Consider the problem \begin{equation} \begin{cases} \frac{\partial v^{\epsilon}(t,x)}{\partial t}= L^{\epsilon}v^{\epsilon}(t,x)+c(x)v^{\epsilon}(t,x)+g(x); \quad t>0, \: x \in \mathbb{R}^r \\ v^{\epsilon}(0,x)=f(x) \\ \end{cases} \end{equation} for $\epsilon > 0$,together with the problem for the first-order operator which is obtained for $\epsilon = 0$: \begin{equation} \begin{cases} \frac{\partial v^{0}(t,x)}{\partial t}= L^{0}v^{0}(t,x)+c(x)v^{0}(t,x)+g(x); \quad t>0, \: x \in \mathbb{R}^r \\ v^{0}(0,x)=f(x) \\ \end{cases} \end{equation} Consider then the SDE: $dX_{t}^{\epsilon,x}=b(X_{t}^{\epsilon, x})dt+\epsilon \sigma(X_{t}^{\epsilon,x})dW_t$, where $W$ is a Wiener process and
$ L^{\epsilon}= \frac{\epsilon^2}{2} \sum_{i,j=1}^{r}a^{i,j}(x)\frac{\partial^2}{\partial x^i \partial x^j} + \sum_{i=1}^{r}b^i(x)\frac{\partial}{\partial x^i} $
with $\sigma(x)\sigma^t(x)=a^{i,j}(x)$.
I proved that \begin{equation*} \begin{split} \lim_{\epsilon \to 0} v^{\epsilon}(t,x) &= f(X_{t}^{0,x})\exp\biggl\{ \int_{0}^{t}c(X_{s}^{0,x})ds \biggr\} \\ &+ \int_{0}^{t} g(X_{s}^{0,x})\exp \biggl\{ \int_{0}^{s}c(X_{v}^{0,x})dv \biggr\} ds \end{split} \end{equation*} and now I have to show that the right-hand side is the solution of the second Cauchy problem (the one with $epsilon = 0$). I know that I have to compute the derivatives and substitute but I'm having some trouble. Here's what I came up with:
$ \frac{\partial v^0}{\partial t}= f(X_{t}^{0,x})c(X_{t}^{0,x})e^{\int_{0}^{t}c(X_{s}^{0,x})ds}+g(X_{t}^{0,x})e^{\int_{0}^{t}c(X_{s}^{0,x})ds} $
$ L^0v^0 = \frac{\partial f}{\partial X}(X_{t}^{0,x})e^{\int c(X_{s}^{0,x})ds}+f(X_{t}^{0,x})c(X_{t}^{0,x})e^{\int c(X_{s}^{0,x})ds}+g(X_{t}^{0,x})e^{\int c(X_{s}^{0,x})ds} $
If I substitute in the second problem and sum all the terms up, I'm left with \begin{equation} \frac{\partial f}{\partial X}(X_{t}^{0,x})e^{\int_{0}^{t} c(X_{s}^{0,x})ds}+f(X_{t}^{0,x})c(X_{t}^{0,x})e^{\int_{0}^{t} c(X_{s}^{0,x})ds}+c(X_{t}^{0,x})\int_{0}^{t}g(X_{s}^{0,x})e^{\int_{0}^{s} c(X_{v}^{0,x})dv}ds+g(X_{t}^{0,x})=0 \end{equation}
Is what I did until now correct? If so how do I conclude? If not how do I fix it to get the desired solution? Thank you in advance.
When you compute $\frac{\partial v^0(t,x)}{\partial t}$, you clearly forgot the term $\nabla f(X^{0,x}_t) \cdot b(X^{0,x}_t) \exp\left[\int_0^t c(X^{0,x}_s) \mathrm{d}s\right]$, where the gradient is taken with respect to the space variable only...