parabolic PDE

Friedman (in his book: PDEs of Parabolic Type) shows how to construct a solution to the Cauchy problem $$\partial_t u(t,x) = b(x) \partial_x u(t,x) + \frac{1}{2} \sigma(x)^2 \partial_{x,x} u(t,x)$$ with initial condition $u(0,x) = f(x)$ using the parametrix method (no probability involved) under the conditions: $b, \sigma, f$ are $\alpha$-Holder continuous. In particular, $u \in C^{1,2}$.

It is also well-know by the Feynman-Kac theorem that if $u$ has polynomial growth, then it has a stochastic representation as $u(t,x) = \mathbb{E}[f(X_t^x)]$, where $X$ solves the SDE $$X_t^x = x + \int_0^t b(X_s^x) \, ds + \int_0^t \sigma(X_s^x) \, dB_s .$$

As Friedman shows (in another book: SDEs and Applications), one can define $v(t,x) := \mathbb{E}[f(X_t^x)]$ and prove directly that $v \in C^{1,2}$ and it solves the above PDE when the following conditions are satisfied: $b, \sigma, f$ are all twice continuously differentiable with bounded derivatives.

My questions:

(1): Why do we make different assumptions about the regularity of $b, \sigma, f$ in each case?

(2): Can one define $v(t,x) := \mathbb{E}[f(X_t^x)]$ and, under the assumptions $b, \sigma, f$ are $\alpha$-Holder continuous (possibly also need linear growth), prove that $v$ solves the PDE?

• of course $b,σ,f$ are all twice continuously differentiable ! see : www1.se.cuhk.edu.hk/~seem5670/lecturenotes/Lecture5.pdf if $b,σ,f$ are not twice differentiable, how to define the stochastic diﬀerential equation ? $dX_{t}=\beta(t,X_{t})du+\gamma(t,X_{t})dW_{t}$ if we can not define the stochastic equation, how to transform it to Cauchy problem ? if you want to use the holder condition to study this stochastic diﬀerential equation, i suggest this paper : math.ucla.edu/~samxu/FeynmanKac.pdf is it helpful ? thank you ! Commented May 8, 2014 at 3:32
• This doesn't seem to have anything to do with the question being asked. Commented May 8, 2014 at 3:37
• Just for clarity: You certainly do not need $b, \sigma$ to be $C^2$ for there to exist a unique (strong) solution to the SDE. There are lots of weaker conditions under which it holds, the most common being $b, \sigma$ globally Lipschitz. Commented May 8, 2014 at 7:42

$$Au:=\frac{1}{2}\operatorname{tr}(a(x)D^{2}u(x))-\langle b(x),Du(x)\rangle+c(x)u(x)$$
and boundary data $$g$$, they obtain for the elliptic case $$f=Au$$
and the parabolic case $$u_{t}=Au-f$$