Reading the Wikipedia page on the heat equation, I've found the following passage:
An additional term may be introduced into the equation to account for radiative loss of heat. According to the Stefan–Boltzmann law, this term is $\mu(u^4-v^4)$, where $v=v(x,t)$ is the temperature of the surroundings, and $\mu$ is a coefficient that depends on the Stefan–Boltzmann constant and the emissivity of the material. The rate of change in internal energy becomes $$\frac{\partial Q}{\partial t}=-\frac{\partial q}{\partial x}-\mu\left(u^{4}-v^{4}\right)$$ and the equation for the evolution of $u$ becomes $$\frac{\partial u}{\partial t}=\frac{k}{c\rho}\frac{\partial^{2}u}{\partial x^{2}}-\frac{\mu}{c\rho}\left(u^{4}-v^{4}\right).$$
For simplicity and a certain convenience with the formulas we're going to use in the future, assume that the coefficients $k$, $c$, and $\mu$ are such that this equation takes the form $$\partial_t u=\frac{1}{2}\Delta_x u-u^4-v^4.$$ Then, assuming that we're in a vacuum and far away from other things, so that $v\approx0$, we can again simplify the equation to \begin{equation} \partial_t u=\frac{1}{2}\Delta_x u-u^4.\tag{1} \end{equation} Now, I've read elsewhere that when the radiative loss is small, it's okay to approximate the quartic term with a linear one, and consider instead the equation \begin{equation} \partial_t u=\frac{1}{2}\Delta_x u-u.\tag{2} \end{equation}
Question. Equation $(2)$ admits a Feynman–Kac solution of the form \begin{align*} u(x,t) &= \mathbb{E}\left[f(B_t+x)e^{-\int^{t}_{0}(B_s+x)\,\mathrm{d}s}\right]\\ &= \int_{\mathcal{C}([0,t],\mathbb{R})}f(\gamma(t)+x)e^{-\int^{t}_{0}(\gamma(s)+x)\,\mathrm{d}s}\,\mathrm{d}W(\gamma). \end{align*} Is there a similar Feynman–Kac formula for Equation $(1)$, the nonlinear, physically more accurate, version of Equation $(2)$?